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Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 7 Wave Optics Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 7 Wave Optics

Question 1.
Name the various theories of the nature of light.
Answer:
The various theories of the nature of light from the 17th century to the modern times are

1. Descartes’ corpuscular theory of light (1637)
2. Newton’s corpuscular theory (1666)
3. Huygens’ wave theory of light propagation (1678), modified, verified and put on a firm mathematical base by Young, Fraunhofer, Fresnel and Kirchhoff (in the 1800s)
4. Maxwell’s electromagnetic theory (1865)
5. the light quantum, i.e., the photon model of the modern quantum theory by Planck (1900) and Einstein (1905).

Question 2.
State the postulates of Newton’s corpuscular theory of light.
Answer:
Sir Isaac Newton developed the corpuscular theory of light proposed by Rene’ Descartes (1596-1650), French philosopher and mathematician. The theory assumed that light consists of a stream of corpuscles emitted by a luminous source.
Postulates of Newton’s corpuscular theory of light:

1. Light corpuscles are minute, light and perfectly elastic particles.
2. A luminous source emits light corpuscles in all directions which then travel at high speed in straight lines in a given medium.
3. The constituent colours of white light are due to different sizes of the corpuscles.
4. The light corpuscles stimulate the sense of sight on their impact on the retina of the eye.
5. A reflective surface exerts a force of repulsion normal to the surface on the light corpuscles when they strike the surface.
6. A transparent medium exerts a force of attraction normal to the surface on the light corpuscles striking the surface. This force is different for different mediums.

Notes :

1. A consequence of the assumption (6) is that, according to the corpuscular theory, the speed of light in a denser medium is greater than that in air and has different values for different mediums.
2. It was known from earliest recorded times that when light is incident on the surface of glass or water, it is partly reflected and partly transmitted, simultaneously. To explain this, Newton postulated that the corpuscles must have fits of easy reflection and fits of easy transmission and must pass periodically from one state to the other.

Question 3.
State the drawbacks of Newton’s corpuscular theory of light.
Answer:
Drawbacks of Newton’s corpuscular theory of light:

1. The theory predicted that the speed of light in a
   denser medium should be greater than that in air. This was disproved when experiment showed that the speed of light in water is less than that in air (carried out in 1850 by French physicist Jean Bernard Leon Foucault). ,
2. The theory could not satisfactorily explain the phenomenon of polarization and the simultaneousness of reflection and refraction.
3. The corpuscular theory failed to explain the phenomena of diffraction and interference.
4. There was no basis for the hypothesis that the constituent colours of white light are due to different sized corpuscles.

Question 4.
What is a ray of light?
Answer:
A ray of light is the path along which light energy is transmitted from one point to another in an optical system.

Question 5.
What is meant by ray optics or geometrical optics?
Answer:
The formation of

1. shadows, and
2. images by mirrors and lenses can be explained by assuming that light propagates in a straight line in terms of rays. The study of optical phenomena under this assumption is called ray optics. It is also called geometrical optics as geometry is used in this study.

Question 6.
Give a brief account of Huygens’ wave theory of light. State its merits and demerits.
Answer:
Huygens’ wave theory of light [Christiaan Huygens (1629-95), Dutch physicist] :

1. Light emitted by a source propagates in the form of waves. Huygens’ original theory assumed them to be longitudinal waves.
2. In a homogeneous isotropic medium, light from a point source spreads by spherical waves.
3. It was presumed that a wave motion needed a medium for its propagation. Hence, the theory postulated a medium called luminiferous ether that exists everywhere, in vacuum as well as in transparent bodies. Ether had to be assigned some extraordinary properties, a high modulus of elasticity (to account for the high speed of light), zero density (so that it offers no resistance to planetary motion) and perfect transparency.
4. The different colours of light are due to different wavelengths.

Merits :

1. Huygens’ wave theory satisfactorily explains reflection and refraction as well as their simultaneity.
2. In explaining refraction, the theory concludes that the speed of light in a denser medium is less than that in a rarer medium, in agreement with experimental findings.
3. The theory was later used by Young in 1800-04, Fraunhofer and Fresnel in 1814 to satisfactorily explain interference, diffraction and rectilinear propagation of light. The phenomenon of polarization could also be explained considering the light waves to be transverse.

Demerits :

1. It was found much later that the hypothetical medium, luminiferous ether, has no experimental basis. Einstein discarded the idea of ether completely in 1905.
2. Phenomena like absorption and emission of light, photoelectric effect and Compton effect, cannot be explained on the basis of the wave theory.

[Note: To decide between the particle and wave theories of light, Dominique Francois Jean Arago (1786-1853), French physicist, suggested the measurement of the speed of light in air and water. The experiment was performed in 1850 by Leon Foucault (1819-68), French physicist, using Arago’s experimental equipment. He found that the speed of light in water is less than that in air.]

Question 7.
What is meant by wave optics?
Answer:
It is not possible to explain certain phenomena of light, such as interference, diffraction and polarization, with the help of ray optics (geometrical optics). The branch of optics which uses wave nature of light to explain these optical phenomena is called wave optics.

Question 8.
What was Maxwell’s concept of light?
Answer:
In 1865, James Clerk Maxwell developed a mathematical theory on the intimate relationship between electricity and magnetism. His theory predicted light to be a high-frequency transverse electromagnetic wave in ether. Electric and magnetic fields in the wave vary periodically in space and time at right angles to each other and to the direction of propagation of the wave.

The speed of the electromagnetic waves in a medium, as calculated on the basis of Maxwell’s theory, was experimentally found to be equal to the measured speed of light in that medium. Maxwell’s electromagnetic theory of light, with addition by others till 1896, could account for all the known phenomena regarding the propagation (or transmission) of light through space and through matter.

Question 9.
What is the photon model or quantum hypothesis of light ?
Answer:
To explain the interaction of light and matter (as in the emission or absorption of radiation), Max Planck, in 1900, and Einstein, in 1905, hypothesized light as concentrated or localized packets of energy. Such an energy packet is called a quantum of energy, which was given the name photon much later, in 1926, by Frithiof Wolfers and Gilbert Newton Lewis. For a radiation of frequency v, a quantum of energy is hv, where h is a universal constant, now called Planck’s constant.

[Note : ‘Localisation’ of energy in a region gives light its particle nature while frequency is a wave characteristic. The complementary properties of particle and wave of light quanta are reconciled as follows : light propagates as wave but interacts with matter as particle.]

Question 10.
Give a brief account of the wave nature of light.
Answer:

1. Light is a transverse, electromagnetic wave.
2. A light wave consists of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of propagation of the wave.
3. Like all other electromagnetic waves, light waves do not require any material medium as they can travel even through vacuum.
4. In a material medium, the speed of light depends on the refractive index of the medium, which, in turn depends on the permeability and permittivity of the medium.

Question 11.
Define absolute refractive index of a medium.
Answer:
The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium.
[Note : Absolute refractive index of a medium (n) =

The absolute refractive index of vacuum is 1 (by definition) and that of air is greater than 1, but very nearly equal to 1.]

Question 12.
State the characteristics of the electromagnetic waves.
Answer:

1. The electromagnetic waves are transverse in nature as they propagate by oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of propagation of the wave.
2. These waves do not require any material medium for their propagation, i.e., they can travel even through vacuum.
3. The wavelength of the electromagnetic waves ranges from very small (< 1 fm) to very large (> 1 km). The waves are classified in the order of increasing wavelength as γ-rays, X rays, ultraviolet, visible, infrared, microwave and radio waves.
4. In vacuum, the speed of electromagnetic waves does not depend on the frequency of the wave. But, in a material medium, it depends on the frequency. For a given frequency, the speed is different in different mediums.
   [Note : 1 fm (femtometre) = 10^-15 m]

Question 13.
Define and explain :
(a) a wave normal
(b) a ray of light.
Answer:
(a) Wavenormal: A wave normal at a point on a wavefront is defined as a line drawn perpendicular to the wavefront in the direction of propagation of the wavefront.

In a homogeneous isotropic medium, a wavefront moves parallel to itself. Thus, at any point in the medium, the direction in which the wavefront moves is always perpendicular to the wavefront at that point. This direction is given by the wave normal at that point.

(b) Ray of light: The direction in which light is propagated is called a ray of light.

This term (ray of light) is also used to mean a narrow beam of light waves. Only in a homogeneous isotropic medium is a ray of light the same as a wave normal. For spherical wavefronts spreading out from a point source, the rays are radially divergent. The rays corresponding to a plane’ wavefront form a parallel beam.

Question 14.
What is a cylindrical wavefront? Draw the corresponding diagram.
Answer:
Cylindrical wavefront: An extended linear source, such as an aperture in the form of a narrow slit, gives rise to cylindrical wavefronts. All the points equidistant from the source lie on the curved surface of a cylinder. Thus, the shape of the wavefront is cylindrical.

Question 15.
What is a plane wavefront? Draw the corresponding diagram.
Answer:
Plane wavefront: It may be treated as a part of a spherical or cylindrical wavefront at a very great distance from the source, such that the wavefront has a negligible curvature.

Question 16.
Draw a neat labelled diagram illustrating spherical wavefronts corresponding to a diverging beam of light.
Answer:

Note : Lenses can be used to obtain

1. a converging beam of light
2. a diverging beam of light.

Question 17.
State Huygens’ principle.
Answer:
Huygens’ principle : Every point on a wavefront acts as a secondary source of light and sends out secondary wavelets in all directions. The secondary wavelets travel with the speed of light in the medium. These wavelets are effective only in the forward direction and not in the backward direction. At any instant, the forward-going envelope or the surface of tangency to these wavelets gives the position of the new wavefront at that instant.

Question 18.
Explain the construction and propagation of a plane wavefront using Huygens’ principle.
Answer:
Huygens’ construction nf a plane wavefront: A plane wavefront may be treated as a part of a spherical or cylindrical wave at a very great distance from a point source or an extended source, such that the wavefront has a negligible curvature. Let A, B, C, D, …, be points on a plane wavefront in a homogeneous isotropic medium in which the speed of light, taken to be monochromatic, is v.

In a time, t = T, secondary wavelets with points A, B, C, D,…, as secondary sources travel a distance vT. To find the position of the wavefront after a time t = T, we draw spheres of radii vT with A, B, C,…, as centres. The envelope or the surface of tangency to these spheres is a plane A’B’C’. This plane, the new wavefront, is at a perpendicular distance vT from the original wavefront in the direction of propagation of the wave. Thus, in an isotropic medium, plane wavefronts are propagated as planes.

Question 19.
Explain the construction and propagation of a spherical wavefront using Huygens’ principle.
Answer:
Huygens’ construction of a spherical wavefront: Consider a point source of monochromatic light S in a homogeneous isotropic medium. The light waves travel with the same speed v in all directions. After time f, the wave will reach all the points which are at a distance vt from S. This is spherical wavefront XY. Let, A, B, C, …,, be points on this wavefront.

To find the new wavefront after time T, we draw spheres of radius vT with A, B, C,…, as centres. The envelope or the surface of tangency of these spheres is the surface A’B’C’. This is the new spherical wavefront X’Y’. Thus, in an isotropic medium, spherical wavefronts are propagated as concentric spheres.

Question 20.
Suppose a parallel beam of monochromatic light is incident normally at a boundary separating two media. Explain what happens to the wavelength and frequency of the light as it propagates from medium 1 to medium 2. What happens when the medium 1 is vacuum ?
Answer:
Consider a parallel beam of monochromatic light incident normally on interface PQ separating a rarer medium (medium 1) and a denser medium (medium 2).

The three successive wavefronts AB, CD and EF are separated by a distance λ₁, the wavelength of light in first medium. The corresponding three wavefronts after refraction, are A’B’, C’D’ and E’F’. Due to the denser medium, the speed of light reduces and hence the wavefronts cover a less distance than that covered in the same time in the first medium. Thus, the wavefronts are comparitively more closely spaced than in the first medium. This distance between successive wavefronts is λ₂, the wavelength of light in the second medium. Thus, λ₂ is less than λ₁,. To find the relation between λ₁ and λ₂, let us consider the wavefront AB reaching

PQ at time t = 0. The next wavefront CD, separated from AB by distance λ₁, will reach PQ at time t = T. Let v₁ and v₂ be the speeds of light in medium 1 and medium 2 respectively. T is the time during which light covers distance λ₁ in medium 1 and λ₂ in medium 2.

This shows how the wavelength of light changes in refraction.

If the first medium is vacuum where the wavelength of light is λ₀ and n is the absolute refractive index of medium 2, then
λ₂ = λ₀\(\left(\frac{v_{2}}{c}\right)\) = \(\frac{\lambda_{0}}{n}\) (as v₁ = c) …(3)
Now, speed = frequency × wavelength.
Hence, the ratio of the frequencies v₁ and v₂, of the wave in the two mediums can be written using
EQ. (2) as, \(\frac{v_{1}}{v_{2}}\) = \(\frac{v_{1} / \lambda_{1}}{v_{2} / \lambda_{2}}\) = 1 …(4)

Thus, the frequency of a wave remains unchanged while going from one medium to another. Thus, v₀ = v₁ = v₂, where v₀ is the frequency of light in vacuum.

Question 21.
The refractive indices of diamond and water with respect to air are 2.4 and 4/3 respectively. What is the refractive index of diamond with respect to water ?
Answer:

is the refractive index of diamond with respect to water.

Question 22.
What is the refractive index of water with respect to diamond ? For data, see Question 21. above.
Answer:

is the refractive index of water with respect to diamond.

Question 23.
What happens to the frequency, wavelength and speed of light as it passes from one medium to another?
Answer:
The wavelength and speed of light change, but the frequency remains the same.

Question 24.
The refractive index of water with respect to air, for light of wavelength Aa in air, is 4/3. What is the wavelength of the light in water?
Answer:

as the wavelength of the light in water.

Question 25.
If the frequency of certain light is 6 × 10¹⁴ Hz, what is its wavelength in free space ? [c = 3 × 10⁸ m/s]
Answer:

is the required wavelength.

Question 26.
If the wavelength of certain light in air is 5000 Å and that in a certain medium is 4000 Å, what is the refractive index of the medium with respect to air ?
Answer:

is the refractive index of the medium with respect to air.

Data : c = 3 × 10⁸ m/s

Question 27.
Solve the following :

Question 1.
If the refractive index of glass is 3/2 and that of water is 4/3 respectively, find the speed of light in glass and in water.
Solution:
Let n_g and n_w be the refractive indices of glass and water respectively. Also let v_a, v_g and v_w be the speeds of light in air, glass and water, respectively.

Question 2.
The refractive indices of water and diamond are 4/3 and 2.42 respectively. Find the speed of light in water and diamond.
Solution :

This is the speed of light in diamond.

Question 3.
The refractive indices of glass and water with respect to air are \(\frac{3}{2}\) and \(\frac{4}{3}\), respectively. Determine the refractive index of glass with respect to water.
Solution :


This is the refractive index of glass with respect to water.

Question 4.
A diamond (refractive index = 2.42) is dipped into a liquid of refractive index 1.4. Find the refractive index of diamond with respect to the liquid.
Solution :
Data : n_d = 2.42, n₁ = 1.4
The refractive index of diamond with respect to liquid,

Question 5.
The refractive index of glycerine is 1.46. What is the speed of light in glycerine ? [Speed of light in vacuum = 3 × 10⁸ m/s]
Solution :
Data : n_g = 1.46, c = 3 × 10⁸ m/s c
n_g = \(\frac{c}{v}\)
v = \(\frac{c}{n}\) = \(\frac{3 \times 10^{8}}{1.46}\) = 2.055 × 10⁸ m/s
This is the speed of light in glycerine.

Question 6.
The refractive indices of glycerine and diamond with respect to air are 1.46 and 2.42 respectively. Calculate the speed of light in glycerine and in diamond. From these calculate the refractive index of diamond with respect to glycerine.
Solution:
Let n_g and n_d be the refractive indices of glycerine and diamond respectively. Also, let v_a, v_g and v_d be the speeds of light in air, glycerine and diamond respectively.
Data : v_a = 3 × 10⁸ m/s, n_g = 1.46, n_d = 2.42
(i) Refractive index of glycerine with respect to air,

(ii) Refractive index of diamond with respect to air,
n_d = \(\frac{v_{\mathrm{a}}}{v_{\mathrm{d}}}\)

(iii) Refractive index of diamond with respect to glycerine,

Question 7.
The wavelengths of a certain light in air and in a medium are 4560 Å and 3648 Å, respectively. Compare the speed of light in air with its speed in the medium.
Solution:
Let v_a and v_m be the speeds of light in air and in the medium respectively and let λ_a and λ_m be the wavelengths of light in air and in the medium respectively. Let v be the frequency of light in air.
When light passes from one medium to another, its frequency remains unchanged.

Question 8.
Monochromatic light of wavelength 632.8 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected and (b) refracted light ? [Given : Refractive index of water = 1.33]
Solution:
Let v be the frequency of the light, and λ₁ and λ₂ the wavelengths of reflected and refracted light respectively.
Data : λ₁ = 632.8 nm = 632.8 × 10^-9 m,
c = 3 × 10⁸ m/s, _an_w = 1.33
(a) For reflected light:
When the wave travels in air, its speed v₁ = c

(b) For refracted light:
Frequency does not change while going from one medium to other.
∴ V = 4.741 × 10¹⁴ Hz

This is the speed of the light in water.

Question 9.
The refractive indices of water for red and violet colours are 1.325 and 1.334, respectively. Find the difference between the speeds of the rays of these two colours in water. [c = 3 × 10⁸ m/s]
Solution :
Data : n_r = 1.325, n_v = 1.334, c = 3 × 10⁸ m/s

Question 10.
If the difference in speeds of light in glass and water is 2.505 × 10⁷ m/s, find the speed of light in air. [Refractive index of glass = 1.5, refractive index of water = 1.333]
Solution :


This is the speed of light in air.

Question 11.
If the difference in speeds of light in glass and water is 0.25 × 10⁸ m/s, find the speed of light in air. [n_g = 1.5 and n_w = \(\frac{4}{3}\)]
Solution :

∴ The speed of light in air, c = 12 (n_w – v_g)
= 12 × 0.25 × 10⁸
= 3 × 10⁸ m/s

Question 12.
Red light of wavelength 6400 Å in air has wavelength 4000 Å in glass. If the wavelength of violet light in air is 4400 Å, what is its wavelength in glass? Assume that the glass has the same refractive index for red and violet colours.
Solution :

[ Note : This problem was asked in Board examination in October 2014. The assumption n_r(glass) = n_v(glass) is manifestly gross. No glass can have the same refractive index for red and violet colours.]

Question 13.
A ray of light passes from air to glass. If the angle of incidence is 74° and the angle of refraction is 37°, find the refractive index of the glass.
Solution :
Data : i = 74°, r = 37°
n = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin 74^{\circ}}{\sin 37^{\circ}}\) = \(\frac{0.9613}{0.6018}\)
∴ n = 1.597
This is the refractive index of the glass.

Question 14.
A ray of light is incident on a glass slab making an angle of 30° with the surface. Calculate the angle of refraction in glass and the speed of light in glass. The refractive index of glass and speed of light in air are 1.5 and 3 × 10⁸ m/s, respectively. Solution :
Data : v_air = 3 × 10⁸ m/s; n = 1.5
The angle of incidence (i) is the angle made by the incident ray with the normal drawn to the refracting surface.
∴ i = 90° – 30° = 60°
(a) Angle of refraction (r) in glass :

(b) Speed of light (v_glass) in glass :

Question 15.
A ray of light is incident on a water surface of refractive index \(\frac{4}{3}\) making an angle of 40° with the surface. Find the angle of refraction.
Solution :

∴ The angle of refraction, r = sin^-1(0.5745) = 35°4′

Question 16.
A ray of light travelling in air is incident on a glass slab making an angle of 30° with the surface. Calculate the angle by which the refracted ray in glass is deviated from its original path and the speed of light in glass [Refractive index of glass = 1.5].
Solution:
Solve for the speed (ug) of light in glass and the angle of refraction (r) in glass as in Solved Problem (14). ug = 2 × 10⁸ m/s and r = 35°16′
The angle of incidence (z), i.e., the angle between the incident ray and the normal to the glass surface, is i = 90° – 30° = 60°.
Hence, the angle by which the refracted ray is deviated from the original path is δ = i – r = 60° – 35°16′ = 24°44′

Question 17.
The wavelength of blue light in air is 4500 A. What is its frequency? If the refractive index of glass for blue light is 1.55, what will be the wavelength of blue light in glass ?
Solution :
Data : λ_a = 4500 Å = 4.5 × 10^-7 m, n_g = 1.55, v_a = 3 × 10⁸ m/s
v_a = v_aλ_a

This is the wavelength of blue light in glass.

Question 18.
White light consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55 ?
Solution:
Let λ₁ and λ₂ be the wavelengths of light in water for 400 nm and 700 nm (wavelengths in vacuum) respectively. Let λ_a be the wavelength of light in vacuum.

The wavelength range seen when white light is passed through the glass would be 258.06 nm to 451.61 nm.

Question 19.
Determine the change in wavelength of electro-magnetic radiation as it passes from air to glass, if the refractive index of glass with respect to air for the radiation under consideration is 1.5 and the frequency of the radiation is 3.5 × 10¹⁴ Hz. [Speed of the radiation in air (c) = 3 × 10⁸ m/s]
Solution:
Let v be the frequency of the electromagnetic radiation and c its speed in air. Let λ_a and λ_b be the wavelengths of the radiation in air and glass, respectively.

Question 20.
Determine the change in wavelength of light during its passage from air to glass, if the refractive index of glass with respect to air is 1.5 and the frequency of light is 5 × 10¹⁴ Hz. [Speed of light in air = (c) = 3 × 10⁸ m/s] Solution:
Let v be the frequency of the electromagnetic radiation and c its speed in air. Let λ_a and λ_b be the wavelengths of the radiation in air and glass, respectively.

λ_a = 6000 Å, λ_g = 4000 Å ∴ λ_a – λ_g = 2000 Å

Question 21.
A light beam of wavelength 6400 Å is incident normally on the surface of a glass slab of thickness 5 cm. Its wavelength in glass is 4000 A. The beam of light takes the same time to travel from the source to the surface as it takes to travel through the glass slab. Calculate the distance of the source from the surface.
Solution :
Data : λ_a = 6400 Å, s_g = thickness of the glass slab = 5 cm, λ_g = 4000 Å
The speeds of light in glass and air are, respectively, v_g = λ_gv and v_a = λ_av
where the frequency of the light v remains un-changed with the change of medium.
The time taken to travel through the glass slab,
t_g = \(\frac{\mathrm{Sg}}{v_{\mathrm{g}}}\) = \(\frac{S_{g}}{\lambda_{g} v}\)
The time taken to travel through air,
t_a = \(\frac{S_{a}}{v_{a}}\) = \(\frac{d}{\lambda_{\mathrm{a}} v}\)
where s_a = d is the distance of the glass surface from the source.
Since t_a = t_g
\(\frac{d}{\lambda_{\mathrm{a}} v}\) = \(\frac{s_{\mathrm{g}}}{\lambda_{\mathrm{g}} v}\)
∴ d = \(\frac{\lambda_{\mathrm{a}}}{\lambda_{\mathrm{g}}} \mathrm{s}_{\mathrm{g}}\)S_g = \(\frac{6400}{4000}\) × 5 = 1.6 × 5 = 8 cm

Question 22.
A parallel beam of monochromatic light is incident on a glass slab at an angle of incidence 60°. Find the ratio of the width of the beam in glass to that in air if the refractive index of glass is \(\frac{3}{2}\).
Data: i = 60, _an_g = 1.5
By Snell’s law, _an_g = \(\frac{\sin i}{\sin r}\) ∴ sin r = \(\frac{\sin i}{\mathrm{a} n_{\mathrm{g}}}\)
Solution:

Question 23.
The width of a plane incident wavefront is found to be doubled in a denser medium if it makes an angle of 70° with the surface. Calculate the refractive index for the denser medium.
Solution :

Data : i = 70°, \(\frac{\mathrm{CD}}{\mathrm{AB}}\) = 2
\(\frac{\cos r}{\cos i}\) = \(\frac{\mathrm{CD}}{\mathrm{AB}}\) = 2
∴ cos r = 2 cos 70° = 2 × 0.3420 = 0.6840
∴ r = cos^-10.6840 = 46°51′
The refractive index of the denser medium relative to the rarer medium
= \(\frac{\sin i}{\sin r}\) = \(\frac{\sin 70^{\circ}}{\sin 46^{\circ} 51^{\prime}}\) = \(\frac{0.9397}{0.7296}\) = 1.288

Question 24.
A ray of light travelling through air falls on the surface of a glass slab at an angle i. It is found that the angle between the reflected and the refracted rays is 90°. If the speed of light in glass is 2 × 10⁸ m/s, find the angle of incidence.
Solution :
Data : c = 3 × 10⁸ m/s, v_g = 2 × 10⁸ m/s, angle between the reflected ray and the refracted ray = 90°
n_g = \(\frac{c}{v_{g}}\) = \(\frac{3 \times 10^{8}}{2 \times 10^{8}}\) = 1.5
The angle between the reflected and refracted rays = (90° – i) + (90° – r) = 180° – (i + r) = 90° (by the data)
i + r = 90° ∴ r = 90° – i
∴ sin r = sin(90° – i) = cos i
n_g = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin i}{\cos i}\) = tan i
∴ The angle of incidence,
i = tan^-1 ng = tan^-1 1.5 = 56°19′

Question 28.
What is meant by polarized light ? How does it differ from unpolarized light?
Answer:
According to the electromagnetic theory of light, a light wave consists of electric and magnetic fields vibrating at right angles to each other and to the direction of propagation of the wave. If the vibrations of the electric field \(\vec{E}\) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

If the vibrations of \(\vec{E}\) in a light wave are in all directions perpendicular to the direction of propagation of the light wave, the light wave is said to be unpolarized. Ordinary light, e.g. that emitted by a bulb, is unpolarized.

According to Biot, unpolarized light may be considered as a superposition of many linearly polarized waves, with random orientations. Also, these component waves are noncoherent, that is, irregular in their phase relationships.

[Note : Ordinary light consists of wave trains, each coming from a separate atom in the source. A beam of ordinary light in a single direction consists of millions of such wave trains from the very large number of atoms in the source radiating in that direction. Hence, the vibrations of \(\vec{E}\) are in all transverse directions with equal probability. Thus, light from an ordinary source is.un-polarized.]

Question 29.
How are polarized light and unpolarized light represented in a ray diagram ?
OR
How will you distinguish between polarized and unpolarized light in a ray diagram?
Answer:
Linearly polarized light is represented in a ray diagram by double-headed arrows or short lines drawn perpendicular to the direction of propagation of light, as shown in below figure.

An unpolarized light beam is represented by both dots and arrows, as shown in above figure. The dots are the ‘end views’ of arrows that are oriented normal to the plane of the diagram.

[Note : The length of an arrow or line represents the amplitude of the electric field (\(\vec{E}\)) in the plane of the diagram and the direction indicates the polarization axis of the beam, i.e., the direction of vibration of \(\vec{E}\). According to Jean Biot (1774-1862), French physicist, unpolarized light may be considered as a superposition of many linearly polarized waves, with random orientations in planes perpendicular to the direction of propagation. In an analytical treatment, the electric vectors may be resolved along two mutually perpendicular directions so that the multiplicity of vectors in can be replaced by the two mutually perpendicular vectors as in, both perpendicular to the direction of

propagation of light. Sir David Brewster (1781-1868), British physicist, conceived the ordinary unpolarized light to consist of two perpendicular, polarized components of equal intensity.]

Question 30.
What is a polarizer?
Answer:
When a beam of unpolarized light is passed through certain types of materials (or devices), these materials (or devices) allow only those light waves to pass through which have their electric field along a particular direction. All the other waves with the electric field in other directions are blocked. A material (or a device) which exhibits this special property is called a polarizer.

Question 31.
Explain the terms :

1. polarizing axis of a polarizer
2. plane of vibration
3. plane of polarization.

Answer:

1. Polarizing axis of a polarizer : When unpolarized light is incident on a polarizer, the particular direction along which the electric field of the emergent wave is oriented is called the polarizing axis of the polarizer.
2. Plane of vibration : The plane of vibration of an electromagnetic wave is the plane of vibration of the electric field vector containing the direction of propagation of the wave. Experiment shows that it is the electric field vector E which produces the optical polarization effects.
3. Plane of polarization : The plane of polarization of an electromagnetic wave is defined as the plane perpendicular to the plane of vibration. It is the plane containing the magnetic field vector and the direction of propagation of the wave.
   

Note :
For vision, photography, action of light on electrons and many other observed effects of light, it is the electric vector that is more important than the magnetic one. Hence, nowadays the plane of polarization is taken to be the plane of vibration. Most authoritative; modern books on Optics do not, therefore, explicitly define the plane of vibration. The above convention is obsolete and redundant.

Question 32.
What is a Polaroid? Explain its construction.
Answer:
A Polaroid is a synthetic dichroic sheet polarizer packed with tiny dichroic crystals oriented parallel to each other such that the transmitted light is plane polarized.

Construction : The first large polarizing sheet filter was made by US inventor Edwin H. Land (1909-91). He used the microscopic needlelike crystals of iodoquinine sulphate (known as herapathite) made into a thick colloidal dispersion in nitrocellulose. This material was squeezed through a long narrow slit which forced the needles to orient parallel to one another. The material was then dried to form a solid plastic sheet.

[Note : The modern version of Polaroid is made from long-chain polymer, polyvinyl alcohol. The transmission axis of a Polaroid is the plane of vibration of a plane polarized light which passes through with minimum absorption.]

If the vibrations of the electric field \(\vec{E}\) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

This phenomenon of restricting the vibrations of light, i.e., of the electric field vector in a particular direction, which is perpendicular to the direction of the propagation of the wave is called polarization of light.

Consider an unpolarized light wave travelling along the x-direction. Let c, v and λ be the speed, frequency and wavelength, respectively, of the wave. The magnitude of its electric field (\(\vec{E}\)) is,
E = E₀ sin (kx – ωt), where E₀ = E_max = amplitude of the wave, ω = 2πv = angular frequency of the wave and k = \(\frac{2 \pi}{\lambda}\) = magnitude of the wave vector or propagation vector.
The intensity of the wave is proportional to |E₀|². The direction of the electric field can be anywhere in the y-z plane. This wave is passed through two identical polarizers as shown.

When a wave with its electric field inclined at an angle φ to the axis of the first polarizer is passed through the polarizer, the component E₀ cos φ will pass through it. The other component E₀ sin φ which is perpendicular to it will be blocked.

Now, after passing through this polarizer, the intensity of this wave will be proportional to the square of its amplitude, i.e., proportional to |E₀ cos φ|².

The intensity of the plane-polarized wave emerging from the first polarizer can be obtained by averaging E₀ cos φ|² over all values of φ between 0 and 180°. The intensity of the wave will be
proportional to \(\frac{1}{2}\) |E₀|² as the average value of cos² φ over this range is \(\frac{1}{2}\). Thus the intensity of an unpolarized wave reduces by half after passing through a polarizer.

When the plane-polarized wave emerges from the first polarizer, let us assume that its electric field (\(\overrightarrow{E_{1}}\)) is along the y-direction. Thus, this electric field is,
\(\overrightarrow{E_{1}}\) = \(\hat{i}\)E₁₀ sin (kx – ωt) … (1)
where, E₁₀ is the amplitude of this polarized wave. The intensity of the polarized wave,
I₁ ∝|E₁₀|² …(2)

Now this wave passes through the second polarizer whose polarization axis (transmission axis) makes an angle θ with the y-direction. This allows only the component E₁₀ cos θ to pass through it. Thus, the amplitude of the wave which passes through the second polarizer is E₂₀ = E₁₀ cos θ and its intensity,
I₂ ∝|E₂₀|²
∴ I₂ ∝ | E₁₀|² cos² θ
∴ I₂ = I₁ cos² θ … (3)

Thus, when plane-polarized light of intensity I₁ is incident on the second identical polarizer, the intensity of light transmitted by the second polarizer varies as cos² θ, i.e., I₂ = I₁ cos² θ, where θ is the angle between the transmission axes of the two polarizers. This is known as Malus’ law.

[Note : Etienne Louis Malus (1775-1812), French military engineer and physicist, discovered in 1809 that light can be polarized by reflection. He was the first to use the word polarization, but his arguments were based on Newton’s corpuscular theory.]

Question 33.
Unpolarized light is passed through two polarizers. Under what condition is the intensity of the emergent light
(i) maximum
(ii) zero?
Answer:
If the angle θ between the axes of polarization of the two polarizers, is 0°, i.e., the polarization axes of the two polarizers are parallel, the intensity of emergent light is maximum. If θ = 90°, i.e., the polarization axes of the two polarizers are perpendicular, no light emerges from the second polarizer, thus the intensity of emergent light is minimum.
[Note: θ = 0° and 90° are known as parallel and cross settings of the two polarizers.]

Question 34.
With a neat labelled diagram, explain the use of a pair of polarizers to vary the intensity of light. What are crossed polarizers?
Answer:
Consider an unpolarized light beam incident perpendicularly on the first polarizing sheet, called the polarizer, whose transmission axis is vertical, say. The light emerging from this sheet is polarized vertically, and the transmitted electric field is \(\overrightarrow{E_{0}}\).

The polarized light beam then passes through a second polarizing sheet, called the analyser, which is placed parallel to the polarizer with its transmission axis at an angle θ to the transmission axis of the polarizer. The component of \(\overrightarrow{E_{0}}\) which is perpendicular to the axis of the analyser is completely absorbed, and the component parallel to that axis is E₀ cos θ.

Since intensity varies as the square of the amplitude, the transmitted intensity varies as I = I₀ cos² θ

where I₀ is the incident light intensity on the analyser. This expression (known as Malus’s law) shows that the transmitted intensity is maximum when the transmission axes are parallel and zero when the transmission axes are perpendicular to each other.

Crossed polarizers are a pair of polarizers with their transmission axes perpendicular to each other so that the transmitted light intensity is zero.

Question 35.
If the angle made by the axis of polarization of the second polarizer to that of the first polarizer is 60°, what can you say about the intensity of the light transmitted by the second polarizer?
Answer:
If I₁ is the intensity of the light incident on the second polarizer and I₂ is the intensity of the light transmitted by the second polarizer,
I₂ = I₁ cos² θ = I₁ cos² 60°
= I₁\(\left(\frac{1}{2}\right)^{2}\)
= \(\frac{\boldsymbol{I}_{1}}{4}\)

Question 36.
If 75% of incident light is transmitted by the second polarizer, what is the angle made by the transmission axis of the second polarizer to the transmission axis of the first polarizer?
Answer:

Question 37.
Explain the phenomenon of polarization of light by reflection.
Answer:
Consider a ray of unpolarized monochromatic light incident at an angle θ_B on a boundary between two transparent media as shown. Medium 1 is a rarer medium with refractive index n₁ and medium 2 is a denser medium with refractive index n₂. Part of incident light gets refracted and the rest

gets reflected. The degree of polarization of the reflected ray varies with the angle of incidence.

The electric field of the incident wave is in the plane perpendicular to the direction of propagation of incident light. This electric field can be resolved into a component parallel to the plane of the paper, shown by double arrows, and a component perpendicular to the plane of the paper shown by dots, both having equal magnitude. Generally, the reflected and refracted rays are partially polarized, i.e., the two components do not have equal magnitude.

In 1812, Sir David Brewster discovered that for a particular angle of incidence. θ_B, the reflected wave is completely plane-polarized with its electric field perpendicular to the plane of the paper while the refracted wave is partially polarized. This particular angle of incidence (θ_B) is called the Brewster angle.

For this angle of incidence, the refracted and reflected rays are perpendicular to each other. For angle of refraction θ_r,
θ_B + θ_r = 90° …… (1)
From Snell’s law of refraction,
∴ n₁ sin θ_B = n₂ sin θ_r … (2)
From Eqs. (1) and (2), we have,

This is called Brewster’s law.

Question 38.
Define polarizing angle. At the polarizing angle, what is the plane of polarization of the reflected ray ?
Answer:
Polarizing angle : The polarizing angle for an interface is the angle of incidence for a ray of unpolarized light at which the reflected ray is completely polarized.

At the polarizing angle, the reflected ray is completely plane polarized in the plane of incidence.

Question 39.
Give one example in which polarization by reflection is used.
Answer:
Polarization by reflection is used to cut out glare from nonmetallic surfaces. Special sunglasses are used for this purpose. Sunglasses fitted with Polaroids reduce the intensity of partially or completely polarized / reflected light incident on the eyes from reflecting surfaces.

Question 40.
State any four uses of a Polaroid.
Answer:

1. A Polaroid lens filter makes use of polarization by reflection. This filter is used in photography to reduce or eliminate glare from reflective nonmetallic surfaces like glass, rock faces, water and foliage. It can also deepen the colour of the skies.
2. Polaroid sunglasses reduce the transmitted intensity by a factor of one half and also reduce or eliminate glare from nonmetallic surfaces such as asphalt roadways and snow fields.
3. Polaroid filters are used in liquid crystal display (LCD) screens.
4. Polaroids are used to produce and show 3-D movies to give the viewer a perception of depth.

[Note :Three-dimensional movies are filmed from two slightly different camera locations and shown at the same time through two projectors fitted with polarizers having different transmission axes. The audience wear glasses which have two Polaroid filters whose transmission axes are the same as those on the projectors.]

Question 41.
Explain the phenomenon of polarization by scattering.
Answer:
When a beam of sunlight strikes air molecules or dust particles whose size is of the order of wavelength of light, the beam gets scattered. The scattered light observed in a direction perpendicular to the direction of incidence, is plane polarized. This phenomenon is called polarization by scattering.

As shown in above figure, a beam of an un-polarized light is incident along the Z-axis on a molecule. Light waves being transverse in nature, all the possible directions of vibration of electric field vector in the unpolarized light are confined to the XY plane. The light incident on the molecule is scattered by the electromagnetic field of the molecule.

When observed along the X-axis, only the vibrations of electric field vector which are parallel to Y-axis can be seen. In the similar manner, when observed along the Y-axis, only the vibrations of electric field vector which are parallel to the X-axis can be seen. Thus, the light scattered in a direction perpendicular to the incident light is plane polarized.

Question 42.
Solve the following.

Question 1.
The angle between the transmission axes of two polarizers is 45°. What will be the ratio of the intensities of the original light and the transmitted light after passing through the second polarizer?
Solution :
Data : θ = 45°
According to Malus’ law,
I₂ = I₂ cos² θ

= 2
This is the required ratio.

Question 2.
Two polarizers are so oriented that the maximum amount of light is transmitted. To what fraction of its maximum value is the intensity of the transmitted light reduced when the second polarizer is rotated through
(a) 300
(b) 600?
Solution:
Data : θ = 30°, 60°,
I₂ = I₂ cos²θ
where I₁ is the intensity of the incident light and I₂ is that of the transmitted light.
(i) For θ = 30°

I₂ = 0.75 I₁
= 75% of I₁

(ii) For θ = 60°
I₂ = I₁ (cos 60°)²
= I₁\(\left(\frac{1}{2}\right)^{2}\) = \(\frac{1}{4}\)I₁
= 0.25 I₁ = 25% of I₁

Question 3.
For a glass plate as a polarizer with refractive index 1.633, calculate the angle of incidence at which reflected light is completely polarized.
Solution :
Data : n = 1.633
tan θ_B = n = 1.633
∴ The polarizing angle,
θ_B = tan^-1 1.633 = 58°31′

Question 4.
Find the refractive index of glass if the angle of incidence at which the light reflected from the surface of the glass is completely polarized is 58°.
Solution :
Data : θ_B = 58°
n_g = tan θ_B = tan 58° .
= 1.6003
This is the refractive index of the glass.

Question 5.
The critical angle for a glass-air interface is sin^-1\(\frac{5}{8}\). A ray of unpolarized monochromatic light in air is incident on the glass. What is the polarizing angle?
Solution :
Data : θ_c = sin^-1 \(\frac{5}{8}\)
∴ sin θ_c = \(\frac{5}{8}\)
∴ n = \(\frac{1}{\sin \theta_{\mathrm{c}}}\) = \(\frac{8}{5}\) = 1.6
Also, n = tan θ_B
∴ θ_B = tan^-1 n
∴ The polarizing angle, θ_B = tan^-1 1.6 = 58°

Question 6.
The refractive index of a medium is \(\sqrt{3}\). What is the angle of refraction, if the unpolarized light is incident on it at the polarizing angle of the medium?
Solution :

This is the angle of refraction.

Question 7.
For a given medium, the polarizing angle is 60°. What is the critical angle for this medium ?
Solution :

This is the critical angle for the medium.

Question 8.
Unpolarized light is incident on a plane glass surface. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other? [Given n = 1.5]
Solution :
Data : Refractive index, n = 1.5
The reflected and refracted rays will be perpendicular to each other, when the angle of incidence = the polarizing angle θ_B,
tan θ_B = n = 1.5
∴ θ_B = tan^-1 (1.5) = 56°19′

Question 9.
If a glass plate of refractive index 1.732 is to be used as a polarizer, what would be the
(i) polarizing angle and
(ii) angle of refraction?
Solution :
Data : n_g = 1.732
n_g = tan θ_B
0B = tan^-1 (1.732) = 60°
This is the polarizing angle,

∴ θ_r = sin^-1\(\left(\frac{0.8660}{1.732}\right)\)
= sin^-1 (0.5) = 30°
This is the angle of refraction.

Question 10.
For a certain unpolarized monochromatic light incident on glass and water, the polarizing angles are 59°32′ and 53°4′, respectively. What would be the polarizing angle for the light if it is incident from water on to the glass?
Solution:

Question 11.
The wavelengths of a certain blue light in air and in water are 4800 Å and 3600 Å, respectively. Find the corresponding Brewster angle.
Solution:

This is the Brewster angle for the given light incident on water surface.

Question 12.
A ray of light is incident on a glass slab at the polarizing angle of 58°. Calculate the change in the wavelength of light in glass.
Solution :


∴ The change in the wavelength of the light in glass, λ_a – λ_b = 0.37514λ_a, i.e., 37.51% of its wavelength in air.

Question 13.
For what angle of incidence will light incident on a bucket filled with liquid having refractive index 1.3 be completely polarized after reflection ?
Solution:
The reflected light will be completely polarized when the angle of incidence is equal to the Brewster’s angle which is given by θ_B = tan^-1 \(\frac{n_{2}}{n_{1}}\), where n₁ and n₂ are refractive indices of the first and the second medium respectively. In this case, n₁ = 1 and n₂ = 1.3.
Thus, the required angle of incidence = Brewster’s angle = tan^-1\(\frac{1.3}{1}\) = 52.26°

Question 43.
State and explain the principle of superposition of waves.
Answer:
Principle of superposition of waves : The dis-placement at a point due to the combined effect of a number of waves arriving simultaneously at the point is the vector sum of the displacements due to the individual waves arriving at the point.

Explanation : This is a general principle of linear systems applied to wave phenomena. When two or more wave trains arrive at a point simultaneously, they interpenetrate without disturbing each other. The resultant displacement at that point is the vector sum of the displacements due to the individual waves arriving at the point. The amplitude and phase angle of the resulting disturbance are functions of the individual amplitudes and phases.

Notes :

1. In the case of mechanical waves, e.g., sound, the displacement is that of a vibrating particle of the medium.
2. When we consider superposition of electromagnetic waves, light in this chapter, the term displacement refers to the electric field component of the electromagnetic wave.

Question 44.
Explain what you understand by interference of light.
Answer:
The phenomenon in which the superposition of two or more light waves produces a resultant disturbance of redistributed light intensity or energy is called the interference of light.

Light waves are transverse in nature. If two monochromatic light waves of the same frequency arrive in phase at a point, the crest of one wave coincides with the crest of the other and the trough of one wave coincides with the trough of the other. Therefore, the resultant amplitude and hence the resultant intensity of light at that point is maximum and the point is bright. This phenomenon is called constructive interference. If two light waves having the same amplitude are in opposite phase, the crest of one wave coincides with the trough of the other. Therefore, the resultant amplitude, and hence the intensity, at that point is minimum (zero) and the point is dark. This phenomenon is called destructive interference. If the amplitudes are unequal, the resultant amplitude is minimum, but not zero. At other points, the intensity of light lies between the maximum and zero.

Question 45.
Explain how the phenomenon of interference can be demonstrated in a ripple tank.
Answer:

1. ‘Two pins, a small distance d apart, are attached to the electrical vibrator or an electrically maintained tuning fork of a ripple tank. The pins are kept vertical with their tips in contact with the surface of water in the ripple tank.
2. When the vibrator is switched on, the two pins vibrate together in phase with the same frequency and the same amplitude. Their tips form the sources S₁ and S₂ of circular waves which spread outward along the water surface.
   
3. The waves from the two sources interfere constructively at points where they meet in phase. Thus, where the crest of a wave from S₁ is superposed on the crest of a wave from S₂ (such as point A), and where the trough of a wave from S₁ is superposed on the trough of a wave from S₂ (such as point B), the water molecules have maximum amplitude of vibration.
4. The waves from the two sources interfere destructively at points where they meet in opposite phase. Thus, where the crest of a wave from S₁ is superposed on the trough of a wave from S₂ (such as point C), and where the trough of a wave from S₁ is superposed on the crest of a wave from S₂ (such as point D), the water molecules have minimum amplitude of vibration.

Question 46.
What are coherent sources? Is it possible to observe interference pattern with light from any two different sources ? Why ?
Answer:
Coherent sources : Two sources of light are said to be coherent if the phase difference between the emitted waves remains constant.

It is not possible to observe interference pattern with light from any two different sources. This is because, no observable interference phenomenon occurs by superposing light from two different sources. This happens due to the fact that different sources emit waves of different frequencies. Even if the two sources emit light of the same frequency, the phase difference between the wave trains from them fluctuates randomly and rapidly, i.e., they are not coherent.

Consequently, the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed.

Question 47.
State the conditions for constructive and destructive interference of light.
Answer:
(1) Constructive interference (brightness) : There is constructive interference at a point and the brightness or intensity is maximum there, if the two waves of light of the same frequency arrive at the point in phase, i.e., with a phase difference of zero or an integral multiple of 2π radians.
A phase difference of 2π radians corresponds to a path difference λ, where λ is the wavelength of light. Since

for constructive interference with maximum intensity of light, phase difference = 0, 2π, 4π, 6π, … rad
= n(2π) rad
or path difference = 0, λ, 2λ, 3λ, …, etc.
= nλ
where n = 0, 1, 2, 3, …, etc.

(2) Destructive interference (darkness) : There is destructive interference at a point and the point is the darkest, i.e., the intensity of light is minimum, if the two waves of light of the same frequency and intensity arrive at the point in opposite phase, i.e., with a phase difference of an odd-integral multiple of π radians. A phase difference 2π radians corresponds to a path difference λ, where λ is the wavelength of light.
∴ For destructive interference with minimum intensity of light, phase difference = π, 3π, 5π, … rad
= (2m – 1)π rad
or path difference = λ/2, 3λ/2, 5λ/2, …, etc.
= (2m – 1)\(\frac{\lambda}{2}\)
where m = 1, 2, 3, …, etc.

Question 48.
In Young’s double-slit experiment using light of wavelength 5000Å, what phase difference corresponds to the 11th dark fringe from the centre of the interference pattern?
Answer:
The required phase difference is (2m – 1)π rad = (2 × 11 – 1)π rad = 21π rad.

Question 49.
In Young’s double-slit experiment, if the path difference at a certain point on the screen is 2.999 λ, what can you say about the intensity of light at that point ?
Answer:
The intensity of light at that point will be close to the maximum intensity and the point will be nearly bright as the path difference = 2.999 λ \(\approx\) 3λ (integral multiple of λ).

Question 50.
In Young’s double-slit experiment, if the path difference at a certain point on the screen is 7.4999 λ, what can you say about the nature of the illumination at that point ?
Answer:
The point will be nearly dark as the path difference = 7.4999 λ \(\approx\) (8 – 0.5) λ, which is of the form
(m – \(\frac{1}{2}\))λ, where m = 1, 2, … …. 8

Question 51.
How is Young’s interference experiment performed using a single source of light?
Answer:
When a narrow slit is placed in front of an intense source of monochromatic light, cylindrical wave-fronts propagate from the slit. In Young’s experiment, two coherent sources are then obtained by wavefront splitting by placing a second screen with two narrow slits at a small distance from the first slit.

Question 52.
State any two points of importance of Young’s experiment to observe the interference of light.
Answer:
Importance of Young’s experiment observe the interference of light:

1. It was the first experiment (1800-04) in which the interference of light was observed.
2. This experiment showed that light is propagated in the form of waves.
3. From this experiment, the wavelength of monochromatic light can be determined.

Question 53.
What is the nature of the interference pattern obtained using white light?
Answer:
With white light, one gets a white central fringe at the point of zero path difference along with a few coloured fringes on both the sides, the colours soon fade off to white.

The central fringe is white because waves of all wavelengths constructively interfere here. For a path difference of \(\frac{1}{2}\)λ_violet, complete destructive interference occurs only for the violet colour; for waves of other wavelengths, there is only partial destructive interference. Consequently, we have a line devoid of violet colour and thus reddish in appearance. A point for which the path difference = \(\frac{1}{2}\)λ_red is similarly devoid of red colour, and appears violettish. Thus, following the white central fringe we have coloured fringes, from reddish to violettish. Beyond this, the fringes disappear because there are so many wavelengths in the visible region which constructively interfere that we observe practically uniform white illumination.

Question 54.
In Young’s double-slit experiment, the second minimum in the interference pattern is exactly in front of one slit. The distance between the two slits is d and that between the source and screen is D. What is the wavelength of the light used ?
Answer:
The distance of the mth minimum from the central fringe is,

This is the wavelength of the light used.

Question 55.
In Young’s double-slit experiment, a glass slide of refractive index n_g and thickness b is placed in front of one of the slits. What happens to the interference pattern and fringe width ? Derive an expression for the positions of the bright fringes in the interference pattern.
Answer:
Suppose, in Young’s double-slit experiment, a glass slide of refractive index n_g and thickness b is placed in front of the slit S₁.

If the experiment is set up in free space, this introduces an additional optical path length (n_g – 1)b in the path S₁P.
Then, the optical path difference to the point P from S₁ and S₂ is,
∆l = S₂P – [S₁P + (n_g – 1)b] = (S₂P – S₁P) – (n_g – 1)b = y\(\frac{d}{D}\) – (n_g – 1)b … (1)
where y = PO’, d is the distance between S₁ and S₂, and D is the distance of the screen from S₁ and S₂. Thus, point P will be bright (maximum intensity) if ∆l = nλ, where n = 0, 1, 2, … . The central bright fringe, corresponding to n = 0, will be obtained at a distance y₀ from O’ such that,
∆l = y_o\(\frac{d}{D}\) – (n_g – 1)b = 0
∴ y_o\(\frac{d}{D}\) = (n_g – 1)b ∴ y₀ = \(\frac{D}{d}\)(ng-l) b …(2)
Therefore, the central bright fringe and the interference pattern will shift up (towards P) by a distance y₀ given by EQ. (2).
The distance of the nth bright fringe from O’ towards P is,

The distance of the (n + l)th bright fringe from O’ towards P is

Therefore, the fringe width,
w = y_n+1 – y_n = \(\frac{\lambda D}{d}\) ….(5)
Thus, the fringe width remains unchanged.

Question 56.
What happens to the interference pattern when the phase difference between the two sources of light changes with time ?
Answer:
If the two sources do not maintain their phase relation during the time required for observation, the intensity of light at any point on the screen and consequently the interference pattern changes rapidly, and hence steady interference pattern is not observed.

Question 57.
Obtain expressions in terms of electric field, for the resultant amplitude and the intensity for the interference pattern produced by monochromatic light waves from two coherent sources.
Answer:
Consider a two-source interference pattern produced by superposition of monochromatic light waves of angular frequency ω, wavelength λ and constant phase difference, φ. At the centre of a bright fringe, there is constructive interference. Here, the amplitude of the resultant wave is double the amplitude of the wave incident on AB. Now, the intensity is proportional to the square of the amplitude of the wave. Hence, the resultant intensity is, I = 4I₀, where I₀ is the intensity of the incident wave. At the centre of a dark fringe, there is destructive interference. Here, the amplitude of the resultant wave and hence the intensity is zero. At other points the intensity is between 4I₀ and zero, depending on the phase difference.

Let the equations of the two waves coming from S₁ and S₂ at some point in the interference pattern be, E₁ = E₂ sin ωt and E₂ = E₀ sin (ωt + φ) respectively, where E₀ is the amplitude of the electric field vector.

By the principle of superposition of waves, the resultant electric field at that point is the algebraic sum, E = E₁ + E₂
∴ E = E₀ sin ωt + E₀ sin (ωt + φ)
= E₀ [sin ωt + sin (ωt + φ)]
= 2E₀ sin (ωt + φ/ 2) cos (φ/2)
The amplitude of the resultant wave is 2 E₀ cos (φ/2).
Therefore, the intensity at that point is I ∝ |2 E₀ cos (φ/2) |²
∴ I = 4I₀ cos² (φ/2)
as I₀ ∝ |E₀|².

Question 58.
Monochromatic light waves of amplitudes E₁₀ and E₂₀ and a constant phase difference φ produce an interference pattern. State an expression for the resultant amplitude at a point in the pattern. Hence, deduce the conditions for
(i) constructive interference with maximum intensity
(ii) destructive interference with minimum intensity. Also show that the ratio of the maximum and minimum intensities is
\(\frac{I_{\max }}{I_{\min }}\) = \(\left(\frac{\boldsymbol{E}_{10}+\boldsymbol{E}_{20}}{\boldsymbol{E}_{10}-\boldsymbol{E}_{20}}\right)^{2}\)
Answer:
Consider a two-source interference pattern produced by monochromatic light waves of angular frequency ω, wavelength λ, amplitudes E₁₀ and E₂₀ and a constant phase difference φ.

Let the individual electric fields along the same line due to the waves at some point in the interference pattern be
E₁ = E₁₀ sin ωt and E₂ = E₂₀ sin (ωt + φ)
By the principle of superposition of waves, the resultant electric field (displacement) at that point is the algebraic sum

Since, the intensity of a wave is proportional to the square of its amplitude, the resultant intensity at P,
I ∝ |R|²

Thus, the intensity depends on cos φ.
The condition for constructive interference with maximum intensity is cos φ is maximum, equal to 1, i.e., φ = 2nπ (n = 0, 1, 2, 3…) … (3)
The condition for destructive interference with minimum intensity is cos φ is minimum equal to

Question 59.
If the amplitude ratio (E₂₀/E₁₀) of two interfering coherent waves producing an interference pattern is 0.5, what is the ratio of the minimum intensity to maximum intensity?
Answer:

Question 60.
If the amplitude ratio (E₂₀/E₁₀) of two interfering coherent waves producing an interference pattern is 0.8, what is the ratio of the maximum intensity to the minimum intensity?
Answer:

Question 61.
Two slits in Young’s experiment have widths in the ratio 2 : 3. What is the ratio of the intensities of light waves coming from them?
Answer:
The required ratio is \(\frac{I_{1}}{I_{2}}\) = \(\frac{w_{1}}{w_{2}}\) = \(\frac{2}{3}\)

Question 62.
Monochromatic light waves of intensities I₁ and I₂, and a constant phase difference φ produce an interference pattern. State an expression for the resultant intensity at a point in the pattern. Hence deduce the expressions for the resultant intensity, maximum intensity and minimum intensity if I₁ = I₂ = I₀.
Answer:
Consider a two-source interference pattern produced by monochromatic light waves of intensities I₁ and I₂, and a constant phase difference φ. The resultant intensity at a point in the pattern is

At a point of constructive interference with maximum intensity, cos φ = 1.
∴ I_max = 2I₀(1 + 1) = 4I₀ …(3)
At point of destructive interference, with minimum intensity, cos φ = – 1.
∴ I_min = 2I₀(1 – 1) = 0… (4)
Notes :

(1) Since 1 + cos φ = 2 cos²φ, EQ. (2) above can be expressed as I = 4I₀ cos²\(\frac{\phi}{2}\) = I_max cos²\(\frac{\phi}{2}\).

(2) The average of cos² \(\frac{\phi}{2}\), averaged over one cycle, is \(\frac{1}{2}\).
Therefore, the average intensity of a bright and dark fringe in an interference pattern is I_av = 4I₀ × \(\frac{1}{2}\) = 2I₀. The graph of l versus φ is shown, in which the dotted line shows I_av.

In the absence of interference phenomenon, the intensity at a point due to two waves each of intensity I₀ is 2I₀, which is the same as the average intensity in an interference pattern. Thus, interference of light is consistent with the law of conservation of energy. Interference produces a redistribution of energy out of regions where it is destructive into the regions where it is constructive.

Question 63.
At a point in an interference pattern, the two interfering coherent waves of equal intensity I₀ have phase difference 60°. What will be the resultant intensity at that point ?
Answer:
Resultant intensity, I₀ = 2I₀ (1 + cos φ)
= 2I₀(1 + cos 60°) = 2I₀(1 + \(\frac{1}{2}\)) = 3I₀.

Question 64.
Solve the following :

Question 1.
Find the ratio of intensities at two points X and Y on a screen in Young’s double-slit experiment where waves from the slits S₁, and S₂ have path difference of 0 and \(\frac{\lambda}{4}\) respectively.
Solution :

Question 2.
Two coherent sources, whose intensity ratio is 81 : 1, produce interference fringes. Calculate the ratio of the intensities of maxima and minima in the fringe system.
Solution :
Data : I₁ : I₂ = 81 : 1
If E₁₀ and E₂₀ are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is

Question 3.
In Young’s double-slit experiment, the ratio of the intensities at the maxima and minima in the interference pattern is 36 : 16. What is the ratio of the widths of the two slits?
Answer:

Question 4.
In Young’s double-slit experiment, the ratio of the intensities of the maxima and minima in an interference pattern is 36 : 9. What is the ratio of the intensities of the two interfering waves?
Answer:

∴ The ratio of the intensities of the two interfering waves is 9 : 1.

Question 5.
Two slits in Young’s double-slit experiment have widths in the ratio 81 : 1. What is the ratio of the amplitudes of light waves coming from them?
Solution:
Data : w₁ : w₂ = 81 : 1
Since the intensity of a wave is directly proportional to the square of its amplitude,
\(\frac{I_{1}}{I_{2}}\) = \(\left(\frac{E_{10}}{E_{20}}\right)\) …. (1)
Also, the intensity of a wave coming out of a slit is directly propotional to the slit width.
∴ \(\frac{I_{1}}{I_{2}}\) = \(\frac{w_{1}}{w_{2}}\) …. (2)
From Eqs. (1) and (2),

The ratio of the amplitudes of light waves from the slits is 9: 1.

Question 6.
Two monochromatic light waves of equal intensities produce an interference pattern. At a point in the pattern, the phase difference between the interfering waves is π/3 rad. Express the intensity at this point as a fraction of the maximum intensity in the pattern.
Solution :
Data : I₁ = I₂ = I₀, φ = π/3 rad
The resultant intensity at the point in the pattern is

Question 7.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ is I. What is the intensity of light at a point where the path difference is λ/3 ?
Solution :
Data : ∆l₁ = λ, I₁ = I, ∆l₂ = λ/3
We assume that light waves coming out of the two slits are of equal intensity I₀.
Then, at a point in the interference pattern where the phase difference between the interfering waves is φ, the resultant intensity is,
I = 2I₀ (1 + cos φ)
Phase difference (φ) = \(\frac{2 \pi}{\lambda}\) × path difference (∆l)


∴ A The intensity of light at a point where the path difference is λ/3 is λ/4.

Question 8.
The optical path difference between identical waves from two coherent sources and arriving at a point is 87λ. What can you say about the resultant intensity at the point? If the path difference is 49.19 μm, calculate the wavelength of light.
Solution :
Data : Path difference = 87λ = 49.19 /rm.
(i) Path difference = 87λ = nλ, where n = 87. As the path difference is an integral multiple of the wavelength λ, the point where the waves interfere is a bright point with maximum intensity.

(ii) 87 λ = 49.19 μn = 49.19 × 10^-6 m .
∴ λ = \(\frac{49.19 \times 10^{-6}}{87}\) = 5.654 × 10^-7 m = 5654 Å

Question 9.
The optical path difference between identical waves from two coherent sources and arriving at a point is 172. What can you say about the resultant intensity at the point? If the path difference is 9.18 μm, calculate the wavelength of light.
Solution :
Data : Path difference = 17λ = 9.18 μm
(i) Path difference = 17λ = nλ, where n = 17
As the path difference is an integral multiple of the wavelength λ, the point where the waves interfere is a bright point with maximum intensity.

(ii) 17λ = 9.18 μm = 9.18 × 10^-6 m

Question 10.
At a point on the two-slit interference pattern obtained using a source of green light of wavelength 5500 Å, the path difference is 4.125 pm. Is the point at the centre of a bright or dark fringe ? Hence, find the order of the fringe.
Solution :
Data : Path difference, ∆l = 4.125 × 10^-6 λ = 5500 Å = 5.5 × 10^-7 m

As the path difference is an odd integral multiple of \(\frac{\lambda}{2}\) the point is at the centre of a dark fringe.
∴ p = 2m – 1 (m = 1, 2, 3…)
∴ 2m – 1 = 15 ∴ m = 8
∴ The order of the fringe is 8 (i.e., the point lies at the centre of the 8th dark fringe.)

Question 11.
The two slits in an interference experiment are illuminated by light of wavelength 5600 A. Deter-mine the path difference and phase difference between the waves arriving at the centre of the eighth dark fringe on the screen.
Solution :
Data : λ = 5600 A = 5.6 × 10^-7 m
Order of the dark fringe is 8, ∴ m = 8
For the fringe to be dark, path difference
= (2m – 1)\(\frac{\lambda}{2}\)
∴ Path difference, ∆l = (2 × 8 – 1)\(\frac{\lambda}{2}\) = 7.5λ
= 7.5 × 5.6 × 10^-7 m
= 4.2 × 10^-6 m
Phase difference, φ = \(\frac{2 \pi}{\lambda}\) × ∆l
= \(\frac{2 \pi}{5.6 \times 10^{-7}}\) × 4.2 × 10^-6 = 15π

Question 12.
In Young’s double-slit experiment, interference fringes are observed on a screen 1 m away from the two slits which are 2 mm apart. A point P on the screen is 1.8 mm from the central bright fringe,
(i) Find the path difference at P.
(ii) If the wavelength of the light used in 4800 Å, what can you say about the illumination at P?
Solution :
Data : D = 1 m, d = 2 mm = 2 × 10^-3 m, y = 1.8 mm = 1.8 × 10^-3 m, λ = 4800 Å = 4.8 × 10^-7 m

∴ Path difference = 15\(\frac{\lambda}{2}\)
As the path difference is an odd integral multiple of \(\frac{\lambda}{2}\), point P is a dark point with minimum intensity.

Question 13.
Two slits 1.25 mm apart are illuminated by light of wavelength 4500 Å. The screen is 1 m away from the plane of the slits. Find the separation between the 2nd bright fringe on one side and the 2nd bright fringe on the other side of the central maximum.
Solution :
Data : D = 1 m, d = 1.25 mm = 1.25 × 10^-3 m,
λ = 4500 Å = 4500 × 10^-10 m = 4.5 × 10^-7 m
Since, it is a second bright fringe, n = 2. If s is the distance between the 2nd bright fringe on one side and the 2nd bright fringe on the other side of the central maximum, then
s = 2y = 2nλ\(\frac{D}{d}\)

= 14.4 × 10^-4 m
W = 1.44 mm

Question 14.
A plane wavefront of light of wavelength 5000 Å is incident on two slits in a screen perpendicular to the direction of light rays. If the total separation of 10 bright fringes on a screen 2 m away is 4 mm, find the distance between the slits.
Solution:
Given : λ = 5000 Å = 5 × 10^-7 m,
D = 2 m and the total separation of 10 fringes = 4 mm = 4 × 10^-3 m

Question 15.
In Young’s double-slit experiment with slits of equal width, a point P on the screen is at a distance equal to one-fourth of the fringe width from the central maximum. If the intensity at the central maximum is I_c, find the intensity at P.
Solution :

Since, the slits have equal width, the intensities of the two interfering waves are equal, say I₀. Then the intensity at a point on the screen is
I = 4I₀ cos² \(\frac{\phi}{2}\)
At the central maximum, φ = 0.

The intensity at P is half that at the central maximum.

Question 16.
In a double-slit experiment, the optical path difference between the waves coming from two coherent sources at a point P on one side of the central bright band is 7.5 × 10^-6 m and that at a point Q on the other side of the central bright band is 1.8 × 10^-6 m. How many bright and dark bands are observed between points P and Q if the wavelength of light used is 6 × 10^-7 m ?
Solution :
Data : ∆l₁ = 7.5 × 10^-6 m, ∆l₂ = 1.8 × 10^-6 m λ = 6 × 10^-7 m
For point P : Let p\(\frac{\lambda}{2}\) = ∆l₁

The path difference ∆l₁ is an odd integral multiple of λ/2 : ∆l₁ = (2m – 1)\(\frac{\lambda}{2}\), where m is an integer,
∴ 2m – 1 = 25 ∴ m = 13
∴ Point P is at the centre of the 13th dark band.
For point Q :
Let q\(\frac{\lambda}{2}\) = ∆l₂

The path difference ∆l₂ is an even integral multiple of \(\frac{\lambda}{2}\) : ∆l₂ = (2n)\(\frac{\lambda}{2}\), where n is an integer
∴ 2n = 6 ∴ n = 3
∴ Point Q is at the centre of the 3rd bright band. Between points P and Q, excluding the respective bands at P and Q, the number of dark bands = 12 + 3 = 15 and the number of bright bands (including the central bright band) = 12 + 2 + 1 = 15

Question 17.
In Young’s double-slit experiment, light waves of wavelength 5.2 × 10^-7 m and 6.5 × 10^-7 m are used in turn keeping the same geometry. Compare the fringe widths in the two cases.
Solution :
Data : λ₁ = 5.2 × 10^-7 m, λ₂ = 6.5 × 10^-7 m
As, W = \(\frac{\lambda D}{d}\) and the geometry is the same, i.e., D and d remain the same,

Question 18.
Two coherent sources are 1.8 mm apart and the fringes are observed on a screen 80 cm away from them. It is found that with a certain source of light, the fourth bright fringe is situated at a distance of 1.08 mm from the central fringe. Calculate the wavelength of light.
Solution :
Data : d = 1.8 mm = 1.8 × 10^-3 m
D = 80 cm = 0.8 m
For fourth bright fringe, n = 4

This is the wavelength of light.

Question 19.
Green light of wavelength 5100 Å from a narrow slit is incident on a double-slit. If the overall separation of 10 fringes on a screen 200 cm away from it is 2 cm, find the slit-separation.
Solution :
Data : λ = 5100 Å = 5.1 × 10^-7 m,
W = \(\frac{2}{10}\) cm = 2 × 10^-3 m,
D = 200 cm = 2 m
W = \(\frac{\lambda D}{d}\)
∴ d = \(\frac{\lambda D}{W}\) = \(\frac{5.1 \times 10^{-7} \times 2}{2 \times 10^{-3}}\)
= 5.1 × 10^-4 m
This is the slit-separation.

Question 20.
Sodium light of wavelength 5.896 × 10^-7 m is passed through two pinholes 0.5 mm apart, and an interference pattern is formed on a screen kept parallel to the plane of the pinholes and 1.2 m from them. Find the distance between
(i) the second and the fifth bright fringes
(ii) the third and the seventh dark fringes on the same side of the central bright point.
Solution :
Data : λ = 5.896 × 10^-7 m, d = 0.5 mm= 0.5 × 10^-3 m, D = 1.2 m
(i) The distance of the nth bright fringe from the central bright point is

The distance between the second and the fifth bright fringes on the same side of the central bright point is 4.245 × 10^-3 m.

(ii) The distance of the mth dark fringe from the central bright point is

= 5.66 × 10^-3 m
The distance between the third and the seventh dark fringes on the same side of the central bright point is 5.66 × 10^-3 m.

Question 21.
In Young’s double-slit experiment, the two slits are 2 mm apart. The interference fringes for light of wavelength 6000Å are formed on a screen 80 cm away from them.
(i) How far is the second bright fringe from the central bright point?
(ii) How far is the second dark fringe from the central bright point?
Solution:
Data: D = 80 cm = 0.8 m,
d = 2 mm = 2 × 10^-3 m, λ = 6000Å = 6 × 10^-7 m
(i) For the second bright fringe from the central bright point, n = 2

This is the required distance.

(ii) For the second dark fringe from the central bright point, m = 2

This is the required distance.

Question 21.
In Young’s double-slit experiment, the distance between two consecutive bright fringes on a screen placed at 1.5 m from the two slits is 0.6 mm. What would be the fringe width, if the screen is brought towards the slits by 50 cm, keeping rest of the setting the same?
Solution:
Let λ be the wavelengths of light used and d the distance between the two sources (i.e., slits), if D is the distance between the sources and the screen, the fringe width is
w = \(\frac{\lambda D}{d}\)
For the same λ and d, W ∝ D.
∴ \(\frac{W_{2}}{W_{1}}\) = \(\frac{D_{2}}{D_{1}}\)
Data:
W₁ = 0.6 mm = 6 × 10^-4 m,
D₁ = 1.5 m, D₂ = 1 m

This is the required fringe width.

Question 22.
On passing light of wavelength 5000 Å through two pinholes 2 mm apart, an interference pattern is formed on a screen kept parallel to the plane of the pinholes and 100 cm from them. Find the distance between the fifth bright band on one side of the central bright band and the sixth dark band on the other side.
Solution :
Data : λ = 5000 Å = 5 × 10^-7 m, d = 2 mm = 2 × 10^-3 m, D = 100 cm = 1 m

∴ y_6d = (6 – \(\frac{1}{2}\)) × 2.5 × 10^-4m = 1.375 × 10^-3 m
∴ y_5b + y_6d = 1.25 × 10^-3m + 1.375 × 10^-3m
= 2.625 × 10^-3 m = 2.625 mm
This is the required distance.

Question 23.
Monochromatic light from a narrow slit illuminates two narrow slits 3 mm apart, producing an interference pattern with bright fringes 0.15 mm apart on a screen 75 cm away from the slits. Find the wavelength of the light. How will the fringe width be altered if
(a) the distance of the screen from the slits is doubled
(b) the separation between the slits is doubled ?
Solution :
Data : d = 3 mm = 3 × 10^-3 m,
W = 0.15 mm = 1.5 × 10^-4 m, D = 75 cm = 0.75 m

Question 24.
In Young’s double-slit experiment the slits are 2 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the ninth bright fringe is at a distance of 2.208 mm from the second dark fringe from the centre of the fringe pattern. Find the wavelength of light used.
Solution :
Data : d = 2 mm = 2 × 10^-3 m,

Question 25.
In Young’s double-slit experiment, the two slits separated by 4 mm are illuminated by light of wavelength 6400 Å. Interference fringes are obtained on a screen placed at a distance of 60 cm from the slits. Find the change in the fringe width if the separation between the slits is
(i) increased by 1 mm
(ii) decreased by 1 mm.
Solution :
Data : d = 4 mm = 4 × 10^-3 m, λ = 6.4 × 10^-7m, D = 0.6 m, d’ = 5 mm = 5 × 10^-3 m, d” = 3 mm = 3 × 10^-3m
Fringe width, W = \(\frac{\lambda D}{d}\) ∴ W ∝ \(\frac{1}{d}\)
(i) Since d’ > d, W’ < W, i.e., the fringe width decreases.
Decrease in the fringe width = W – W

(ii) Since d” < d, W” > W, i.e., the fringe width increases.
Increase in the fringe width = W” – W

Question 26.
In Young’s double-slit experiment, the separation between the slits is 3 mm and the distance between the slits and the screen is 1 m. If the wavelength of light used is 6000 Å, calculate the fringe width. What will be the change in the fringe width if the entire apparatus is immersed in a liquid of refractive index \(\frac{4}{3}\)?
Solution :
Data : d = 3 × 10^-3 m, D = 1 m, λ = 6 × 10^-7 m, n = \(\frac{4}{3}\)

Question 27.
In Young’s double-slit experiment using monochromatic light, the fringe pattern shifts by certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the slits-to-screen distance is doubled. It is found that the distance between successive maxima now is the same as the observed fringe shift with the mica sheet. Calculate the wavelength of the monochromatic light used.
Solution :
Data : n_m = 1.6, b = 1.964 microns
= 1.964 × 10^-6 m,
D₂ = 2D₁, W₂ = y₀
The fringe shift with the mica sheet,
y₀ = \(\frac{D_{1}}{d}\)(n_m – 1)b
Subsequent to the removal of the mica sheet and doubling the slits-to-screen distance, the new fringe width is,

Question 28.
What must be the thickness of a thin film which, when kept near one of the slits shifts the central fringe by 5 mm for incident light of wavelength 5890 Å in Young’s double-slit interference experiment ? The refractive index of the material of the film is 1.1 and the distance between the slits is 0.5 mm.
Solution :
Data : λ = 5890 Å, nm = 1.1, the shift of the central bright fringe = 5 mm

Let t be the thickness of the film and P the point on the screen where the central fringe has shifted. Suppose the film is kept in front of slit S1. Due to the film, the optical path travelled by the light passing through it increases by (1.1 – 1)f = 0.1t. Thus, the optical paths between the two beams passing through the two slits are not equal at the midpoint of the screen but are equal at P, 5 mm away from the centre. At this point the distance travelled by light from the other slit S₂ to the screen is larger than that travelled by light from S₁ by 0.1t.

The difference in distances, S₂P – S₁P = yλ/d, where y is the distance along the screen = 5 mm = 5 × 10^-3m and d = 0.5 mm = 5 × 10^-4 m.

This has to be equal to the difference in optical paths introduced by the film.
Thus, 0.1t = 5 × 10^-3 × 5890 × 10^-10/5 × 10^-4.
∴ t = 5890 × 10^-8m = 5.89 × 10^-5 m = 0.0589 mm

Question 29.
In a biprism experiment, the eyepiece is placed at a distance of 1.2 metres from the source. The distance between the virtual sources was found to be 7.5 × 10^-4 m. Find the wavelength of light if the eyepiece is to be moved transversely through a distance of 1.888 cm for 20 fringes.
Solution :
Data : D = 1.2 m, d = 7.5 × 10^-4 m,
20 W = 1.888 cm = 1.888 × 10^-2 m

Question 30.
A biprism is placed 5 cm from the slit illuminated by sodium light of wavelength 5890 Å. The width of the fringes obtained on a screen 75 cm from the biprism is 9.424 × 10^-2 cm. What is the distance between the two coherent sources?
Solution :
Data : D = 5 cm + 75 cm = 80 cm = 0.8 m,
λ = 5890 Å = 5.890 × 10^-7 m,
W = 9.424 × 10^-2 cm = 9.424 × 10^-4 m
Fringe width, W = \(\frac{\lambda D}{d}\)
∴ The distance between the two coherent sources,

= 5 × 10^-4 m = 0.5 mm

Question 31.
In a biprism experiment, the distance of the 20th bright band from the centre of the interference pattern is 8 mm. Calculate the distance of the 30th bright band from the centre.
Solution :
Data : y₂₀ = 8 mm (bright band)
The distance of the nth bright band from the centre of the interference pattern,

The distance of the 30th bright band from the centre of the interference pattern is 12 mm.

Question 32.
In a biprism experiment, a source of light having wavelength 6500 Å is replaced by a source of light having wavelength 5500 Å. Calculate the change in the fringe width, if the screen is at a distance of 1 m from the sources which are 1 mm apart.
Solution :

The fringe width decreases by 1 × 10^-4 m = 0.1 mm.

Question 33.
In a biprism experiment, light of wavelength 5200 Å is used to get an interference pattern on the screen. The fringe width changes by 1.3 mm when the screen is moved towards the biprism by 50 cm. Find the distance between the two virtual images of the slit.
Solution :
Data : λ = 5200 Å = 5.2 × 10^-7 m,
W₁ – W₂ = 1.3 mm = 1.3 × 10^-3 m,
D₁ – D₂ = 50 cm = 0.5 m

This is the distance between the two virtual sources.

Question 34.
In a biprism experiment, the 10th dark band is observed at 2.09 mm from the central bright point on the screen with red light of wavelength 6400 Å. By how much will the fringe width change if blue light of wavelength 4800 A is used with the same setting?
Solution :

Question 35.
In a biprism experiment, the slit is illuminated by red light of wavelength 6400 Å, and the cross wire of the eyepiece is adjusted at the centre of the 3rd bright band. On using blue light, it is found that the 4th bright band is on the cross wire. Find the wavelength of blue light.
Solution :
Data : λ_r = 6400 Å, y₃ (red, bright)

This is the wavelength of blue light.

Question 36.
In a biprism experiment, the wavelength of red light used is 6000 Å and the nth bright band is obtained at a point P on the screen. Keeping the same setting, the source is replaced by a source of green light of wavelength 5000 Å and the (n + 1)th bright band of green light coincides with point P. Find n.
Solution :

Question 37.
In a biprism experiment, the slit is illuminated by light of wavelength 4800 Å. The distance between the slit and the biprism is 15 cm and that between the biprism and the eyepiece is 85 cm. If the distance between the virtual sources is 3 mm, determine the distance between the 4th bright band on one side and the 4th dark band on the other side of the central band.
Solution :
Data : λ = 4800 Å = 4.8 × 10^-7 m,
d = 3 mm = 3 × 10^-3 m,
D = distance between the slit and the biprism + distance between the biprism and the eyepiece = 15 + 85 = 100 cm = 1 m The distance of the nth bright band from the central band is

This is the required distance.

Question 38.
An isosceles prism of refracting angle 179° and refractive index 1.6 is used as a biprism by keeping it 10 cm away from a slit, the edge of the biprism being parallel to the slit. The slit is illuminated by a light of wavelength 600 nm and the screen is 90 cm away from the biprism. Calculate the location of the centre of the 10th dark band from the centre of the interference pattern and the path difference at this location.
Solution :

Question 39.
In a biprism experiment, the distance between the second and tenth dark bands on the same side of the central bright band is 0.12 cm, that between the slit and the biprism is 20 cm and that between the biprism and the eyepiece is 80 cm. If the slit images given by the lens in the two positions are 4.5 mm and 2 mm apart, find the wavelength of light used.
Solution :
The distance between the second and tenth dark bands on the same side of the central band is equal to 8 times the fringe width (W).
∴ 8 W = 0.12 cm (by the data)
∴ W = \(\frac{0.12}{8}\) cm = 0.015 cm = 0.015 × 10^-2 m
The distance (D) between the slit and the eyepiece is equal to the sum of the distance between the slit and the biprism and the distance between the biprism and the eyepiece.
∴ D = 20 + 80 = 100 cm = 1 m (by the data)
Also, d₁ = 4.5 mm and d₂ = 2 mm
∴ The distance (d) between the virtual images of the slit is
d = \(\sqrt{d_{1} d_{2}}\) = \(\) mm = 3 mm
= 3 × 10^-3 m
∴ The wavelength of light,

Question 40.
In a biprism experiment, the slit and the eyepiece. are 10 cm and 80 cm away from the biprism. When a convex lens was interposed at 30 cm from the slit, the separation of the two magnified images of the slit was found to be 4.5 mm. If the wavelength of the source is 4500 Å, calculate the fringe width.
Solution:
Data : d₁ = 4.5 mm = 4.5 × 10^-3 m,
λ = 4500 Å = 4.5 × 10^-7 m,
distance between the slit and the eyepiece (D) = distance between the slit and the biprism + distance between the biprism and the eyepiece = 10 cm + 80 cm = 90 cm = 0.9 m, u₁ = 30 cm = 0.3 m v₁ = D – u₁ = 0.9m – 0.3m = 0.6 m
Linear magnification of a lens,

Question 65.
Describe with a neat labelled ray diagram the Fraunhofer diffraction pattern due to a single slit. Obtain the expressions for the positions of the intensity minima and maxima. Also obtain the expression for the width of the central maximum.
Answer:
When a parallel beam of monochromatic light of wavelength λ illuminates a single slit of finite width a, we observe on a screen some distance from the slit, a broad pattern of alternate dark and bright fringes. The pattern consists of a central bright fringe, with successive dark and bright fringes of diminishing intensity on both sides. This is called ‘ the diffraction pattern of a single slit.

Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance D (» a) from the slit and at the focal plane of the convex lens.

We can imagine the single slit as being made up of a large number of Huygens’ sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens’ wavelets.
Now, imagine the single slit as made up of two adjacent slits, each of width a/2. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will therefore also in phase at the point P₀ on the screen, where P₀ is equidistant from all the Huygens sources. At P₀, then, we get the central maximum.

For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and O (or O and B) is λ/2, which is the condition for destructive interference. Suppose, the nodal line OP for the first minimum subtends an angle θ at the slit; θ is very small. With P as the centre and PA as radius, strike an arc intersecting PB at C. Since, D » a, the arc AC can be considered a straight line at right angles to PB. Then, ∆ ABC is a right-angled triangle similar to A OP₀P.
This means that, ∠BAC = θ
∴ BC = a sin θ
∴ Difference in path length,
BC = PB – PA = (PB – PO) + (PO – PA)

(∵ θ is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the with minimum (wi = +1, ±2, ±3, …).
θ_m = \(\frac{m \lambda}{a}\) (with minimum) … (2)
as θ_m is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of \(\frac{\lambda}{2}\) :

(mth secondary maximum) … (3)
Width of the central maximum :
Equation (1) gives the angular half width of the central maximum. Therefore, the angular width of the central maximum is,
2θ = \(\frac{2 \lambda}{a}\) … (4)
From ∆OP₀P, P₀P = D tan θ \(\simeq\) D sin θ
(∵ θ is very small and in radian)
∴ y₁ = P₀P = \(\frac{D \lambda}{a}\) [from Eq. (1)] … (5)
This is the distance of the first minimum from the centre of the central maximum.
∴ Width of the central bright fringe :
W_c = 2y_1d = 2W = 2\(\left(\frac{\lambda D}{a}\right)\) …(6)

The central bright fringe is spread between the first dark fringes on either side. Thus, the width of the central bright fringe is the distance between the centres of the first dark fringe on either side.

If the lens is very close to the slit, D is very nearly equal to f, where f is the focal length of the lens. Then W_c = 2\(\left(\frac{\lambda D}{a}\right)\) = 2\(\left(\frac{\lambda f}{a}\right)\) … (7)

Question 66.
Represent graphically intensity distribution in
(a) Young’s double-slit interference
(b) single- slit diffraction and
(c) double-slit diffraction
Answer:

Question 67.
State the characteristics of a single-slit diffraction pattern.
Answer:
Characteristics of a single-slit diffraction pattern :

1. The image cast by a single-slit is not the expected purely geometrical image.
2. For a given wavelength, the width of the diffraction pattern is inversely proportional to the slit width.
3. For a given slit width a, the width of the diffraction pattern is proportional to the wavelength.
4. The intensities of the non-central, i.e., secondary, maxima are much less than the intensity of the central maximum.
5. The minima and the non-central maxima are of the same width, Dλ/a.
6. The width of the central maximum is 2Dλ/a. It is twice the width of the non-central maxima or minima.

Question 68.
Explain briefly the double-slit diffraction pattern.
Answer:
The double-slit diffraction pattern is determined by the diffraction patterns due to the individual slits, and by the interference between them.

We can see narrow interference fringes similar to those obtained in Young’s double-slit experiment. These fringes vary in brightness and the shape of their envelope as that of the single-slit diffraction pattern.

Question 69.
What should be the order of the size of an obstacle or aperture to produce diffraction of light?
Answer:
For pronounced diffraction, the size of an obstacle or aperture should be of the order of the wavelength of light or greater.
[Note : For diffraction from a single slit of width a with monochromatic light of wavelength λ, the condition for first minimum (dark fringe) is
sin θ₁ = \(\frac{\lambda}{a}\)
When a = λ, θ₁ = 90° and the central maximum spreads over 180°; then, while the diffraction is maximum, no fringe pattern is seen at all.

When a » X (say, a is of the order of a centimetre or more), θ₁ is so small that there is practically no diffraction and the illuminated region on the screen is almost as given by geometrical optics. However, diffraction pattern due to a straight-edge will always be seen at the edge of the illuminated region. Hence, for an observable fringe pattern due to a single slit, a should be of the order of X with a > λ..]

Question 70.
Solve the following.

Question 1.
Plane waves of light from a sodium lamp are incident on a slit of width 2 μm. A screen is located 2 m from the slit. Find the spacing between the first secondary maxima of two sodium lines as measured on the screen.
(Given : λ₁ = 5890 Å and ₂ = 5896 Å)
Solution:


This is the required spacing.

Question 2.
The diffraction pattern of a single slit of width 0. 5 cm is formed by a lens of focal length 40 cm. Calculate the distance between the first dark and the next bright fringe from the axis. The wavelength of light used is 4890 Å.
Solution :
Data : a = 0.5 cm = 0.5 × 10^-2 m,
D \(\simeq\) f = 40 cm = 0.4 m, λ = 4890 Å = 4.890 × 10^-7 m
The distance between the first dark fringe and the next bright fringe = \(\frac{\lambda D}{2 a}\)

Question 3.
Light of wavelength 6000 Å is incident on a slit of width 0.3 mm. A screen is placed parallel to the slit 2 m away from the slit. Find the position of the first dark fringe from the centre of the central maximum. Also, find the width of the central maximum.
Solution :
Data : λ = 6 × 10^-7 m, a = 0.3 mm = 3 × 10^-4 m,
D = 2 m
The distance y_m of the m th minimum from the centre of the central maximum is

The first dark fringe is 4 mm from the centre of the central fringe.
∴ Half-width of the central maximum = 4 mm
∴ The width of the central maximum = 2 × 4 mm = 8 mm

Question 4.
In a single-slit diffraction pattern, the distance between the first minimum on the right and the first minimum on the left is 5.2 mm. The screen on which the pattern is displayed is 80 cm from the slit and the wavelength of light is 5460 Å. Calculate the slit width.
Solution :
Data : y₁ (minimum, right) + y₁ (minimum, left)

Question 5.
In a single-slit diffraction pattern, the distance between the first minimum on the right and the first minimum on the left of the central maximum is 4 mm. The screen is 2 m from the slit and the wavelength of light used is 6000 Å. Calculate the width of the slit and the width of the central maximum.
Solution :
Data : y₁ (minimum, right) + y₁ (minimum, left)
= 4 mm = 4 × 10^-3 m, D = 2 m, λ = 6 × 10^-7 m

Question 6.
Determine the angular spread between the central maximum and first order maximum of the diffraction pattern due to a single slit of width 0.25 mm, when light of 6650 Å is incident on it normally.
Solution:
Data : λ = 6650 Å = 6650 × 10^-10m, a = 0.25 mm

This is the required angular spread.

Question 71.
Explain and define the resolving power of an optical instrument.
Answer:
Resolving power of an optical instrument:
The primary aim of using an optical instrument is to see fine details, whether observing a star system through a telescope or a living cell through a microscope. After passing through an optical system, light from two adjacent parts of the object should produce sharp, distinct (separate) images of those parts. The objective lens or mirror of a telescope or microscope acts like a circular aperture. The diffraction pattern of a circular aperture consists of a central bright spot (called the Airy disc and corresponds to the central maximum) and concentric dark and bright rings.

Light from two close objects or parts of an object after passing through the aperture of an optical system produces overlapping diffraction patterns that tend to obscure the image. If these diffraction patterns are so broad that their central maxima overlap substantially, it is difficult to decide if the intensity distribution is produced by two separate objects or by one.

The resolving power of an optical instrument, e.g., a telescope or microscope, is a measure of its ability to produce detectably separate images of objects that are close together.

Definition : The smallest linear or angular separation between two point objects which appear just resolved when viewed through an optical instrument is called the limit of resolution of the instrument and its reciprocal is called the resolving power of the instrument.

Question 72.
State and explain Rayleigh’s criterion for minimum resolution.
Answer:
Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.

The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.

Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.

Question 73.
Explain the Rayleigh criterion for the limit of resolution for
(i) two linear objects
(ii) a pair of point objects.
Answer:
(i) The Rayleigh criterion for the limit of resolution for two linear objects : Consider, two self luminous objects or slits separated by some distance. Let λ be the wavelength of the light and a the width of the slits. As per the Rayleigh criterion, the first minimum of the diffraction pattern of one of the sources should coincide with the central maximum of the other. Thus, it is at the just resolved condition.
The angular separation dθ (position) of the first principal minimum is,
dθ = \(\frac{\lambda}{a}\) …(1)
This angular separation between the two objects must be minimum as this minimum coincides with the central maximum of the other. This is called the limit of resolution of that instrument. It is written as,
limit of resolution, dθ = \(\frac{\lambda}{a}\)
Minimum separation between the two linear objects that are just resolved, at distance D from the instrument is, ni
y = D(dθ) = \(\frac{D \lambda}{a}\) …(2)
It is the distance of the first minimum from the centre.

(ii) The Rayleigh criterion for the limit of resolution for a pair of point objects : The objects to be viewed through a microscope are often of the point-size. The diffraction pattern of such objects consists of a central bright spot called the Airy disc and corresponds to the central maximum surrounded by concentric dark and bright rings called Airy rings.

If a red laser beam passes through a 90 µm pinhole aperture, the Airy disc and several orders (rings) of diffraction are as shown.

According to Lord Rayleigh, for such objects to be just resolved, the first dark ring of the diffraction pattern of the first object should be formed at the centre of the diffraction pattern of the second object and vice versa. Thus, the minimum separation between the images on the screen should be equal to the radius of the first dark ring.
This is applicable to the eye, microscope, telescope, etc.

Question 74.
Define and explain the resolving power of a microscope. State the expressions for the resolving power of
(i) a microscope with a pair of non-luminous objects
(ii) a microscope with self luminous point objects.
OR
What is meant by the limit of resolution and the resolving power of a microscope?
Answer:
Definition : The limit of resolution of a microscope is the least separation between two-closely spaced points on an object which are just resolved when viewed through the microscope.

Definition : The resolving power of a microscope is defined to be the reciprocal of its limit of resolution.

In a compound microscope, the objective lens forms a real, magnified image of an object placed just beyond the focal length of the lens. The objective has a short focal length (for greater magnification) and is held close to the object so that it gathers as much of the light scattered by the object as possible.

Let a be the least separation between two point objects O and O’ viewed through an objective AB of a compound microscope. The medium between the object and the objective has a refractive index n. The images of the objects O and O’ are I and I’ respectively. In this case, the angular separation between the objects, at the objective is 2α. D is the diameter of the objective AB.

According to the Rayleigh criterion, the first dark ring due to O’ should coincide with I and that of O should coincide with I’. The nature of illumination at a point on the screen is determined by the effective path difference at that point. Let us consider point I to be symmetric with respect to O. Paths of the extreme rays reaching I from O’ are O’AI and O’BI. The paths AI and BI are equal. Thus, the actual path difference is O’B – O’A.

The enlarged view of the region around O and O’ is as shown.
From this,
path difference = DO’ + O’C
= 2a sin α

(i) Microscope with a pair of non-luminous objects (dark objects):
In actual practice, the objects O and O’ viewed through a microscope are illuminated by the same source. Often the eyepiece of the microscope is filled with some transparent material of refractive index n. Then the wavelength of light in this material is
λ_n = \(\frac{\lambda}{n}\) where λ is the wavelength of light in air.

In such a set up the path difference at the first dark ring in λ_n. Thus, from eq (1),

The factor n sin a is called numerical aperture (NA). The resolving power of the microscope is
R = \(\frac{1}{a}\) = \(\frac{2 \mathrm{NA}}{\lambda}\) …. (3)

(ii) Microscope with self luminous point objects : Applying Abbe’s theory of Airy discs and rings to Fraunhofer diffraction due to a pair of self luminous point objects, the path difference between the extreme rays, at the first dark ring is 1.22 λ, thus, for the requirement of just resolution,

The resolving power of the microscope is,
R = \(\frac{1}{a}\) = \(\frac{\mathrm{NA}}{0.61 \lambda}\) …. (5)
When the value of a is minimum, the quality of resolution is high.
Notes:

1. If f and D are the focal length and the diameter of the microscope objective.
   sin i_max \(\simeq \frac{D / 2}{f}\) so that Eq. (3) can be written as resolving power = \(\frac{n D}{f \lambda}\)
2. Ernst Abbe(1840 -1905), German physicist and developer of optical instruments.

Question 75.
Explain why microscopes of high magnifying power have oil filled (oil-immersion) objectives.
Answer:
Higher angular magnification of a high magnifying power microscope is of little use if the finer details in a tiny object are obscured by diffraction effects. Hence, a microscope of high magnifying power must also have a high resolving power.
Resolving power ot a microscope = \(\frac{2 n \sin \alpha}{\lambda}\)
Where α ≡ the half angle of the angular separation between the objects, at the objective lens. n ≡ the refractive index of the medium between the object and the objective, λ ≡ the wavelength of the light used to illuminate the object.

The factor n sin α is called the numerical aperture of the objective and the resolving power increases with increase in the numerical aperture. To increase α the diameter of the objective would have to be increased. But this increase in aperture would degrade the image by decreasing the resolving power. Hence, in microscopes of high magnifying power, the object is immersed in oil that is in contact with the objective. Usually cedarwood oil having a refractive index 1.5 (close to that of the objective glass) is used. Closeness of the refractive indices also reduces loss of light by reflection at the objective lens.

Question 76.
On what factors does the resolving power of a microscope depend? How can it be increased?
Answer:
Resolving power of a microscope
= \(\frac{2 n \sin \alpha}{\lambda}\) = \(\frac{2 \mathrm{NA}}{\lambda}\)

where, α ≡ the half angle of the angular separation between the objects at the objective lens. n ≡ the refractive index of the medium between the object and the objective, λ ≡ the wavelength of the light used to illuminate the object, NA = n sin α = the numerical aperture of the objective.

Thus, the resolving power of a microscope depends directly on the NA and inversely on λ.
The resolving power is increased by,

1. increasing the numerical aperture using oil-immersion objective.
2. illuminating the object with smaller wavelength radiation. But our eyes are not very sensitive to the shorter wavelength blue end of the visible spectrum. Hence, ultraviolet radiation is used for illumination with quartz lenses, but then photographs must be taken to examine the image.

Question 77.
With a neat ray diagram, explain the resolving power of a telescope. On what factors does it depend ?
OR
What is meant by the angular limit of resolution and resolving power of a telescope?
Answer:
The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.

Consider two stars seen through a telescope. The diameter (D) of the objective lens or mirror corresponds to the diffracting aperture. For a distant point source, the first diffraction minimum is at an angle θ away from the centre such that

Dsin θ = 1.22 λ
where λ is the wavelength of light. The angle 9 is usually so small that we can substitute sin θ \(\approx\) θ (θ in radian). Thus, the Airy disc for each star will be spread out over an angular half-width θ = 1.22 λ/D about its geometrical image point. The radius of the Airy disc at the focal plane of the objective lens is r = fθ = 1.22fλ/D, where f is the focal length of the objective.

When observing two closely-spaced stars, the Rayleigh criterion for just resolving the images as that of two point sources (instead of one) is met when the centre of one Airy disc falls on the first minimum of the other pattern. Thus, the angular limit (or angular separation) of resolution is
θ = \(\frac{1.22 \lambda}{D}\) … (1)

and the linear separation between the images at the focal plane of the objective lens is
y = fθ … (2)
∴ Resolving power of a telescope.
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\) … (3)

It depends

1. directly on the diameter of the objective lens or mirror,
2. inversely on the wavelength of the radiation.

Question 78.
How can the resolving power of a telescope be increased?
Answer:
The resolving power of an astronomical telescope depends directly on the diameter of the objective lens or mirror, and inversely on the wavelength of radiation. Hence, the resolving power can be increased by

1. using an objective lens/mirror of larger diameter
2. observing a celestial object at smaller wavelengths.

Question 79.
Define the resolving power of a telescope and state its formula. What are the advantages of using a large objective lens in an astronomical telescope:
Answer:
Definition: The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.
Formula : Resolving power of a telescope
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\)
where θ ≡ the minimum angular separation of two closely-spaced celestial objects or the angular limit of resolution, D ≡ the diameter of the objective lens of the telescope, λ ≡ the wavelengh of light.

Advantages of a large objective lens in an astronomical telescope :

1. The resolving power is directly proportional to the diameter of the objective lens. Hence, a large objective lens results in a smaller Airy disc and a sharper image.
2. It collects more of the incident radiation from a distant object which results in a brighter image.

Question 80.
What is the resolving power of a telescope if the diameter of the objective of the telescope is 1.22 m and the wavelength of light is 5000 Å ?
Answer:
Resolving power of the telescope = \(\frac{D}{1.22 \lambda}\)
= \(\frac{1.22}{1.22 \times 5 \times 10^{-7}}\)
= \(\frac{10 \times 10^{6}}{5}\) = 2 × 10⁶ rad^-1

Question 81.
Briefly explain about radio telescope.
Answer:
A radio telescope is an astronomical instrument consisting of a radio receiver and an antenna system used to detect radio-frequency radiation emitted by extraterrestrial sources. The wavelength of radio-frequency radiation is very long. Hence, it is expressed in metre. In order to attain the resolution of an optical telescope, the radio telescope should be very large. A single disc of such large diameters is impracticable. In such cases arrays of antennae spread over several kilometres are used. Giant Metrewave Radio Telescope (GMRT) located at Narayangaon, Pune, Maharashtra uses such an array consisting of 30 dishes of diameter 45 m each. It spreads over 25 km. It is the largest distance between two of its antennae. The highest angular resolution achievable ranges from about 60 arcsec at the lowest frequency of 50 MHz to about 2 arcsec at 1.4 GHz.

Question 82.
Solve the following :

Question 1.
Light of wavelength 650 nm falls on the objective of a microscope having total angle of angular w separation as 60°. Calculate the resolving power of the microscope. [Refractive index of air = 1]
Solution :
Data : λ = 650 nm = 6.5 × 10^-7m, n = 1,
2α = 60°
∴ α = 30°
Resolving power of the microscope = \(\frac{2 n \sin \alpha}{\lambda}\)
= \(\frac{2 \times 1 \times \sin 30^{\circ}}{6.5 \times 10^{-7}}\) = \(\frac{10}{6.5}\) × 10⁶

Question 2.
The semi vertical angle of the cone of the rays incident on the objective of a microscope is 20°. If the wavelength of incident light is 6600 Å, calculate the smallest distance between two points which can be just resolved.
Solution :
Data : α = 20°, λ = 6600 Å = 6.6 × 10^-7 m, n = 1 (air) resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.
Formula : Resolving power of a telescope
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\)
where θ ≡ the minimum angular separation of two closely-spaced celestial objects or the angular limit of resolution, D ≡ the diameter of the objective lens of the telescope, λ ≡ the wavelengh of light.

Advantages of a large objective lens in an astronomical telescope :

1. The resolving power is directly proportional to the diameter of the objective lens. Hence, a large objective lens results in a smaller Airy disc and a sharper image.
2. It collects more of the incident radiation from a distant object which results in a brighter image.

Question 83.
What is the resolving power of a telescope if the diameter of the objective of the telescope is 1.22 m and the wavelength of light is 5000 Å ?.
Answer:
Resolving power of the telescope = \(\frac{D}{1.22 \lambda}\)

Question 84.
Briefly explain about radio telescope.
Answer:
A radio telescope is an astronomical instrument consisting of a radio receiver and an antenna system used to detect radio-frequency radiation emitted by extraterrestrial sources. The wavelength of radio-frequency radiation is very long. Hence, it is expressed in metre. In order to attain the resolution of an optical telescope, the radio telescope should be very large. A single disc of such large diameters is impracticable. In such cases arrays of antennae spread over several kilometres are used. Giant Metrewave Radio Telescope (GMRT) located at Narayangaon, Pune, Maharashtra uses such an array consisting of 30 dishes of diameter 45 m each. It spreads over 25 km. It is the largest distance between two of its antennae. The highest angular resolution achievable ranges from about 60 arcsec at the lowest frequency of 50 MHz to about 2 arcsec at 1.4 GHz.

Question 85.
Solve the following :

Question 1.
Light of wavelength 650 nm falls on the objective of a microscope having total angle of angular separation as 60°. Calculate the resolving power of the microscope. [Refractive index of air = 1]
Solution :
Data : λ = 650 nm = 6.5 × 10^-7m, n = 1, 2α = 60°
∴ α = 30°
Resolving power of the microscope = \(\frac{2 n \sin \alpha}{\lambda}\)

= 1.538 × 10⁶ m^-1

Question 2.
The semi vertical angle of the cone of the rays incident on the objective of a microscope is 20°. If the wavelength of incident light is 6600 Å, calculate the smallest distance between two points which can be just resolved.
Solution :
Data : α = 20°, λ = 6600 Å = 6.6 × 10^-7 m, n = 1 (air)
The numerical aperture, NA = n sin α
= 1 × sin 20° = 0.3420
The limit of resolution for an illuminated object,

Question 3.
What is the minimum distance between two objects which can be resolved by a microscope having the visual angle of 30° when light of wavelength 600 nm is used ?
Solution :

Question 4.
The two headlights of an approaching automobile are 1.22 m apart. At what maximum distance will eye resolve them? Assume a pupil diameter of 5.0 mm and λ = 5500 Å. Assume also that this distance is determined only by the diffraction effect at the circular aperture.
Solution:
Data : y = 1.22 m, diameter D = 5 × 10^-3 m,

This is the required distance.

Question 5.
What is the minimum angular separation between two stars if a telescope is used to observe them with an objective of aperture 20 cm? The wavelength of light is 5900 Å.
Solution :
Data : D = 20 cm = 0.2 m,
λ = 5900 Å = 5.9 × 10^-7 m

This is the required angular separation.

Question 6.
The diameter of the objective of a telescope is 10 cm. Find the resolving power of the telescope is 10 cm. Find the resolving power of the telescope if the wavelength of light is 5000 Å.
Solution :
Data : D = 10 cm = 0.1 m, λ = 5000 Å = 5 × 10^-7 m
The resolving power of the telescope D 0.1

Question 7.
A telescope has an objective of diameter 2.44 m. What is its angular resolution when it observes at 5500 A?
Solution :
Data : λ = 5500A = 5.5 × 10^-7 m,
D = 2.44 m
Angular resolution. ∆θ = 1.22λ/D, D being the diameter of the aperture.
∴ ∆θ = 1.22 × 5.5 × \(\frac{10^{-7}}{2.44}\)= 2.75 × 10^-7 rad
= 0.0567 arcsec

Question 8.
The minimum angular separation between two stars is 4 × 10^-6 rad when a telescope is used to observe them with an objective of aperture 16 cm. Find the wavelength of the light.
Solution :
Data : θ = 4 × 10^-6 rad, D = 16 cm = 0.16 m
θ = \(\frac{1.22 \lambda}{D}\)
∴ The wavelength of the light used,

Question 9.
Estimate the smallest angular separation of two stars which can be just resolved by the telescope having objective of diameter 25 cm. The mean wavelength of light is 555 nm.
Solution :
Data : λ = 555 nm 555 × 10^-9 m
D = 25 cm = 25 × 10^-2 m

= 2.708 × 10^-6 rad
This is the required angular separation.

Question 86.
Huygens’ wave theory could not explain
(A) interference
(B) reflection
(C) photoelectric effect
(D) refraction.
Answer:
(C) photoelectric effect

Question 87.
The wavefront originating from a point source of light at finite distance is a wavefront.
(A) circular
(B) spherical
(C) plane
(D) cylindrical
Answer:
(B) spherical

Question 88.
In an isotropic medium, the secondary wavelets centred on every point of a given wavefront are all
(A) spherical
(B) cylindrical
(C) oval
(D) of arbitrary shape.
Answer:
(A) spherical

Question 89.
Consider a medium through which light is propagating with a speed v. Given a wavefront, in order to determine the wavefront after a time interval ∆t, the secondary wavelets are drawn with
a radius.
(A) of unit length
(B) v ∆ t
(C) \(\frac{\Delta t}{v}\)
(D) \(\frac{v}{\Delta t}\)
Answer:
(B) v ∆ t

Question 90.
Two points, equidistant from a point source of light, are situated at diametrically opposite positions in an isotropic medium. The phase difference between the light waves passing through the two points is
(A) zero
(B) π/2 rad
(C) π rad
(D) finite, but not these.
Answer:
(A) zero

Question 91.
Wavenormals to spherical wavefronts can be
(A) only diverging
(B) only converging
(C) parallel to each other
(D) diverging or converging.
Answer:
(D) diverging or converging.

Question 92.
Huygens’ principle is used to
(A) obtain the new position of wavefront geometrically
(B) explain the principle of superposition of waves
(C) explain the phenomenon of interference
(D) explain the phenomenon of polarization.
Answer:
(A) obtain the new position of wavefront geometrically

Question 93.
When a ray of light enters into water from air,
(A) its wavelength decreases
(B) its wavelength increases
(C) its frequency increases
(D) its frequency decreases.
Answer:
(A) its wavelength decreases

Question 94.
A parallel beam of light travelling in water is incident obliquely on a glass surface. After refraction, its width
(A) decreases
(B) increases
(C) remains the same
(D) becomes zero.
Answer:
(B) increases

Question 95.
Light of a certain colour has 1800 waves to the millimetre in air. What is its frequency in water?
[n = \(\frac{4}{3}\) for water]
(A) 1.67 × 10⁶ Hz
(B) 4.05 × 10¹⁴ Hz
(C) 5.4 × 10¹⁴ Hz
(D) 7.2 × 10¹⁴ Hz.
Answer:
(C) 5.4 × 10¹⁴ Hz

Question 96.
A ray of light, in passing from vacuum into a medium of refractive index n, suffers a deviation d equal to half the angle of incidence. Then, the refractive index is
(A) sin δ
(B) 2 sin δ
(C) cos δ
(D) 2 cos δ.
Answer:
(D) 2 cos δ.

Question 97.
A ray of light passes from vacuum to a medium of refractive index n. The angle of incidence is found to be twice the angle of refraction. The angle of incidence is
(A) cos^-1 \(\left(\frac{n}{2}\right)\)
(B) cos^-1 (n)
(C) 2cos^-1 \(\left(\frac{n}{2}\right)\)
(D) 2sin^-1 \(\left(\frac{n}{2}\right)\).
Answer:
(C) 2cos^-1 \(\left(\frac{n}{2}\right)\)

Question 98.
If the polarizing angle for a given medium is 60°, then the refractive index of the medium is
(A) \(\frac{1}{\sqrt{3}}\)
(B) \(\frac{\sqrt{3}}{2}\)
(C) 1
(D) \(\sqrt{3}\)
Answer:
(D) \(\sqrt{3}\)

Question 99.
A narrow beam of light in air is incident on glass at an angle of incidence of 58°. If the reflected beam is completely plane polarized, the refractive index of the glass is
(A) 1.9
(B) 1.8
(C) 1.7
(D) 1.6.
Answer:
(D) 1.6.

Question 100.
Polarization of light CANNOT be produced by
(A) reflection
(B) double refraction
(C) dichroism
(D) diffraction.
Answer:
(D) diffraction.

Question 101.
A glass plate of refractive index 1.732 is to be used as a polarizer. Its polarizing angle is
(A) 30°
(B) 45°
(C) 60°
(D) 90°.
Answer:
(C) 60°

Question 102.
Light transmitted through a Polaroid P₁ has an intensity I and is incident on a crossed Polaroid P₂. The intensity of the light transmitted by P₂ is
(A) zero
(B) \(\frac{1}{2}\)I
(C) I
(D) 2I.
Answer:
(A) zero

Question 103.
At points of constructive interference with maximum intensity of two coherent monochromatic waves (wavelength λ), the path difference between them is
(A) zero or an integral multiple of λ
(B) zero or an integral multiple of λ/2
(C) zero or an even integral multiple of λ/2
(D) an odd integral multiple of λ/2.
Answer:
(A) zero or an integral multiple of λ

Question 104.
Two sources of light are said to be coherent if light from them have
(A) the same speed and the same phase
(B) the same phase and the same or nearly the same amplitude
(C) constant phase difference and nearly the same frequency
(D) zero, or some constant, phase difference.
Answer:
(D) zero, or some constant, phase difference.

Question 105.
In a two-source interference pattern, the phase difference between the waves reaching a dark point in radian is (m = 1, 2, 3, …)
(A) 0
(B) mπ
(C) (2m – 1)\(\frac{\pi}{2}\)
(D) (2m – 1)π
Answer:
(D) (2m – 1)π

Question 106.
For destructive interference, the phase difference (in radian) between the two waves should be
(A) 0, 2π, π, …
(B) 0, 2π, 4π, …
(C) π, 3π, 5π, …
(D) \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\), \(\frac{5 \pi}{2}\), …..
Answer:
(C) π, 3π, 5π, …

Question 107.
For constructive interference, the phase difference (in radian) between the two waves should be
(A) 0, \(\frac{\pi}{2}\), π, …
(B) 0, 2π, 4π, …
(C) π, 3π, 5π, …
(D) \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\), …..
Answer:
(B) 0, 2π, 4π, …

Question 108.
If λ is the wavelength of light used in Young’s double-slit experiment, the path difference for a phase difference of 11π rad is
(A) 23 λ
(B) 11 λ
(C) 11\(\frac{\lambda}{2}\)
(D) 23\(\frac{\lambda}{2}\)
Answer:
(C) 11\(\frac{\lambda}{2}\)

Question 109.
The fringe width in an interference pattern is W. The distance between the 6th dark fringe and the 4th bright fringe on the same side of the central bright fringe is
(A) 1.5 W
(B) 2 W
(C) 2.5 W
(D) 10.5 W.
Answer:
(A) 1.5 W

Question 110.
Which of the following graphs shows the variation of the fringe width with the frequency of light in a two-source interference pattern ?

Answer:

Question 111.
In an interference pattern using two coherent sources of light, the fringe width is
(A) directly proportional to the wavelength
(B) inversely proportional to the square of the wavelength
(C) inversely proportional to the wavelength
(D) directly proportional to the square of the wavelength.
Answer:
(A) directly proportional to the wavelength

Question 112.
Two slits, 2 mm apart, are placed 300 cm from a screen. When light of wavelength 6000 Å is used, the separation (in mm) between the successive bright lines of the interference pattern is
(A) 0.9
(B) 4.5
(C) 6
(D) 9.
Answer:
(A) 0.9

Question 113.
In two separate setups of Young’s double-slit experiment, the wavelengths of the lights used are in the ratio 1 : 2 while the separation between the slits are in the ratio 2 : 1. If the fringe widths are equal, the ratio of the distances between the slit and the screen is
(A) 1 : 4
(B) 1 : 2
(C) 2 : 1
(D) 4 : 1.
Answer:
(D) 4 : 1.

Question 114.
In Young’s double-slit experiment, the slit separation is reduced to half while the distance of the screen from the slits is increased by 50%. In terms of the initial fringe width, W, the new fringe width is,
(A) \(\frac{1}{4}\) W
(B) \(\frac{3}{4}\) W
(C) \(\frac{3}{2}\) W
(D) 3 W.
Answer:
(D) 3 W.

Question 115.
A pair of slits 1.5 mm apart is illuminated with monochromatic light of wavelength 5500 Å and the interference pattern is obtained on a screen 1.5 m from the slits. The least distance of a point from the central maximum where the intensity is minimum is
(A) 0.275 mm
(B) 0.55 mm
(C) 2.75 mm
(D) 5.5 mm.
Answer:
(A) 0.275 mm

Question 116.
In Young’s double-slit experiment, if a thin transparent sheet covers one of the slits, the optical path of the wave from that slit
(A) increases
(B) decreases
(C) changes by \(\frac{D}{d}\) (n_m – 1)t
(D) is not affected.
Answer:
(A) increases

Question 117.
In Young’s double-slit experiment, if a thin mica sheet of thickness t and refractive index n_m covers one of the slits, the optical path of the wave from that slit
(A) increases by (n_m – 1 )t
(B) decreases by (n_m – 1) t
(C) changes by \(\frac{D}{d}\) (n_m – 1) t
(D) is not affected.
Answer:
(A) increases by (n_m – 1 )t

Question 118.
In a two-slit intereference experiment, if a thin transparent sheet of thickness f and refractive index n_m covers both the slits, the optical path difference between the two interfering waves
(A) increases by (n_m – 1)t
(B) decreases by (n_m – 1)t
(C) changes by \(\frac{D}{d}\)(n_m – 1)t
(D) is not affected.
Answer:
(D) is not affected.

Question 119.
In a biprism experiment two interfering waves are produced by division of
(A) amplitude
(B) wavefront
(C) amplitude and wavefront
(D) neither wavefront nor amplitude.
Answer:
(B) wavefront

Question 120.
In Fresnel’s biprism experiment, with the eyepiece 1 m from the two coherent sources, the fringe width obtained is 0.4 mm. If just the eyepiece is moved towards the biprism by 25 cm, then the fringe width
(A) decreases by 0.01 mm
(B) decreases by 0.1 mm
(C) increases by 0.01 mm
(D) increases by 0.1 mm.
Answer:
(B) decreases by 0.1 mm

Question 121.
In a biprism experiment, keeping the experimental setup unchanged, the fringe width
(A) increases with increase in wavelength
(B) decreases with increase in wavelength
(C) increases with decrease in wavelength
(D) remains unchanged with change in wavelength.
Answer:
(A) increases with increase in wavelength

Question 122.
In finding the distance between the two coherent sources in Fresnel’s biprism experiment by the conjugate foci method, one uses
(A) a long focus convex lens that forms real images of the virtual sources
(B) a short focus concave lens that forms real images of the virtual sources
(C) a short focus convex lens that forms virtual images of the virtual sources
(D) a short focus convex lens that forms real images of the virtual sources.
Answer:
(D) a short focus convex lens that forms real images of the virtual sources.

Question 123.
Using a light of wavelength 4800 Å in Fresnel’s biprism experiment, 21 fringes are obtained in a given region. If light of wavelength 5600 Å is used, the number of fringes in the same region will be
(A) 14
(B) 18
(C) 21
(D) 24.
Answer:
(B) 18

Question 124.
To obtain pronounced diffraction with a single slit illuminated by light of wavelength λ, the slit width should be
(A) of the same order as λ
(B) considerably larger than λ
(C) considerably smaller than λ
(D) exactly equal to λ/2.
Answer:
(B) considerably larger than λ

Question 125.
In single-slit diffraction, which of the following are equal ?
(A) Widths of all bright and dark fringes
(B) Intensities of non-central bright fringes
(C) Widths of non-central bright fringes
(D) Both widths and intensities of noncentral bright fringes.
Answer:
(C) Widths of non-central bright fringes

Question 126.
In single slit diffraction (at a narrow slit of width a), the intensity of the central maximum is
(A) independent of a
(B) proportional to a
(C) proportional to a²
(D) inversely proportional to a.
Answer:
(D) inversely proportional to a.

Question 127.
For a single slit of width a, the diffraction pattern minima are located at angles θ_m, where m is a positive, non-zero integer. Which of the following expressions is most correct ?
(A) a sin θ_m = mλ
(B) a sin θ_m = \(\frac{m \lambda}{2}\)
(C) a sin θ_m = (2m + 1)\(\frac{\lambda}{2}\)
(D) a sin θ_m = (2m – 1)\(\frac{\lambda}{2}\)
Answer:
(A) a sin θ_m = mλ

Question 128.
In a diffraction pattern due to a single slit of width a with incident light of wavelength λ, at an angle of diffraction θ, the condition for the first minimum is
(A) λ sin θ = a
(B) a cos θ = λ
(C) a sin θ = λ
(D) λ cos θ = a.
Answer:
(C) a sin θ = λ

Question 129.
The fringes produced in a diffraction pattern are of
(A) equal width with the same intensity
(B) unequal width with varying intensity
(C) equal intensity
(D) equal width with varying intensity.
Answer:
(B) unequal width with varying intensity

Question 130.
For a single slit of width a, the first diffraction maximum with light of wavelength λ subtends an angle θ such that sin θ is equal to
(A) \(\frac{\lambda}{2 a}\)
(B) \(\frac{\lambda}{a}\)
(C) \(\frac{1.5 \lambda}{a}\)
(D) \(\frac{2 \lambda}{a} .\)
Answer:
(C) \(\frac{1.5 \lambda}{a}\)

Question 131.
Fraunhofer diffraction pattern of a parallel beam of light (wavelength λ) passing through a narrow slit (width a) is observed on a screen using a convex lens (focal length f). The angular half-width of the central fringe is
(A) \(\frac{2 \lambda f}{a}\)
(B) \(\frac{\lambda f}{a}\)
(C) \(\frac{2 \lambda}{a}\)
(D) \(\frac{\lambda}{a}\)
Answer:
(D) \(\frac{\lambda}{a}\)

Question 132.
A parallel beam of light (λ = 5000 Å) is incident normally on a narrow slit of width 0.2 mm. The angular separation between the first two minima is
(A) 2.5 × 10^-3 degree
(B) 2.5 × 10^-3 rad
(C) 5 × 10^-3 degree
(D) 5 × 10^-3 rad.
Answer:
(B) 2.5 × 10^-3 rad

Question 133.
A parallel beam of light (λ = 5000 Å) is incident normally on a narrow slit of width 0.2 mm. The Fraunhofer diffraction pattern is observed on a screen placed at the focal plane of a convex lens (f = 20 cm). The first two minima are separated by
(A) 0.005 cm
(B) 0.05 cm
(C) 2.5 mm
(D) 5 mm.
Answer:
(B) 0.05 cm

Question 134.
A plane wave of wavelength 5500 Å is incident normally on a slit of width 2 × 10^-2 cm. The width of the central maximum on a screen 50 cm away is
(A) 2.50 × 10^-3 cm
(B) 2.75 × 10^-3 cm
(C) 2.75 × 10^-3 m
(D) 5.50 × 10^-3 m.
Answer:
(D) 5.50 × 10^-3 m.

Question 135.
If NA is the numerical aperture of a microscope objective, then its resolving power with an illumination of wavelength λ is
(A) \(\frac{0.5 \lambda}{\mathrm{NA}}\)
(B) \(\frac{0.61 \lambda}{\mathrm{NA}}\)
(C) \(\frac{1.22 \mathrm{NA}}{\lambda}\)
(D) \(\frac{2 \mathrm{NA}}{\lambda}\)
Answer:
(A) \(\frac{0.5 \lambda}{\mathrm{NA}}\)

Question 136.
A microscope with numerical aperture 0.122 is used with light of wavelength 6000 A. The limit of resolution is
(A) 3.33 × 10⁶ m
(B) 3.33 mm
(C) 3 × 10^-6 m
(D) 3 × 10^-7 m.
Answer:
(C) 3 × 10^-6 m

Question 137.
The objective lens of a telescope has a diameter D. The angular limit of resolution of the telescope for light of wavelength λ is
(A) \(\frac{D}{1.22 \lambda}\)
(B) \(\frac{1.22 \lambda}{D}\)
(C) \(\frac{D}{0.61 \lambda}\)
(D) \(\frac{0.61 \lambda}{D} .\)
Answer:
(B) \(\frac{1.22 \lambda}{D}\)

Question 138.
Two closely-spaced distant stars are just resolved when seen through a telescope with an objective lens of diameter D and focal length f. The separation between their images is given by.
(A) \(\frac{D}{1.22 \lambda f}\)
(B) \(\frac{f \mathcal{D}}{1.22 \lambda}\)
(C) \(\frac{1.22 \lambda f}{D}\)
(D) \(\frac{1.22 D \lambda}{f}\)
Answer:
(C) \(\frac{1.22 \lambda f}{D}\)

Question 139.
High magnifying power microscopes have oil-immersion objectives
(A) to increase the fringe width
(B) to increase the numerical aperture of the objective
(C) to decrease the wavelength of light
(D) because oil does not damage the observed sample.
Answer:
(B) to increase the numerical aperture of the objective

Question 140.
If the numerical aperture of a microscope is increased, then its
(A) resolving power decreases
(B) limit of resolution decreases
(C) resolving power remains constant
(D) limit of resolution increases
Answer:
(B) limit of resolution decreases

Question 141.
The numerical aperture of the objective of a microscope is 0.12. The limit of resolution, when light of wavelength 6000 A is used to view an object, is
(A) 0.25 × 10^-7 m
(B) 2.5 × 10^-7m
(C) 25 × 10^-7 m
(D) 250 × 10^-7 m.
Answer:
(C) 25 × 10^-7 m

Question 142.
The resolving power of a refracting telescope is increased by
(A) using oil-immersion objective
(B) increasing the diameter D of the objective lens
(C) resorting to short-wavelength radiation
(D) increasing D and using smaller λ.
Answer:
(D) increasing D and using smaller λ.

Question 143.
The resolving power of a telescope of aperture 100 cm, for light of wavelength 5.5 × 10^-7 m, is
(A) 0.149 × 10⁷
(B) 1.49 × 10⁷
(C) 14.9 × 10⁷
(D) 149 × 10⁷
Answer:
(A) 0.149 × 10⁷

Question 144.
The resolving power of a telescope depends upon the
(A) length of the telescope
(B) focal length of the objective
(C) diameter of the objective
(D) focal length of the eyepiece.
Answer:
(C) diameter of the objective

Question 145.
If a is the aperture of a telescope and 2 is the wavelength of light then the resolving power of the telescope is
(A) \(\frac{\lambda}{1.22 a}\)
(B) \(\frac{1.22 a}{\lambda}\)
(C) \(\frac{1.22 \lambda}{a}\)
(D) \(\frac{a}{1.22 \lambda}\)
Answer:
(D) \(\frac{a}{1.22 \lambda}\)

Question 146.
Using a monochromatic light of wavelength 2 in Young’s double-slit experiment, the eleventh dark fringe is obtained on the screen for a phase difference of
(A) \(\frac{11}{2}\) π rad
(B) \(\frac{21}{2}\) π rad
(C) 13 π rad
(D) 21 π rad
Answer:
(D) 21 π rad

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 15 Structure of Atoms and Nuclei Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 15 Structure of Atoms and Nuclei

Question 1.
What was Dalton’s atomic theory of chemistry?
Answer:
John Dalton (1766-1844), British meteorologist, in his atomic theory of chemistry (1808-1810) proposed the following postulates :
(1) Matter consists of very small indivisible particles called atoms.
(2) Each element consists of a characteristic kind of identical atoms. There are consequently as many different kinds of atoms as there are elements.
(3) When different elements combine to form a compound, the smallest unit of the compound consists of a definite number of atoms of each element. These ‘compound atoms’ are now called molecules.
(4) In chemical reactions, atoms are neither created nor destroyed, but only rearranged.

Question 2.
Explain Thomson’s model of the atom. What are its drawbacks?
Answer:
The first model of the atom with a sub-structure was put forward in 1898 by Sir J.J. Thomson (1856-1940), a British physicist. According to this model, an atom consists of a sphere with a uniform distribution of positive charge and electrons embedded in it such that the atom is electrically neutral and stable.

Drawbacks : This model, known as the plumpudding model, failed to account for the observed scattering of α-particles and spectra of various elements.
[Note : It can be shown that the Thomson atom cannot be stable.]

Question 3.
Who suggested the famous a-particle scattering experiment? Why?
Answer:
Sir Ernest Rutherford (1871-1937), New Zealand- British physicist, following his pioneering work on radioactivity and properties of a-particles, had noted that a narrow stream of α-particles gets somewhat broadened or scattered on passing through a thin metal foil or mica sheet. Since most of the α-particles remained undeviated, this suggested that atoms could not be solid spheres as proposed in Thomson’s model and kinetic theory of gases. To probe into the effects of the distribution of an atom’s mass and charge on the a-particles, he suggested his collaborator Geiger and the latter’s student Marsden to see if any α-particles are scattered through a large angle.

Question 4.
With the help of a neat labelled diagram, describe the Geiger-Marsden experiment.
Answer:
The Geiger-Marsden a-scattering (or gold foil) experiment (1908) : Geiger and Marsden made a stream of a-particles strike a very thin gold foil about 40 jum thick. Their apparatus is shown schematically in figure.

Apparatus: A radium compound, an intense source of a-particles, was placed in the lead enclosure B, provided with a small hole. The stream of α-particles was collimated by lead bricks. The number of particles scattered through each angle θ were counted by a rotatable detector. The detector consisted of a small zinc sulphide screen S at the focus of a low power microscope M. Each incidence produced a scintillation-a momentary pinpoint of fluorescence. These scintillations were observed and counted using the microscope.

Geiger-Marsden experiment of scattering of a-particles by a gold foil

Observations: Most of the α-particles passed through the foil almost undeviated, with less than 0.2% deflected by more than 1°. Still smaller fractions were found to be deflected by 90° or more, sometimes almost straight back towards the source.

Rutherford quantitatively accounted for the distribution of small and large angle scattering by considering each scattering to be a single collision of an a-particle with a positive ‘central charge’ Ne concentrated at a point. Since the probability of an a-particle coming very close to such a point charge was small, this explained the very small number of a-particles deflecting through large angles.

Scattering of a-particles by a gold foil

[Notes : (1) Hans Wilhelm Geiger (1882-1945), German physicist and Sir Ernest Marsden (1889-1970), English-New Zealand physicist. (2) In 1908 and 1909, they conducted a series of a-scattering experiments with gold and silver foils of different thicknesses and a thick platinum plate. Rutherford reported (in 1911) that “about 1 in 20000 were turned through 90° on passing through a gold foil about 40 nm thick.” The number ‘1 in 8000’ was reported (in 1909) by Geiger and Marsden for reflection off a thick platinum plate ‘at large angle.]

Question 5.
Explain Rutherford’s model of the atom.
Answer:
Rutherford’s model of the nuclear atom (1911) :
(1) An atom has a very small nucleus which contains all the positive charge and almost all the mass of the atom.
(2) The nuclear size (radius about 10^-14 m) is very small compared to the atomic size (radius about 10^-10m), about 10000 times smaller.
(3) Electrons revolve in circular orbits around the nucleus. The electrostatic force (Coulomb force) of attraction between the positively charged nucleus and the negatively charged electron is the centripetal force required for the orbital motion of the electron.
(4) Since an atom as a whole is electrically neutral, the positive charge on the nucleus must be equal to the total negative charge of all the orbiting electrons.
As this model resembles the solar system, it is known as the planetary atom model.

Question 6.
Solve the following :
(1) An a-particle having a kinetic energy of 8 MeV is projected directly toward the nucleus of an atom of \(\begin{gathered}
208 \\
82
\end{gathered}\)Pb. Find the distance of closest approach.
(e = 1.6 × 10^-19 C, ε₀ = 8.85 × 10^-12 C²/N∙m²)
Solution :
Data : e = 1.6 × 10^-19 C, Z(Pb) = 82, ε₀ = 8.85 × 10^-12 C²/N∙m², (KE)_α = 8 MeV
If d is the distance of closest approach, we must have, by the principle of conservation of energy, initial kinetic energy of the a-particle = potential energy of the a-particle when it is at the distance d from the centre of the nucleus of
\(\begin{gathered}
208 \\
82
\end{gathered}\) Pb.

(2) An α-particle projected directly toward the nucleus of an atom of \(\begin{gathered}
208 \\
82
\end{gathered}\)Pb comes to rest momentarily at the surface of the nucleus. Find the initial kinetic energy and speed of the α-particle.
[m(α) = 6.68 × 10^-27 kg, r₁(α) = 1.8 × 10^-15 m, r₂(Pb) = 7.11 × ^-15 m, e = 1.6 × 10^-19 C, \(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 109 N-m2/C2]
Solution:
For reference, see the solved problem (1) above.

This is the initial speed of the α-particle.
[Note : 3.564 × 10⁷ m/s is about 12% of the speed of light in vacuum.]

Question 7.
What is the emission spectrum of a substance? Explain in brief.
Answer:
The emission spectrum of a substance is the distribution of electromagnetic radiations emitted by the substance when it is heated, or bombarded by electrons, or ions or photons. The distribution is arranged in order of increasing (or decreasing) frequency (or wavelength) and is characteristic of the substance unless the temperature of the substance is very high when the distribution is continuous. The spectrum may be a line spectrum or band spectrum. The intensities corresponding to different frequencies are different.
[Note : The absorption spectrum is formed by absorption of electromagnetic radiation when the substance is exposed to radiation of all frequencies.]

Question 8.
State and explain the formula that gives wavelengths of lines in the hydrogen spectrum.
Answer:
Formula : \(\frac{1}{\lambda}=R\left[\frac{1}{n^{2}}-\frac{1}{m^{2}}\right]\), where λ is the wavelength of a line in the hydrogen spectrum, R is a constant, now called the Rydberg constant, and n and m are integers with n = 1,2,3,… and m = n + 1, n + 2, n + 3,
For a fixed value of n, λ decreases as m increases and has minimum value as m → ∞ λ = \(\frac{n^{2}}{R}\) as

Lyman, Balmer and Paschen series in the hydrogen spectrum (For reference only)
[Note : A line spectrum is atomic in origin. It is a signature of the element, i.e., we can determine the elements present in a mixture of elements by studying the line spectrum of the mixture.]

Question 9.
State the equations corresponding to Bohr’s atomic model.
Answer:

Here, m_e is the mass of the electron, e is the electron charge, Z is the atomic number of the atom, r_n is the radius of the nth stable orbit, v_n is the speed of the electron in the n^th orbit and ε₀ is the absolute premittivity of free space. In hydrogen, Z = 1.
m_e v_n r_n = n\(\frac{h}{2 \pi}\) …………… (2)
[Angular momentum of the electron]

where n ( = 1, 2, 3, …) is the positive integer, called the principal quantum number, and h is Planck’s constant, n denotes the number of the orbit.
E_m – E_n = hv …………. (3)

Here, E_m is the energy of the electron in the m^th orbit, E_n is the energy of the electron in the nth orbit (m > n), hv is the energy of the photon emitted and v is the frequency of the electromagnetic radiation emitted.
[Note : Niels Bohr (1885-1962), Danish theoretical physicist, made significant contribution to atomic and
nuclear physics.]

Question 10.
What is the minimum angular momentum of the electron in an hydrogen atom?
Answer:
\(\frac{h}{2 \pi}\).

Question 11.
Which physical quantity of an atomic electron has the dimensions same as that of h?
Ans.
Angular momentum.

Question 12.
What is meant by a stationary orbit?
Answer:
In the Bohr model of the hydrogen atom, a stationary orbit refers to any of the discrete allowed orbits such that the electron does not radiate energy while it is in such orbits.

Question 13.
Derive an expression for the linear speed of an electron in a Bohr orbit. Hence, show that it is inversely proportional to the principal quantum number.
Answer:
Consider an electron revolving in the nth Bohr orbit around the nucleus of an atom with the atomic number Z. Let m and – e be the mass and charge of the electron, r the radius of the orbit and v the linear speed of the electron.

According to Bohr’s first postulate, centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus.
∴ \(\frac{m v^{2}}{r}=\frac{Z e^{2}}{4 \pi \varepsilon_{0} r^{2}}\) ………….. (1)
where ε₀ is the permittivity of free space.
∴mv² = \(\frac{Z e^{2}}{4 \pi \varepsilon_{0} r}\) ………….. (2)
According to Bohr’s second postulate, the orbital angular momentum of the electron is quantized :
mvr = \(\frac{n h}{2 \pi}\) ………… (3)
where h is Planck’s constant and n is the principal quantum number which takes integral values 1, 2, 3, …, etc.
∴ r = \(\frac{n h}{2 \pi m v}\)
Substituting this expression for r in Eqn (2), we get,

as Z, e, ε₀ and h are constants.
[Note : In this topic, unless stated otherwise, m = m_e, r = r_n, and v = v_n.]

Question 14.
What is the angular momentum of the electron in the third Bohr orbit in the hydrogen atom ?
[\(\frac{h}{2 \pi}\) = 1.055 × 10^-34 kg∙m²/s
Answer:
Angular momentum, L = mvr = \(\frac{nh}{2 \pi}\)
For n = 3, L = 3(\(\frac{h}{2 \pi}\)) = 3 (1.055 × 10^-34)
= 3.165 × 10^-34 kg∙m²/s

Question 15.
Derive an expression for the radius of the nth Bohr orbit in an atom. Hence, show that the radius of the orbit is directly proportional to the square of the principal quantum number.
Answer:
Consider an electron revolving in the nth orbit around the nucleus of an atom with the atomic number Z. Let m and – e be the mass and charge of the electron, r the radius of the orbit and v the linear speed of the electron.

According to Bohr’s first postulate, centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus.

where h is Planck’s constant and n is the principal quantum number which takes integral values 1, 2, 3, …… etc.

Since, ε₀, h, Z, in and e are constants, it follows that r ∝ n², i.e., the radius of a Bohr orbit of the electron in an atom is directly proportional to the square of the principal quantum number.

Question 16.
The radius of the first Bohr orbit in the hydro gen atom is 0.5315 Å. What is the radius of the second Bohr orbit in the hydrogen atom?
Ans.
In the usual notation,

∴ r₂ = 4r₁ = 4 × 0.5315 = 2.125 Å is the required radius.
[Note : r₁ is also denoted by a₀.]

Question 17. Show that the angular speed of an electron in the nth Bohr model is ω = \(\frac{\pi m e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}\) and the corresponding frequency of the revolution of the electron is f = \(\) .
Answer:
The radius of the nth Bohr orbit is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) ………….. (1)
and the linear speed of an electron in this orbit is
v = \(\frac{Z e^{2}}{2 \varepsilon_{0} n h}\) ………….. (2)
where ε₀ ≡ permittivity of free space, h ≡ Planck’s constant, n ≡ principal quantum number, m ≡ electron mass, e ≡ electronic charge and Z ≡ atomic number of the atom.

Since angular speed ω = \(\frac{v}{r}\), then from Eqs. (1) and (2) we get,

which gives the required expression for the angular speed of an electron in the nth Bohr orbit.
From Equestion (3), the frequency of revolution of the electron,

Question 18.
The angular speed of an electron in the first orbit in H-atom is 4.105 × 10¹⁶ rad/s. Find the angular speed of the electron in the second orbit.
Answer:

= 5.131 × 10¹⁵ rad/s This is the required quantity.

Question 19.
The frequency of revolution of the electron in the second Bohr orbit in the hydrogen atom is 8.158 × 10¹⁴ Hz. What is the frequency of revolution of the electron in the fourth Bohr orbit in the hydrogen atom ?
Ans.
In the usual notation,

Question 20.
Show that the energy of the electron in the nth stationary orbit in the hydrogen atom is
E_n = -Rch/n².
Answer:
The energy of the electron in the nth stationary orbit in the hydrogen atom is
E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\)
and the Rydberg constant is
R = \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\)
where m = mass of electron, e = electronic charge, h = Planck’s constant, c = speed of light in free space and i, = permittivity of free space.

Question 21.
State the limitations of Bohr’s atomic model.
Answer:
Limitations of Bohr’s atomic model:

1. The model cannot explain the relative intensities of spectral lines even in the hydrogen spectrum.
2. The model cannot explain the atomic spectra of many-electron atoms of higher elements.
3. The model cannot account for the Zeeman effect and Stark effect (fine structure of spectral lines as revealed in the presence of strong magnetic field and electric field, respectively).

[Note : In 1896, Pieter Zeeman (1865-1943), Dutch physicist, discovered the splitting of spectral lines by magnetic field. In 1913, Johannes Stark (1874-1957), German physicist, discovered the splitting of spectral lines by electric field.]

Question 22.
Draw a neat, labelled energy level diagram for the hydrogen atom. Hence explain the different series of spectral lines for hydrogen.
Answer:
According to Bohr’s model of the hydrogen atom, an atom exists most of the time in one of a number of stable and discrete energy states. The various states arranged in order of increasing energy constitute the energy level diagram of the atom, as shown in below figure for the hydrogen atom. Here, the higher (less negative) energies are at the top while the lower (more negative) energies are toward the bottom.

According to Bohr’s theory, electromagnetic radiation of a particular wavelength is emitted when there is a transition of the electron from a higher energy state to a lower energy state. Let the quantum number n = m represent a higher energy state and n = n represent a lower energy state (m > n). The formation of the different series of spectral lines is explained from the energy level diagram.

(1) Lyman series : This series in the far UV region of the spectrum arises due to the transitions of the electron to n = 1 from m = 2, 3, 4, …, etc. The wavelengths (λ) of the spectral lines in this series are given by
\(\frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\) (m = 2, 3, 4, ……….,∞)
where R is the Rydberg constant.

(2) Balmer series : This series in the visible region of the spectrum arises due to the transitions to n = 2 from m = 3, 4, 5, …, etc. The wavelengths (λ) of the spectral lines in this series are given by
\(\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{m^{2}}\right)\) (m = 3, 4, 5, …. ∞)

(3) Paschen, Brackett and Pfiind series : These three series in the infrared region of the spectrum arise due to the transitions to n = 3, 4 and 5, respectively from m=n + 1, n + 2, etc. The wavelengths (λ) of the lines are given by

In each series, the smallest quantum of radiation (smallest frequency, longest wavelength) arises from the transition from m = n +1 to n, and the largest quantum (highest frequency, shortest wavelength — short wavelength limit or series limit) is for the transition from m = ∞ to n.
[Notes :
(1) The lines in the Lyman and Balmer series, beginning with the longest wavelength, are labelled with the Greek letters α, β, γ, δ, ε, …. Thus, the lines in the Lyman series are called L_α, L_β, L_γ, L_δ … lines while those in the Balmer series are called H_α, H_β, H_γ, H_δ, … lines. Thus, the L_α, line has a wavelength λ, where

(2) The first four prominent hydrogen lines (H_α, H_β, H_γ, H_δ) lie in the visible region and were discovered in the solar spectrum by Anders Angstrom (the first three in 1862 and the fourth in 1871); he also measured their wavelengths to high accuracy.

Johann Jakob Balmer (1825 – 98), Swiss mathema-tician, discovered in 1884 that their wavelengths fitted
the relation λ = \(\frac{B m^{2}}{m^{2}-4}\), where m has integral values 3, 4, 5 and 6 for successive lines and B here is a constant equal to 3645.6 Å. This is Balmer’s formula; originally empirical, it pointed to the need to find an explanation. This led through Rydberg’s work to Bohr’s theory.

(3) Lyman series was discovered between 1906-14, Paschen series in 1908, Brackett series in 1922 and Pfiind series in 1924. A sixth, and the last of the named series in the hydrogen spectrum, is the Humphreys series which results from transitions to n = 6 from m = 7, 8, 9, … etc. The Paschen series lies in the near-infrared region while Pfiind and Humphreys series lie in the far-infrared region.

Further series are unnamed but follow the same pattern. Their lines are increasingly faint showing that they correspond to increasingly rare atomic events.]

Question 23.
Obtain the expressions for longest and shortest wavelengths of spectral lines in ultraviolet region for hydrogen atom.
Answer:
For hydrogen,

where R is the Rydberg constant, and m and n are the principal quantum numbers of the initial and final energy states. The Lyman series in the far UV region of the spectrum arises due to the transitions of the electron to n = 1 from,m = 2, 3,4, …, etc.
For the longest wavelength λ_Lα in the Lyman series, m = 2.

Question 24.
How many spectral series are possible in the hydrogen spectrum?
Answer:
Infinite. The last (sixth) of the named series in the hydrogen spectrum is the Humphreys series which results from transitions to n_f = 6 from n_i = 7, 8, 9, … etc. in the far-infrared region. Further series are unnamed but follow the same pattern. Their lines are increasingly faint showing that they correspond to increasingly rare atomic events.
[Note : There are infinite number of lines in each series. The spacing between adjacent lines decreases with decreasing wavelength, converging to the so-called series limit or short-wavelength limit.]

Question 25.
The energy of the electron in the first Bohr orbit in the hydrogen atom is -13.6 eV. What is its energy in the second and third Bohr orbit?
Answer:

Question 26.
The potential energy of the electron in the first Bohr orbit in the hydrogen atom is -27.2 eV. What is its kinetic energy and binding energy in the same orbit?
Answer:
Kinetic energy = –\(\frac{\text { potential energy }}{2}=-\frac{27.2}{2}\)
= 13.6 eV and
binding energy = – total energy – (potential energy + kinetic energy)
= – (-27.2 + 13.6) = 13.6 eV

Question 27.
Obtain the ratio of the longest wavelength of spectral line in the Paschen series to the longest wavelength of spectral line in the Brackett series.
Answer:
The Bohr formula for the wave number of a spectral line in the hydrogen spectrum is
\(\bar{v}=\frac{1}{\lambda}=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
where is the wavelength of the emitted radiation when the electron jumps from a higher energy state of principal quantum number m to a lower energy state of principal quantum number n. R is the Rydberg constant.

The Paschen series and Brackett series of spectral lines arise due to the transitions to n = 3 and m = 4, respectively. The longest wavelength lines (λ_Pα and λ_Bx) in these series arise due to the transitions from m =4 and m = 5, respectively.

Question 28.
Obtain the ratio of the wavelength of the H_α line to the wavelength of the H_γ. line in the Balmer series.
Ans.
The Bohr formula for the wave number of a spectral line in the hydrogen spectrum is
\(\bar{v}=\frac{1}{\lambda}=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
where λ is the wavelength of the emitted radiation when the electron jumps from a higher energy state of principal quantum number m to a lower energy state of principal quantum number n. R is the Rydberg constant.

The Balmer series of spectral lines arises due to the transitions to n = 2. The H_α and H_γ lines in this series arise due to the transitions from m = 3 and 5, respectively.

Question 29.
On the basis of the de Broglie hypothesis, obtain Bohr’s condition of quantization of angular momentum.
Answer:
By the de Broglie equation, the wavelength associated with an electron having momentum p = mv is
λ = \(\frac{h}{p}=\frac{h}{m v}\) ……….. (1)
In a hydrogen atom in its ground state, the de Broglie wavelength associated with the electron is the same as the circumference of the first Bohr orbit. Therefore, the electron orbit in a hydrogen atom in its ground state corresponds to one complete electron wave joined on itself.

Thus, a stationary orbit can be intepreted as one which can accommodate an integral number of de Broglie wavelength so that the associated matter wave will be in phase with itself and constructive interference will allow a standing wave along the orbit.

Therefore, for a stationary Bohr orbit of circumference 2πr,
2πr = nλ
where n is a positive integer.
∴ 2πr = \(\frac{n h}{m v}\) …………. [From Eqn (1)]
∴ Angular momentum, L = mvr = n(\(\frac{h}{2 \pi}\))
which is just the Bohr condition of angular momentum quantization for stable or allowed orbits.

30. Solve the following :
Data: e = 1.6 × 10^-19 C, m_e = 9.1 × 10^-31 kg, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s. I eV = 1.6 × 10^-19 J, ε₀ = 8.85 × 10^-12 F/m, R = 1.097 × 10⁷ m 1, 1/4πε₀ = 9 × 10⁹ N∙m²/C²

Question 1.
The radius of the first orbit of the electron in a hydrogen atom is 0.53 Å. Find the centripetal force acting on the electron.
Solution:
Data : r = 5.3 × 10^-11 m
The centripetal force on the electron = the electrostatic force between the proton and electron

Question 2.
Calculate the angular momentum of the electron in the third Bohr orbit of hydrogen atom.
Solution:
Data: n = 3
The angular momentum, L = \(\frac{n h}{2 \pi}\)
= \(\frac{3\left(6.63 \times 10^{-34}\right)}{2(3.142)}\) = 3.165 × 10^-34 kg.m²/s

Question 3.
Calculate the potential energy of the electron in the second Bohr orbit of hydrogen atom in electron volt. The radius of the Bohr orbit is 2.12 Å.
Answer:
Data : r = 2.12 × 10^-10 m
The potential energy of the electron

Question 4.
Calculate the radius of the first Bohr orbit in the hydrogen atom. [ε₀ = 8.85 × 10^-12 C²/N∙m² e = 1.6 × 10^-19 C, m(electron) = 9.1 × 10^-31 kg, h = 6.63 × 10^-34 J∙s]
Solution:
Data : ε₀ = 8.85 × 10^-12 C²/N∙m² e = 1.6 × 10^-19 C,
h = 6.63 × 10^-34 J∙s, m = 9.1 × 10^-31 kg
The radius of a Bohr orbit,

This is the radius of the third Bohr orbit in a hydrogen atom.

Question 5.
Find the ratio of the diameter of the first Bohr orbit to that of the fourth Bohr orbit in a hydrogen atom.
Solution:
The radius of a Bohr orbit,

Question 6.
Calculate the frequency of revolution of the electron in the second Bohr orbit of the hydrogen atom. The radius of the orbit is 2.14 Å and the speed of the electron in the orbit is 1.09 × 10⁶ m/s.
Solution:
Data: r = 2.14Å = 2.14 × 10^-10 m, v = 1.09 × 10⁶ m/s
v = ωr = (2πf)r
∴ The frequency of revolution of the electron in the second Bohr orbit,

Question 7.
The speed of the electron in the first Bohr orbit (in H atom) of radius 0.5 Å is 2.3 × 10⁶ m/s. Calculate the period of revolution of the electron in this orbit.
Solution:
Data: r = 0.5 Å = 5 × 10¹¹ m, v = 2.3 x 10⁶ m/s
Period of revolution of the electron,

Question 8.
The energy of the electron in the ground state of the hydrogen atom is -13.6 eV. Find its KE and PE in the same state.
Solution:

Question 9.
An electron is orbiting in the 3rd Bohr orbit in H atom. Calculate the corresponding ionization energy, if the ground state energy is – 13.6 eV.
Solution:
Data : E₁ = -13.6 eV, n₁ = 1, n₃ = 3, E_∞ = 0 eV
The energy of the electron in the nth Bohr orbit,

Question 10.
Find the energy of the electron in the fifth Bohr orbit of the hydrogen atom. [Energy of the electron in the first Bohr orbit -13.6 eVI
Solution:
Data : E₁ = -13.6 eV

Question 11.
Calculate the energy of the electron in the ground state of the hydrogen atom. Express it in joule and in eV.
[m_electron = 9.11 × 10^-31 kg, e = 1.6 × 10^-19 C, h = 6.63 × 10^-34 J∙s, ε₀ = 8.85 × 10^-12 C²/N∙m²]
(3 marks)
Solution:
Data: m = 9.11 × 10^-31 kg, e = 1.6 × 10^-19 C, h = 6.63 × 10^-34 J∙s, ε₀ = 8.85 × 10^-12 C²/N∙m²
The energy of electron in nth Bohr orbit is

Question 12.
The energy of the electron in an excited hydrogen atom is – 0.85 eV. Find the corresponding angular momentum of the electron.
[h = 6.63 × 10^-34 J∙s, π = 3.142, E₁ = -13.6 eV]
Solution :
Data : E_n = – 0.85 eV, h = 6.63 × 10^-34 J∙ s, π = 3.142, E₁ = -13.6 eV

Question 13.
A photon of energy 12.75 eV is absorbed by an electron in the ground state of a hydrogen atom and raises it to an excited state. Find the quantum number of this state.
Solution:
Data : hv = 12.75 eV
The energy of the electron in the ground state of the hydrogen atom, E₁ = -13.6 eV
On absorbing a photon of energy hv = 12.75 eV,
its final energy state is
E₁ = E_i + hv = – 13.6.+ 12.75 = -0.85 eV
E_n = \(-\frac{13.6}{n^{2}}\) eV
∴ E_f = -0.85 = \(-\frac{13.6}{n^{2}}\)
∴ n = \(\sqrt{\frac{13.6}{0.85}}\) = 4
∴ The principal quantum number of the final state of the electron = 4

Question 14.
Determine the linear momentum of the electron in the second Bohr orbit in a hydrogen atom. Hence determine the linear momentum in the third Bohr orbit.
Solution:
The linear speed of the electron in the nth Bohr orbit,
v_n = \(\frac{e^{2}}{2 \varepsilon_{0} n h}\)
∴ The linear momentum of the electron in the second orbit,

Question 15.
A quantum of monochromatic radiation of wavelength λ is incident on an hydrogen atom that takes it from the ground state to the n = 3 state. Find λ and the frequency of the radiation.
[E₁ = -13.6 eV, E₃ = -1.51 eV]
Solution :
Data : n_i = 3, n_f = 1, E₁ = -13.6 eV, E₃ = 1.51 eV, h = 6.63 × 10^-34 J∙s, e = 1.602 × 10^-19 C, c = 3 × 10⁸ m/s
The energy of the incident radiation,
hv = E₃ – E₁
∴ The frequency of the incident radiation,

Question 16.
A hydrogen atom undergoes a transition from a state with n = 4 to a state with n = 1. Calculate the change in the angular momentum of the electron and the wavelength of the emitted radiation.
[h = 6.63 × 10^-34 J∙s, R = 1.097 × 10⁷ m^-1]
Solution :
Data : h = 6.63 × 10^-34 J∙s, R = 1.097 × 10⁷ m^-1
(i) The angular momentum of the electron in the nth orbit of the hydrogen atom is
L_n = \(\frac{n h}{2 \pi}\)
The change in the angular momentum when the electron jumps from the 4th orbit to the 1st orbit is

Question 17.
Find the Rydberg constant given the energy of the electron in the second orbit in hydrogen atom is -3.4 eV.
Solution:
Data: E₂ = -3.4 eV = -3.4 × 1.6 × 10^-19 J, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s

Question 18.
Find the energy of the electron in eV in the third Bohr orbit of the hydrogen atom.
[R = 1.097 × 10⁷ m ^-1, c = 3 × 10⁸ m/s, h = 6.63 × 10^-19 J∙s, e = 1.6 × 10^-19 C]
Solution:
Data : n = 3, R = 1.097 × 10⁷ m ^-1, c = 3 × 10⁸ m/s, h = 6.63 × 10^-19 J∙s, e = 1.6 × 10^-19 C

Question 19.
Calculate the wavelength of the first two lines of the Balmer series in the hydrogen spectrum.
Solution :

∴ The wavelength of the second Balmer line,
λ_Hβ = \(\frac{16}{3.291}\) × 10^-7 = 4.862 × 10^-7 m = 4862 Å

Question 20.
The wavelength of H line of the Balmer series is 4860Å. Calculate the wavelength of the Balmer H_x line.
Solution:
Data : λ_β = 4860 Å

Question 21.
The short wavelength limit of the Lyman series is 911.3 Å. Compute the short wavelength limit of the Balmer series.
Solution:
Data : λ_∞L = 911.3 Å
The wavelengths of the lines in the Lyman series are given by
\(\frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\) (m = 2, 3, 4, ……..)
For the shortest wavelength line, m = ∞. Therefore, the short wavelength limit of the Lyman series is given by

Question 22.
The second member of the Balmer series for the hydrogen atom has wavelength 4860 Å. Calculate Rydberg’s constant.
Solution:

Question 23.
Find the shortest wavelength of the Paschen series, given that the longest wavelength of the Balmer series in the hydrogen spectrum is 6563 Å.
Solution:
Let λ_Bα and λ_P∞ be the wavelength of the first line, i.e., the longest wavelength line, of the Balmer series and the shortest wavelength of the Paschen series, respectively.
Data : λ_Bα = 6560 Å

Question 24.
Find the longest wavelength in the Paschen series. [R = 1.097 × 10⁷ m^-1]
Solution:
Data: R = 1.097 × 10⁷ m^-1
\(\frac{1}{\lambda}=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
For the longest wavelength line in the Paschen series, m = 4 and n = 3.

Question 25.
Find the ratio of the longest to shortest wavelengths in the Paschen series.
Solution:

Question 26.
Find the ratio of the longest wavelength in the Paschen series to the shortest wavelength in the Balmer series.
Solution:

Question 27.
Compute the shortest and the longest wavelength in the Lyman series of hydrogen atom.
Solution:

Question 28.
The wavelength of the first line of the Balmer series is 6563 Å. Calculate the wavelength of the first line of (i) the Lyman series (ii) the Paschen series.
Solution:

∴ The wavelength of the first line of the Lyman series,

Question 31.
Name the constituents of an atomic nucleus. What is a nucleon?
Answer:
The constituents of an atomic nucleus are (1) the proton, a positively charged particle (2) the neutron, a neutral (uncharged) particle. The term nucleon (nuclear constituent) refers to a proton as well as a neutron.

[Note : The electron was discovered by J. J. Thomson in 1897. The proton was discovered by Ernest Rutherford in 1919. The neutron was discovered in 1932 by James Chadwick (1891-1974), British physicist. The existence of the neutron and the deuteron was predicted by Rutherford in 1920. The proton has a mass about 1836 times that of the electron, but the magnitude of the electric charge is the same for both. The mass of the neutron is slightly more than that of the proton.]

Question 32.
Define (i) atomic number (ii) mass number. Give their symbols.
Answer:

1. The number of protons in the nucleus of an atom of an element is called the atomic number of the element. It is also known as the proton number.
   It is denoted by Z.
2. The number of nucleons (protons and neutrons) in the nucleus of an atom is called the mass number or the atomic mass number. It is denoted by A.

[Notes : (i) The number of neutrons in the nucleus of an atom is known as the neutron number, denoted by N. It is usually greater than Z, with the exceptions of helium (2 protons, 2 neutrons) and hydrogen (1 proton, no neutron) (ii) A = Z + N.]

Question 33.
Write the atomic symbol for an element giving the atomic number and mass number. Give two examples. Which of the two numbers is characteristic of the element? Why?
Answer:
An atom is represented as \(\frac{A}{Z}\)X, where X is the chemical symbol for the element, Z is the atomic number and A is the mass number.
Examples : Fluorine, \(\begin{aligned}
&19 \\
&9
\end{aligned}\)F; Phosphorus, \(\begin{aligned}
&31 \\
&15
\end{aligned}\)P; Gold, \(\begin{aligned}
&197 \\
&79
\end{aligned}\)Au.

The atomic number of an element, which is the number of protons in the nucleus of an atom of the element, is characteristic of the element. It equals the number of electrons in the atom and hence determines the chemical properties of the element and its place in the modern periodic table.

Question 34.
What are isotopes? Give an example.
Answer:
Atoms of an element having the same atomic number (Z) but different mass numbers (A) are called the isotopes of that element.

Isotopes of an element have different neutron numbers but the same chemical properties.

Example : Hydrogen has three isotopes, namely, hydrogen \( (\begin{aligned}
&1 \\
&1
\end{aligned}\) H), deuterium \( (\begin{aligned}
&2 \\
&1
\end{aligned}\) D) and tritium \( (\begin{aligned}
&3 \\
&1
\end{aligned}\) T). Deuterium and tritium have one and two neutrons in their nuclei, respectively, in addition to the single proton (Z = 1).

Question 35.
What are isobars? Give an example.
Answer:
Atoms of different elements that have the same mass number (A) but different atomic numbers (Z) are called isobars.

Although isobars have the same mass number, they are different elements because the chemical nature of an element is determined by its atomic number. Isobars have different neutron numbers.

Example : \(\begin{array}{r}
13 \\
6
\end{array}\)C and \(\begin{array}{r}
13 \\
7
\end{array}\)N are isobars. They have
the same mass number, A (viz., 13), but their different proton numbers, Z (6 and 7) make them different elements.

Question 36.
What are isotones? Give an example.
Answer:
Atoms of different elements that have the same neutron number (N) but different atomic numbers (Z) are called isotones.

Although isotones have the same neutron number, they are different elements because the chemical nature of an element is determined by its atomic number.

Question 37.
What is unified atomic mass unit? Express it in J/c² and MeV/c².
Answer:
The unified atomic mass unit is an accepted but non-SI unit of mass. It is defined to be equal to \(\frac{1}{12}\) of the mass of a free atom of the isotope of carbon with mass number 12 which is at rest and in its ground state. It is denoted by u.

Its value in SI unit is obtained experimentally. 1 u = 1.660538782 (83) × 10^-27 kg, with the standard uncertainty in the last two digits given in v parenthesis.
Taking, 1 u = 1.660538782 × 10^-27 kg
c = 2.99792458 × 10⁸ m/s,
e = 1.602176462 × 10^-19 C and
using the relation E = me² { ≡ m ≡ E / c²),
we get, 1 u = 1.49241783 × 10^-10 J/c²
≅ 1.492 × 10^-10 J/c²
and 1 u = 931.494042 MeV / c²
≅ 931.5 MeV/c²

Question 38.
How is the nuclear size determined ? State the relation between nuclear size (radius) and mass number.
Answer:
The nuclear size is determined from particle scattering experiments using fast electrons or neutrons. The de Broglie wavelength of the bombarding electrons or neutrons should be less than the radius of the nucleus under study.

It is found that the volume of a nucleus is directly proportional to the mass number A, i.e., to the number of nucleons in the nucleus. For many purposes, nuclei may be assumed to be spherical.
Thus, a nucleus of radius R has a volume \(\frac{4}{3}\) πR³
∴ R³ ∝ A or R ∝ A^{\(\frac{1}{3}\)}
∴ R = R₀A^{\(\frac{1}{3}\)}
where R₀ ≈ 1.2 × 10^-15 m = 1.2 fm.
∴ R ≈ 1.2 AA^{\(\frac{1}{3}\)} fm

[Note : A nucleus does not have a sharp boundary. Also, electron scattering and neutron scattering yield slightly different values of R₀. Hence, the relation above is only representative of effective nuclear size. 1 femtometre or 1 fm = 10^-15 m; an earlier non-SI unit of the same value called fermi, in honour of Enrico Fermi (1901-54), Italian-US nuclear physicist, is no longer accepted in SI.]

Question 39.
Nuclear density is essentially the same for all nuclei. Justify.
Given 1 u = 1.66 × 10^-27 kg and R₀ ≈ 1.2 fm, estimate the nuclear density.
Answer:
Relative atomic mass rounded to the nearest integral value equals the atomic mass number A. Thus, ignoring the masses of the atomic electrons and binding energies, nuclear mass expressed in unified atomic mass unit = A u. Also, it is experimentally found that the volume of a nucleus is directly proportional to the mass number A. As both nuclear mass and volume are proportional to the mass number, nuclear density is essentially the same for all nuclei.

The order of nuclear density is 10¹⁷ kg/m³.

[Note : Some stars, with masses between 1.4\(M_{\odot}\) and 3\(M_{\odot}\), where \(M_{\odot}\) denotes the mass of our Sun, undergo supernova at the end of their active life and collapse into neutron stars of densities comparable with nuclear v density.]

Question 40.
Define mass defect and state an expression for it.
Answer:
The difference between the sum of the masses of all the individual nucleons in a nucleus and the mass of the nucleus is known as the mass defect.

Mass defect, ∆m = (Zm_p + Nm_n) – M where mp is the proton mass, mn is the neutron mass, M is the mass of the nucleus, Z is the atomic number and N = A – Z is the neutron number.

Question 41.
Explain the term nuclear binding energy and express it in terms of mass defect. What is binding energy per nucleon? Write the expression for it.
Answer:
In the atomic nucleus, the protons and the neutrons are bound together by a strong, short range and charge independent, attractive force called the nuclear force. It is necessary to supply energy to break the nucleus. The minimum energy required to separate a nucleus into its free constituents, i.e., protons and neutrons, is known as the nuclear binding energy. It is the mass energy of the nucleons minus the mass energy of the nucleus.

The mass defect of a nucleus, of mass M, mass number A, proton number Z and neutron number N = A – Z, is
∆m = (Zm_p + Nm_n) – M
where, mp is the proton mass and mn is the neutron mass. Then, from Einstein’s mass-energy relation,
nuclear binding energy = ∆mc²
= [(Zm_p -(- Nm_n) – M] c² (in joule)
= [(Zm_p + Nm_n) – M] \(\frac{c^{2}}{e}\) (in eV)

where c is the speed of light in free space and e is the elementary charge. In calculations using the above equations, mp is replaced by mH (mass of hydrogen atom). M is taken to be the atomic mass and not nuclear mass because the electron masses cancel out and the difference in electronic binding energies can be ignored as these are 106 times smaller than the nuclear binding energy.
∴ Nuclear binding energy
≅ [(Zm_H + Nm_n) – M_atom] c² (in joule)
The minimum energy required on the average to separate a nucleon from a given nucleus is called the binding energy per nucleon for that nucleus. It is the nuclear binding energy for a nucleus divided by its mass number.
∴ Binding energy per nucleon = \(\frac{\Delta m c^{2}}{A}\)
= [(Zm_p + Nm_n) – M] \(\frac{c^{2}}{A}\)
≅ [(Zm_H + Nm_n) – M_atom] \(\frac{c^{2}}{A}\)

Question 42.
What is the significance of binding energy per nucleon?
Answer:
The greater the binding energy per nucleon in a nucleus, the greater is the minimum energy needed to remove a nucleon from the nucleus. Thus, binding energy per nucleon indicates the stability of a nucleus.

[Note : The binding energy per nucleon is high when both Z and N are even numbers, and such nuclei are most common. Nuclei with both Z and N odd are very rare.]

Question 43.
What are stable nuclei? What decides nuclear stability? What are the properties of nuclear force?
Answer:
Those nuclei which for certain combinations of neutrons and protons do not spontaneously disintegrate are called stable nuclei. There are two aspects that decide the stability of a nucleus. Firstly, the existence of nuclear energy levels implies certain configurations to achieve potential energy minimum, and secondly, the balance of forces.

Just like energy levels in atoms, nuclear energy levels are filled in sequence obeying the exclusion principle. Thus, there is a tendency for N to equal Z, or to have both even Z and even N.
Properties of the nuclear force :
(1) The nucleons in a nucleus are held .together by the attractive strong nuclear force. This force is much stronger than gravitational force and electromagnetic force.

(2) Nucleons interact strongly only with their nearest neighbours because the nuclear force has an extremely short range. Gravitational force and electromagnetic force are long range forces. They tend to zero only when the separation between two particles tends to infinity.

(3) Inside a nucleus, this force appears to be the same between two protons, a proton and a neutron, and two neutrons. However, between two protons there is also Coulomb repulsion which has a much longer range and, therefore, has appreciable magnitude throughout the entire nucleus. In nuclei having 2 ≤ Z ≤ 83, with neutrons present, the nuclear force is strong enough to overcome the Coulomb repulsion.

For light nuclei (A < 20), N ≥ Z, but is never smaller (except in \(\begin{aligned} &1 \\ &1 \end{aligned}\)H and \(\begin{aligned} &3 \\ &2 \end{aligned}\)H). However, with more than about 10 protons, an excess of neutrons is required to form a stable nucleus; for high atomic numbers, N/Z = 1.6. For Z > 83, even an excess of neutrons cannot prevent spontaneous disintegration and there are no stable nuclei.
[Note : The strength of the nuclear force is evident from the nuclear binding energy.]

Question 44.
Draw a neat labelled graph showing the variation of binding energy per nucleon as a function of mass number. What can we infer from the
graph?
Answer:
Figure shows the plot of the binding energy per nucleon (BE/A, in MeV per nucleon) against mass number A.

From Figure, we can draw the following inferences :
(1) The greater the binding energy per nucleon, the more stable is the nucleus because greater is the minimum energy needed to remove a nucleon. Thus, the nuclei appearing high on the plot are more tightly bound.

(2) BE/A has a maximum value of about 8.8 MeV per nucleon at A = 56 (⁵⁶Fe nuclide) and then decreases to 7.6 MeV per nucleon at A = 238 (²³⁸U nuclide). Thus, the iron nuclide ⁵⁶Fe has the maximum binding energy per nucleon and is the most stable nuclide. Also, the peak at A = 4 shows that the 4He nucleus (α particle) has higher binding energy per nucleon compared to its neighbours in the periodic table and is exceptionally stable.

(3) The increase in BE /A as A decreases from 240 to 60 shows that if a heavy nucleus splits into two medium-sized fragments, each of the new nuclei will have more BE / A than the original nucleus. The binding energy difference, which can be very large, will then be released. The process of splitting a heavy nucleus is called nuclear fission. The energy released in a fission of ²³⁵U nucleus is about 200 MeV.

(4) Joining together, or fusing, two very light nuclei to form a single nucleus will also lead to larger BE/A in the new heavier nucleus. Again, the binding energy difference will be released. This process,which is called nuclear fusion, is also a very effective way of obtaining energy.

45. Solve the following :
Question 1.
Find the nuclear radii of ²⁰⁶Pb and ²⁰⁸Pb.
Solution:
Data : R₀ = 1.2 × 10^-15 m
Nuclear radius, R=R₀A^{\(\frac{1}{3}\)}
(i) For ²⁰⁶Pb, A = 206
∴ R = (1.2 × 10^-15) (206)^{\(\frac{1}{3}\)}
= 1.2 × 10^-15 × 5.906
= 7.087 × 10^-15 m = 7.087 fm

(ii) For ²⁰⁸ Pb, A = 208
∴ R = (1.2 × 10^-15)(208)^{\(\frac{1}{3}\)}
= 1.2 × 10^-15 × 5.925
= 7.11 × 10^-15 m = 7.11 fm

Question 2.
Given the atomic mass of the isotope of iron 56Fe is 55.93 u, find its nuclear density.
Solution :
Data : A = 56, m = 55.93 u = 55.93 × 1.66 × 10^-27 kg, R₀ = 1.2 × 10^-15 m .

Question 3.
Given the nuclear radius of ¹⁶O is 3.024 fm, find that of ²³⁵U.
Solution:
Data : A₁ = 16 and R₁ = 3.024 fm (for ¹⁶O),
A₂ = 235 (for ²³⁵U)

Question 4.
The mass defect of He nucleus is 0.0304 u. Calculate its binding energy.
Solution:
Data: ∆m = 0.0304 u, 1 u = 931.5 \(\frac{\mathrm{MeV}}{c^{2}}\)
The binding energy of He nucleus
= ∆mc²
= 0.0304 × 931.5
= 28.32 MeV

Question 5.
Calculate the mass defect and binding energy of \(\begin{aligned}
&59 \\
&27
\end{aligned}\) Co which has a nucleus of mass 58.933 u. [Take m^p = 1.0078 u, mⁿ = 1.0087 u]
Solution:
Data: m^p = 1.0078 u, mⁿ = 1.0087 u, m^Co = 58.933 u,
1 u = 931.5 \(\frac{\mathrm{MeV}}{c^{2}}\)
For \(\begin{aligned}
&59 \\
&27
\end{aligned}\)Co, A = 59, Z = 27
∴ N = A – Z = 59 – 27 = 32
The mass defect,
∆m=(Zm_p + N_n) – m_Co
= (27 × 1.0078 + 32 × 1.0087) – 58.933
= (27.2106 + 32.2784) – 58.933
= 59.4890 – 58.933 = 0.556 u
∴ The binding energy
= ∆mc²
= 0.556 uc² × 931.5 \(\frac{\mathrm{MeV}}{uc^{2}}\) = 517.8 MeV

Question 6.
Find the mass energy of a particle of mass 1 u in joule and electronvolt.
Solution:
Data : m = 1 u = 1.66 × 10^-27 kg, c = 3 × 10⁸ m/s,
1 eV = 1.602 × 10^-19 J
The mass energy,
E = mc²
= (1.66 × 10^-27) (3 × 10⁸)⁸
= 1.66 × 9 × 10^-11 n
= 1.494 × 10^-10 J
= \(\frac{1.494 \times 10^{-10}}{1.602 \times 10^{-19}}\) eV
= 9.326 × 10⁸ eV or 932.6 MeV
[Note : The answer differs from the accepted value of about 931.5 MeV because of the rounding off errors in the data used.]

Question 7.
Find the mass energy of a proton at rest in MeV. [m_p = 1.673 × 10^-27kg]
Solution:
Data : m_p = 1.673 × 10^-27 kg, c = 3 × 10⁸ m/s,
1 eV = 1.602 × 10^-19 J
∴ 1 MeV = 10⁶ × 1.602 × 10^-19 J = 1.602 × 10^-13 J
The mass energy of a proton at rest,

Question 8.
Find the mass energy of a particle of mass 1 g in joule.
Solution:
Data : m = 1 g = 1 × 10^-3 kg, c = 3 × 10⁸ m/s,
1 eV = 1.602 × 10^-19 J
The mass energy,
E = mc²
= (1 × 10^-3)(3 × 10⁸)²
= 9 × 10¹³ J

Question 46.
What is radioactivity? OR Define radioactivity.
Answer:
Radioactivity is the phenomenon in which unstable nuclei of an element spontaneously distintegrate into nuclei of another element by emitting a or β particles accompanied by γ-rays. Such transformation is known as radioactive transformation or radioactive decay.

Question 47.
Who discovered radioactivity? Give a brief account of the first observation of radioactivity.
Answer:
Antoine Henri Becquerel (1852-1908), French physicist, discovered radioactivity in 1896.

He had kept photographic plates wrapped in a thick black paper in a drawer of his desk. Later, he also kept uranium salts near the photographic plates. After some days he developed the photographic plates and was surprised to find that they were fogged although he had protected them from light. Becquerel concluded that uranium salts must be emitting some invisible rays which affected the photographic plates. In this way, radioactivity was discovered by Becquerel.

Question 48.
What is a radioactive element? Give two examples of radioactive elements.
Answer:
The element which exhibits the property of radioactivity, i.e., spontaneous disintegration of unstable nuclei of the element by emission of α or β particles accompanied by γ-rays is called a radioactive element. Examples : Uranium, thorium, polonium, radium, actinium.

Question 49.
Name the three radioactive decay processes. State the nature of particle/radiation emitted in each process. What is meant by the Q value or Q factor of the decay?
Answer:
The processes by which a radioactive element can decay are :
(1) α-decay, by emission of an α-particle (\(\begin{aligned}
&4 \\
&2
\end{aligned}\) He nucleus),
(2) β-decay, by emission of an electron (\(\begin{aligned}
&0 \\
&-1
\end{aligned}\)e) or a positron \(\begin{aligned}
&0 \\
&1
\end{aligned}\)e), and

(3) γ-decay, by emitting electromagnetic radiations (γ-rays) of very short wavelength of about 10^-12 m to 10^-14 m.

In a radioactive decay, the difference in the energy equivalent of the mass of the parent atom and that of the sum of the masses of the products is called the Q value or Q factor of the decay. It is also called disintegration energy.

Question 50.
State the observations which lead to the conclusion that radioactivity is a nuclear phenomenon.
Answer:
The rate of disintegration of a radioactive material is not affected by changes in physical and chemical conditions such as (1) temperature and pressure (2) action of electric and magnetic fields (3) chemical composition of the material. The above changes affect the orbital electrons, but not the nucleus. Therefore, we conclude that radioactivity is a nuclear phenomenon.

Question 51.
State the nature and properties of a-particles.
Answer:
Nature of α-particles :
(1) An alpha particle is a helium nucleus, i.e., a doubly ionized helium atom. It consists of two protons and two neutrons.
(2) Mass of the α-particle ≅ 4u
Charge on the α-particle = 2 × charge on the proton
Properties of α-particles :
(1) Of the three types of radioactive radiations, α- particles have the maximum ionizing power. It is about 100 times that of β-particles and 104 times that of γ-rays.

(2) They have the least penetrating power, about 100 times less than that of β-particles and 10⁴ times less than that of γ-rays. They can pass through very thin sheets of paper but are scattered by metal foils and mica. Since α-particles produce intense ionization in a medium, they lose their kinetic energy quickly. As a result, they do not penetrate more than a few centimetres (about 2.7 cm to 8.6 cm) in air under normal conditions.

(3) They are deflected by electric and magnetic fields since they are charged particles. Their deflection is less than that of β-particles in the same field.

(4) They affect photographic plates.
(5) They cause fluorescence in fluorescent materials such as zinc sulphide.
(6) They emerge from the nuclei with tremendous speeds in the range of \(\frac{1}{100}\)th to \(\frac{1}{10}\)th of the speed of light in free space.
(7) They destroy living cells.
[Note : α-rays and β-rays were discovered by Henri BecquereL]

Question 52.
State the nature and properties of β-particles.
Answer:
Nature of β-particles : A β-particle is an electron or a positron.
Properties of β-particles :

1. β-particles have a moderate ionizing power. It is about 100 times less than that of α-particles, but 100 times more than that of γ-rays.
2. They have a moderate penetrating power. It is about 100 times more than that of α-particles, but 100 times less than that of γ-rays.
3. They are deflected by electric and magnetic fields. Their deflection is more than the deflection of α-particles in the same field but in the opposite direction.
4. They affect photographic plates.
5. They cause fluorescence in a fluorescent material such as zinc sulphide.
6. Their energies and speeds are very high. Their speed is of the order of 10⁸ m/s. Some β-particles have speeds of the order of 0.99 c, where c is the speed of light in free space.
7. They cause more biological damage than α-particles, because they have more penetrating power.

[Note : When a nucleus emits an electron, one of its neutrons changes to a proton; the electron is accompanied by a neutral and almost massless particle called antineutrino \(\bar{v}_{\mathrm{e}}\) with which the electron shares its energy and momentum. Hence, β-particles are emitted with speeds ranging from about 0 to 0.99 c.

The properties of a positron are identical to those of an electron except that it carries a positive charge of the same magnitude as an electron. A positron emission in a β⁺ decay is accompanied by a neutrino v_e. A positron is an anti-particle of an electron, and an antineutrino is an antiparticle of neutrino.

The existence of a small neutral particle, emitted simultaneously with the electron in β^–-decay, was proposed in 1931 by Wolfgang Pauli (1900 -1958), Austrian-US theoretical physicist. It was confirmed experimentally in 1956 by Frederic Reines and Clyde Lorrain Cowan, Jr., US physicists. This neutral particle, that appears in β⁺ -decay, was called the neu-trino. It travels with a speed very close to that of light in free space. The existence of positron (in 1928) and other antiparticles was predicted by Paul Adrien Maurice Dirac (1902 – 84), British theoretical physicist. All these predictions were eventually confirmed experimentally; the positron was discovered in 1932 by Carl David Anderson, US physicist.]

Question 53.
State the nature and properties of γ-rays.
Answer:
Nature of γ-rays :

1. γ-rays are electromagnetic waves of very short wavelength (about 10^-12 m to 10^-14 m).
2. They are uncharged.

Properties of γ-rays :

1. γ-rays produce feeble ionization. Their ionizing power is 104 times less than that of a-particles and 100 times less than that of β-particles.
2. They have the maximum penetrating power. It is about 100 times that of β-particles and 10⁴ times that of α-particles.
3. They are not deflected by electric and magnetic fields as they are uncharged.
4. They affect photographic plates.
5. They cause fluorescence in a fluorescent material such as zinc sulphide.
6. Their speed in free space is 3 × 10⁸ m/s( the same as that of light waves and X-rays in free space).
7. γ-rays can be diffracted by crystals. In recent times, γ-ray diffraction has emerged as a powerful tool in structural and defect studies of crystals.
8. They destroy living cells and tissues and are used for destroying cancer cells.

[Note : γ-rays (not established clearly in Berquerel’s work) were discovered in 1900 by Paul Villard (1860-1934), French physicist.]

Question 54.
(a) What is α-decay? What is the consequence of an α-decay on a radioactive element? What is the Q value or Q factor in this case ?
Q = [m_U – m_Th – m_α]c²
(b) What is β-decay ? What is the consequence of a β-decay on a radioactive element? What is the Q value or Q factor in this case ?
Answer:
(a) A radioactive transformation in which an a-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{38} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_U – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β -decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{23} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the antineutrino emitted to conserve the momentum, energy and spin.
Q = [m_Th – m_Pa – m_e]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \({ }_{15}^{30} \mathrm{P} \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

Question 55.
Why are α- and β-particle emissions often accompanied by γ-rays?
Answer:
A given nucleus does not emit α- and βparticles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α- and β-particle emissions are often accompanied by γ-rays.

Question 56.
State the law of radioactive decay and express it in the exponential form.
OR
State the law of radioactive decay. Hence derive the relation N = N₀e^-λt, where the symbols have their usual meanings.
OR
Show that the number of nuclei of a radioactive material decreases exponentially with time.
Answer:
Law of radioactive decay : At any instant, the rate of radioactive distintegration is directly proportional to the number of nuclei of the radioactive element present at that instant.

Derivation : Let N₀ be the number of nuclei present at time t = 0, and N the number of nuclei present at time t.

where λ is a constant of proportionality called the radioactive decay constant or the distintegration constant. It is a constant for a particular radioactive element. The minus sign indicates that N decreases as t increases.
Integrating Eqn (1),

This is the exponential form of the law of radioactive decay. It shows that the number of nuclei present decreases exponentially with time.
[Note : This equation is also written in the form
N(t) = N₀e^-λt]

Question 57.
If the number of nuclei of a radioactive substance becomes \(\frac{1}{e}\) times the initial number in 10 days, what is the decay constant of the substance ?
Answer:

Question 58.
What is meant by the activity of a sample of radioactive element? State the expression for it and also the units.
Answer:
The rate of disintegration of a sample of a radioactive element is called its activity. Let A denote the activity at time t and A₀ the initial activity. Then,

The SI unit of activity is called the becquerel (Bq) in honour of Henri Becquerel. 1 Bq = 1 disintegration per second.

The earlier unit, the curie (Ci) was based on the activity of 1 gram of ²²⁶Ra. 1 Ci = 3.7 × 10¹⁰ Bqn It was named after Marie Curie, Polish-bom French chemist.
[Note : Eqn (1) is also written in the form A (t) = A₀e^-λt]

Question 59.
Define half-life a radioactive element and obtain the relation between half-life and decay constant.
Answer:
The half-life of a radioactive element is defined as the average time interval during which half of the initial number of nuclei of the element disintegrate.

Let N₀ be the number of nuclei of a radioactive element present at time t = 0 and N, the number of nuclei present at time t. From the law of radioactive decay,
N = N₀e^-λt
where λ is the decay constant of the element.
If T is the half-life of the element, then, N = \(\frac{N_{0}}{2}\) when t = T.

This is the relation between the half-life and the decay constant of a radioactive element.

[Note : Radioactive decay law is statistical in nature and applicable only when the number of nuclei in the sample under consideration is very large. λ gives the probability that a nucleus of the element will decay in one second. We cannot know the exact number of nuclei that would decay in a given time interval. Hence, the use of the term average in the definition of half-time.]

Question 60.
What is meant by average life or mean life of a radioactive species ? How is it related to the half-life?
Answer:
Let N₀ = number of nuclei present at time t = 0 and λ = decay constant of a radioactive species.
| dN | = | λNdt |. ∴ The number of nuclei decaying between time tand t + dt is λN₀e^-λtdt. The life time of these nuclei is t. The average life or mean life of a radioactive species is denoted by t and is, by definition,

Question 61.
Define decay constant or distintegration constant of a radioactive element. If λ is the decay constant of a radioactive element, show that about 37% of the original nuclei remains undecayed after a time interval of λ^-1.
Answer:
The decay constant or disintegration constant of a radioactive element is defined as the ratio of the disintegration rate at an instant to the number of undecayed nuclei of the element present at that instant.

Let N₀ be the number of nuclei of a radioactive element present at time t = 0 and N, the number of undecayed nuclei at time t. From the radioactive law,
N = N₀e^-λt
where λ is the decay constant. At t = λ^-1, the fraction of undecayed nuclei is
\(\frac{N}{N_{0}}\) = e^{-λ × λ-1} = e^-1 = \(\frac{1}{e}\)
Since e ≅ 2.718,
\(\frac{N}{N_{0}}=\frac{1}{2.718}\) = 0.3679
Therefore, about 36.79% ≈ 37% of the original nuclei remains undecayed after a time λ^-1. Since λ is the probability that a nucleus of the element will decay in one second, λ^-1 gives the mean-life or the mean life time τ of the radioactive element measured in second; τ = λ^-1.

Question 62.
Show graphically how the number of nuclei (N) of a radioactive element varies with time (t).
Answer:

Question 63.
The half-life of a radioactive material is 4 days. Find the time required for 1/4 of the initial number of radioactive nuclei of the element to remain undisintegrated.
Answer:
For t = nT, N = N₀/ 2ⁿ. In this case, n = 2.
∴ f = 2T = 2 × 4 = 8 days is the required time.

Question 64.
A radioactive sample with half-life 2 days has initial activity 32 μCi. What will be its activity after 8 days?
Answer:
Here, t = 8d and T = 2d
∴ t =4T
For t = nT, N = N₀/ 2ⁿ and activity ∝ N
∴ A = A₀/2⁴ = A₀/16 = 32 /16
= 2 μCi is the required activity.

Question 65.
In successive radioactive decay if the decrease in mass number is 32 and the decrease in atomic number is 8, how many (i) α particles (ii) β particles are emitted in the process ?
Answer:

1. Number of α-particles emitted = 32/4 = 8
2. Number of β-particles emitted = 16 – 8 = 8.

66. Solve the following :
Question 1.
The decay constant of a radioactive substance is 4.33 × 10^-4 per year. Calculate its half-life and average life.
Solution:
Data : λ = 4.33 × 10^-4 per year
The half-life period of the radioactive substance.

Question 2.
The half-life of \(\begin{array}{r}
226 \\
88
\end{array}\)Ra is 1620 y. Find its decay constant in SI unit.
Solution:
Data: T = 1620 y = 1620 × 365 × 8.64 × 10⁴ s
= 5.109 × 10¹⁰ s
The decay constant of \(\begin{array}{r}
226 \\
88
\end{array}\)Ra is
λ = \(\frac{0.693}{T}=\frac{0.693}{5.109 \times 10^{10}}\) = 1.356 × 10^-11 s^-1

Question 3.
The half-life of \(\begin{array}{r}
210 \\
84
\end{array}\)Po is 138 d. Find the time required for 75% of the initial number of radio active nuclei of \(\begin{array}{r}
210 \\
84
\end{array}\)Po to disintegrate.
Solution:
Data: T = 138 d, = \(\frac{N}{N_{0}}=\frac{1}{4}\) (since only 25% of the
initial number of nuclei remains undisintegrated)
In one half-life (t = T), \(\frac{N}{N_{0}}=\frac{1}{2}\). In two half-lives (t = 2T), \(\frac{N}{N_{0}}=\frac{1}{4}\)
∴ The time for 75% of \(\begin{array}{r}
210 \\
84
\end{array}\)Po nuclei to disintegrate is 2T = 2 × 138 = 276 d.

Question 4.
Protactinium \(\begin{array}{r}
233 \\
91
\end{array}\)Pa decays to \(\frac{1}{5}\)th of its initial quantity in 62.7 days. Calculate its decay constant, mean-life and half-life.
Solution:
Data : \(\frac{N}{N_{0}}=\frac{1}{5}\), t = 62.7d

Question 5.
The half-life of \(\begin{array}{r}
210 \\
82
\end{array}\)Pb is 2.2.3 y. How long will it take for its activity to reduce to 30% of the initial activity?
Solution:
Data : T = 22.3 y, A = 0.3 A₀
By the radioactive decay law,
N = N₀e^-λt
∴ λN = λN₀e^-λt
∴ λ = A₀e^-λt
where A₀ = AN₀ is the initial activity and A = λN is the activity at time t.

∴It will take 38.75 y for the activity of \(\begin{array}{r}
210 \\
82
\end{array}\)Pb to reduce to 30% of the initial activity.

Question 6.
Radioactive sodium \(\begin{aligned}
&24 \\
&11
\end{aligned}\)Na has half-life of 15 h. Find its decay constant and mean-life. How much of 10 g of \(\begin{aligned}
&24 \\
&11
\end{aligned}\)Na will be left after 24 h?
Answer:
Data : T = 15 h, m₀ = 10 g, t = 24 h

since mass of a sample is directly proportional to the number of atoms or nuclei present.

Question 7.
Thorium \(\begin{gathered}
232 \\
90
\end{gathered}\)Th disintegrates into lead \(\begin{gathered}
200 \\
82
\end{gathered}\)Pb. Find the number of α and β particles emitted in the disintegration.
Solution:
An α-particle is a helium nucleus with mass number 4 and atomic number 2. In an α-decay, the mass number of the disintegrating nucleus decreases by 4 and its atomic number decreases by 2.

A β^–-particle is an electron with mass number 0 and atomic number – 1. In a β^–-decay, the mass number of the disintegrating nucleus remains unchanged and its atomic number increases by 1.

Let x α-particles and y ß-particles be emitted in the disintegration of \(\begin{gathered}
232 \\
90
\end{gathered}\)Th into \(\begin{gathered}
200 \\
82
\end{gathered}\)Pb.
\({ }_{90}^{232} \mathrm{Th} \stackrel{x \alpha+y \beta^{-}}{\longrightarrow}{ }_{82}^{200} \mathrm{~Pb}\)
∴ 232 – 4x – 0(y) = 200
∴ 4x = 232 – 200 = 32
∴ x = 8
Also, 90 – 2x + 1(y) = 82
∴ 90 – 2(8) + y = 82
∴ y = 82 + 16 – 90 = 8
∴ 8 α-particles and 8 ß^–-particles are emitted in the decay series of ²³²Th to ²⁰⁰Pb.

Question 67.
What is nuclear energy?
Answer:
Energy released in a nuclear reaction such as a spontaneous or induced nuclear fission, or nuclear fusion, or in interaction of two nuclei, is called nuclear energy.

[Note : It is far greater from that released in a chemical reaction.]

Question 68.
What is a nuclear reaction? Give one example.
Answer:
A reaction between the nucleus of an atom and a bombarding particle leading to the production of a new nucleus and, in general, the ejection of one or more particles is known as a nuclear reaction.

Question 69.
What are the quantities conserved in a nuclear reaction?
Answer:
The total momentum, energy, spin, charge and number of nucleons are conserved in a nuclear reaction.

Question 70.
What is fission? Who discovered nuclear fission?
Answer:
Nuclear fission is a nuclear reaction in which a heavy nucleus of an atom, such as that of uranium, splits into two or more fragments of comparable size, either spontaneously or as a result of bombardment of a neutron on the nucleus (induced fission). It is followed by emission of two or three neutrons.

The mass of the original nucleus is more than the sum of the masses of the fragments. This mass difference is released as energy, which can be enormous as in the fission of ²³⁵U.

Nuclear fission was discovered by Lise Meitner, Otto Frisch, Otto Hahn and Fritz Strassmann in 1938.

[Notes : (1) Lise Meitner (1878-1968), Austrian- Swedish physicist and radiochemist. Otto Frisch (1904-79), Austrian-British physicist and Meitner’s nephew. Otto Hahn (1879-1968), German radiochemist. Friedrich Wilhelm “Fritz” Strassman (1902-1980), German chemist. (2) In 1934, Enrico Fermi bombarded uranium with neutrons to produce nuclear reactions leading to formation of transuranic elements (Z > 92). In the process, he had carried out nuclear fission, but he misinterpreted the results. The experiments and analysis carried out by Meitner, Hahn and Strassman, and the theoretical work by Meitner and Frisch led to the discovery of fission in 1938-39.

Frisch named the phenomenon fission. Fermi, for his work in the area of nuclear science, was awarded the 1938 Nobel Prize for physics. Hahn was awarded the 1944 Nobel Prize for chemistry; it was not shared by Meitner as her important role in the discovery of fission came to light much later. (3) The most abundant urnanium isotope ²³⁸U (abundance 99.28%) can be fissioned by neutrons with high kinetic energy (called fast neutrons), at least 1.3 MeV. ²³⁵U (abundance 0.72%) can be fissioned by thermal (or low energy) neutrons having kinetic energy about 0.025 eV. Some types of nuclear reactors require the natural uranium to be enriched to increase its ²³⁵U content to about 3%.]

Question 71.
What are the products of the fission of uranium 235 by thermal neutrons?
Answer:
The products of the fission of ²³⁵U by thermal neutrons are not unique. A variety of fission fragments are produced with mass number A ranging from about 72 to about 138, subject to the conservation of mass-energy, momentum, number of protons (Z) and number of neutrons (N). A few typical fission equations are

Question 72.
What is nuclear fusion? Give one example with an equation.
Answer:
A type of nuclear reaction in which lighter atomic nuclei (of low atomic number) fuse to form a heavier nucleus (of higher atomic number) with the’ release of enormous amount of energy is called nuclear fusion.

Very high temperatures, of about 10⁷ K to 10⁸ K, are required to carry out nuclear fusion. Hence, such a reaction is also called a thermonuclear reaction.

Example : The D-T reaction, being used in experimental fusion reactors, fuses a deuteron and a triton nuclei at temperatures of about 10⁸ K.

(2) The value of the energy released in the fusion of two deuteron nuclei and the temperature at which the reaction occurs mentioned in the textbook are probably misprints.]

Question 73.
Give one area of application of (i) nuclear fission (ii) nuclear fusion.
Answer:
(i) Nuclear fission is used in (1) a nuclear reactor as very efficient and the least-polluting source of energy to generate electricity (2) atomic bombs.

(ii) Nuclear fusion is used in (experimental) fusion reactors to generate electricity without the hazards of radioactive radiations and radioactive pollution which happens with fission reactors. Nuclear fusion reactions in the interior of stars are the source of their energy output and the means of synthesis of higher elements like carbon, nitrogen and silicon from hydrogen and helium.

Question 74.
Explain the basic exothermic reaction in stars.
Answer:
The fusion of hydrogen nuclei into helium nuclei results in release of energy. It is the basic exothermic reaction in stars.
(1) The proton-proton cycle :


The total energy released is 24.7 MeV.

Question 75.
What is a nuclear reactor?
Answer:
A nuclear reactor is a device in which a nuclear fission chain reaction is used in a controlled manner (i) to produce energy in form of heat which is then converted into electricity or (ii) to produce radioisotopes or (iii) to produce new nuclides using a suitable fissionable material such as uranium or plutonium.
In a uranium reactor, \(\begin{gathered}
235 \\
92
\end{gathered}\)U is bombarded by slow neutrons to produce \(\begin{gathered}
235 \\
92
\end{gathered}\)U which undergoes fission.

Question 76.
What is a chain reaction? How is it produced?
Answer:
A chain reaction is a self-multiplying process in which neutrons ejected in a nuclear fission strike neighbouring nuclei of fissionable material and cause more fissions.
A fission of \(\begin{gathered}
235 \\
92
\end{gathered}\)U nucleus by a thermal neutron leads to ejection of two or three neutrons (2.7 neutrons on an average) having high kinetic energy of about 2 MeV. The kinetic energy of at least one of these neutrons is lowered to about 0.025 eV by a suitable moderator and the neutron is used to cause further fission. The process continues and hence it is called a chain reaction.

Multiple Choice Questions

Question 1.
The linear momentum of an electron in a Bohr orbit of an H-atom (principal quantum number n) is proportional to
(A) \(\frac{1}{n^{2}}\)
(B) \(\frac{1}{n}\)
(C) n
(D) n².
Answer:
(B) \(\frac{1}{n}\)

Question 2.
The angular momenta of the electron in successive Bohr orbits differ by
(A) \(\frac{n h}{2 \pi}\)
(B) h
(C) \(\frac{h}{2 \pi}\)
(D) (n – 1)\(\frac{h}{2 \pi}\)
Answer:
(C) \(\frac{h}{2 \pi}\)

Question 3.
The angular momentum of the electron in the second Bohr orbit of hydrogen atom is l. Its angular moementum in the third Bohr orbit is
(A) \(\frac{2}{3}\) l
(B) \(\frac{3}{2}\) l
(C) 3l
(D) \(\frac{4}{3}\) l.
Answer:
(B) \(\frac{3}{2}\) l

Question 4.
The time taken by an electron moving with a speed of 2.18 × 10⁶ m/s to complete one revolution in the first orbit (radius 0.53 A) of hydrogen atom is
(A) 1.527 × 10^-15 s
(B) 1.527 × 10^-16 s
(C) 1.527 × 10^-17 s
(D) 1.527 × 10^-18 s.
Answer:
(B) 1.527 × 10^-16 s

Question 5.
If the electron in a hydrogen atom is raised to one of its excited energy states, the electron’s
(A) potential energy increases and kinetic energy decreases
(B) potential energy decreases and kinetic energy increases
(C) potential energy increases with no change in kinetic energy
(D) potential energy decreases but kinetic energy remains constant.
Answer:
(A) potential energy increases and kinetic energy decreases

Question 6.
The potential energy of the electron in a hydrogen atom in its ground state is
(A) – 6.8 eV
(B) – 13.6 eV
(C) – 27.2 eV
(D) 13.6 eV.
Answer:
(C) – 27.2 eV

Question 7.
The kinetic energy of the orbital electron in a hydrogen atom in the excited state corresponding to n = 2 is
(A) 3.4 eV
(B) 6.8 eV
(C) 13.6 eV
(D) 27.2 eV.
Answer:
(A) 3.4 eV

Question 8.
The ratio of the kinetic energy of an electron in a Bohr orbit to its total energy in the same orbit is
(A) -1
(B) 2
(C) \(\frac{1}{2}\)
(D) -0.5.
Answer:
(A) -1

Question 9.
The energy of an electron in the nth Bohr orbit is proportional to
(A) n²
(B) n
(C) \(\frac{1}{n}\)
(D) \(\frac{1}{n^{2}}\)
Answer:
(D) \(\frac{1}{n^{2}}\)

Question 10.
The radius of the first Bohr orbit is 0.53 Å and that of nth orbit is 212 Å. The value of n is
(A) 2
(B) 12
(C) 20
(D) 400.
Answer:
(C) 20

Question 11.
The radius and the energy of the first Bohr orbit in a hydrogen atom are r₁ and E₁. If the orbital electron makes a transition to a orbit of radius 4rv the energy of the electron changes to
(A) \(\frac{E_{1}}{4}\)
(B) \(\frac{E_{1}}{2}\)
(C) 2E₁
(D) 4E₁
Answer:
(A) \(\frac{E_{1}}{4}\)

Question 12.
A hydrogen atom in its ground state is excited to the state of energy E₃ by an electron colliding with it. The minimum energy that the colliding electron must have is
(A) 10.2 eV
(B) 12.09 eV
(C) 12.5 eV
(D) 13.6 eV.
Answer:
(B) 12.09 eV

Question 13.
The energy of the electron in a hydrogen atom is raised from a state of energy E2 to that of energy E4. In the process, its
(A) energy doubles
(B) angular momentum doubles
(C) velocity doubles
(D) linear momentum doubles.
Answer:
(B) angular momentum doubles

Question 14.
Given that R is the Rydberg constant for hydrogen, the H_α line in the hydrogen spectrum has a wavelength
(A) \(\frac{1}{6 R}\)
(B) 6R
(C) \(\frac{5 R}{36}\)
(D) \(\frac{36}{5 R}\)
Answer:
(D) \(\frac{36}{5 R}\)

Question 15.
When the electron in a hydrogen atom jumps from the second orbit to the first orbit, the wavelength of the emitted rediation is λ. When the electron jumps from the third orbit to the first orbit, the wavelength of the emitted radiation would be
(A) \(\frac{27}{32}\) λ
(B) \(\frac{32}{27}\) λ
(C) \(\frac{2}{3}\) λ
(D) \(\frac{3}{2}\) λ.
Answer:
(A) \(\frac{27}{32}\) λ

Question 16.
In hydrogen atom, the Balmer series is obtained when the electron jumps from
(A) a higher orbit to the first orbit
(B) the first orbit to a higher orbit
(C) a higher orbit to the second orbit
(D) the second orbit to a higher orbit.
Answer:
(C) a higher orbit to the second orbit

Question 17.
The nuclei having the same number of protons but different number of neutrons are called
(A) isobars
(B) α-particles
(C) isotopes
(D) γ-particles.
Answer:
(C) isotopes

Question 18.
The nuclear volume of \(\begin{aligned}
&8 \\
&4
\end{aligned}\) Be is ………. that of \(\begin{aligned}
&1 \\
&1
\end{aligned}\) H.
(A) equal to
(B) two times
(C) four times
(D) eight times.
Answer:
(D) eight times.

Question 19.
The nuclear radius of the tungsten nuclide ₇₄W is twice that of the sodium nuclide \(\begin{aligned}
&23 \\
&11
\end{aligned}\) Na. The neutron number of the tungsten nuclide is
(A) 82
(B) 100
(C) 110
(D) 184.
Answer:
(C) 110

Question 20.
When a β-particle is emitted by a nucleus, its mass number
(A) decreases
(B) remains the same
(C) increases
(D) may decrease or increase.
Answer:
(B) remains the same

Question 21.
When an α-particle is emitted by a nucleus, its mass number
(A) increases by 4
(B) decreases by 4
(C) increases by 2
(D) decreases by 2.
Answer:
(B) decreases by 4

Question 22.
When a γ-ray photon is emitted by an unstable nucleus,
(A) Z increases
(B) Z decreases
(C) A increases
(D) Z and A remain the same.
Answer:
(D) Z and A remain the same.

Question 23.
In a radioactive transformation, a change in the mass number occurs with
(A) α- or β-decay
(B) β-decay
(C) γ-decay
(D) α-decay.
Answer:
(D) α-decay.

Question 24.
The half-life of radium is 1600 y. How much of 1 μg of radium will remain undistintegrated after 8000 y?
(A) \(\frac{1}{8}\) μg
(B) \(\frac{1}{16}\) μg
(C) \(\frac{1}{32}\) μg
(D) \(\frac{1}{64}\) μg.
Answer:
(C) \(\frac{1}{32}\) μg

Question 25.
In one mean lifetime of a radioactive element, the fraction of the nuclei that has disintegrated is [e is the base of natral logarithm.]
(A) \(\frac{1}{e}\)
(B) 1 – \(\frac{1}{e}\)
(C) e
(D) e – 1.
Answer:
(B) 1 – \(\frac{1}{e}\)

Question 26.
The decay constant of a radioactive element is λ After a time 2λ^-1 of the original number of radioactive nuclei about ……………….. remains undecayed.
(A) 37%
(B) 27%
(C) 25%
(D) 13.7%
Answer:
(D) 13.7%

Question 27.
Complete the following fission reaction :

Answer:
(D) \({ }_{35}^{85} \mathrm{Br}+3{ }_{0}^{1} \mathrm{n}\)

Question 28.
The energy generated in the stars is because of
(A) radioactivity
(B) nuclear fission
(C) nuclear fusion
(D) photoelectric phenomenon.
Answer:
(C) nuclear fusion

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 16 Semiconductor Devices Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 16 Semiconductor Devices

Question 1.
What is a PN-junction diode? What is a depletion region? What is barrier potential in a PN-junction?
Answer:
PN-junction diode: A two-terminal semiconductor device consisting of a PN-junction is called a PN- junction diode.
Depletion region: The neighbourhood of the junction between a p-type layer and an n-type layer within a single semiconducting crystal is depleted of free (or mobile) charge carriers. This is called the depletion region or depletion layer.

Barrier potential: The electric potential difference across the PN-junction is called the potential barrier or barrier potential.

[Note: Under the open-circuit condition (no applied potential difference), the width of the depletion region and the height of the potential barrier have their equilibrium values. The width of the depletion region in an unbiased pn-junction diode ranges from 0.5 pm to 1 pm and depends on the dopant concentrations. The barrier potential is about 0.3 V for Ge junction diode and about 0.7 V for Si junction diode.]

Question 2.
Explain the forward bias and reverse bias conditions of a diode.
Answer:
Forward-biased state : When the positive terminal of a cell is connected to the p side of the junction and the negative terminal to the n side, the diode is said to be forward biased. When forward biased, the depletion region narrows and, consequently, the potential barrier is lowered. This causes the majority charge carriers of each region to cross into the other region. This way the diode conducts when forward biased; the total current across the junction is called the forward current and is due to both electron and hole currents. Because of the narrowing of the depletion region, a forward-biased junction diode has a very low resistance and acts as a closed switch.

Reverse-biased state : A pn-junction diode is said to be reverse biased when the positive terminal of a cell or battery is connected to the n side of the junction, and the negative terminal to the p side. When reverse biased, the depletion region widens and the potential barrier is increased, the majority charge carrier concentration in each region decreases against the equilibrium values and the reverse-biased junction diode has a high resistance. The diffusion current across the junction becomes zero. Thus, the diode does not conduct when reverse biased and is said to be in a quiescent or non-conducting state, i.e., it acts as an open switch (almost).

Question 3.
What is rectification? What is a rectifier? How does a pn-junction diode act as a rectifier?
Answer:

1. The process of converting an alternating voltage (or current) to a direct voltage (or current) is called rectification.
   A circuit or device that is used to convert an alternating voltage (or current) to a direct voltage (or current) is called a rectifier. A rectifier produces a unidirectional but pulsating voltage from an alternating voltage.
2. When an alternating voltage is applied across a pn-junction diode, the diode is forward-biased and reverse-biased during alternate half cycles.
3. During the half cycle when the diode is forward- – biased, it conducts. Therefore, there is a current through it from the p-region to the n-region.
4. During the next half cycle, it is reverse-biased and does not conduct. Therefore, current passes only in one direction through the circuit. This way, a pn-junction diode acts as a rectifier.

Question 4.
Explain the need for rectification/rectifiers.
Answer:
Nowadays electrical energy is generated, transmitted and distributed in the form of alternating voltage because it is simpler and more economical than direct current transmission and distribution. Another important reason for the widespread use of alternating voltage in preference to direct voltage is the fact that alternating voltage can be conveniently changed in magnitude by means of a transformer.

However, most electrical and electronic systems need a dc voltage to work. Since the transmitted voltage is very high and alternating, we need to reduce the ac line voltage and then convert it to a relatively constant dc output voltage. The power-line voltage is sequentially stepped down at the distribution substations. At the consumer end, the ac voltage is rectified using junction diodes to dc voltage.

Question 5.
Draw a neat block diagram of a dc power supply and state the function of each part.
OR
With the help of a block diagram, explain the scheme of a power supply for obtaining dc output voltage from ac line voltage.
Answer:
A consumer electronic system called a dc power supply produces a fairly constant dc voltage from ac supply voltage. Below figure shows a functional block diagram of the circuits within a power supply.

Block diagram of dc power supply with waveforms at each stage

The ac supply voltage is usually stepped-down by a transformer and its secondary voltage is converted to a pulsating dc by a diode rectifier. By the superposition theorem, this rectifier output can be looked upon as having two different components : a dc voltage (the average value) and an ac voltage (the fluctuating part). The filter circuit smooths out the pulsating dc. It blocks almost all of the ac component and almost all of the dc component is passed on to the load resistor. Figure shows the filtered output for a rectified full-wave dc. The only deviation from a perfect dc voltage is the small ac load voltage called ripple. A well-designed filter circuit minimizes the ripple. In this way, we get an almost perfect dc voltage, one that is almost constant, like the voltage out of a battery.

The regulation of a power supply is its ability to hold the output steady under conditions of changing input or changing load. As power supplies are loaded, the output voltage tends to drop to a lower value. Nowadays, an integrated circuit (I_C) voltage regulator is connected between a filter and the load resistor, especially in low-voltage power supplies. This device not only reduces the ripple, it also holds the output voltage constant under varying load and ac input voltage.

Question 6.
Distinguish between a half-wave rectifier and full-wave rectifier.
Answer:

Half-wave rectifier	Full-wave rectifier
1. A device or a circuit which rectifies only one-half of each cycle of an alternat­ing voltage is called a half-wave rectifier.	1. A device or a circuit which rectifies both halves of each cycle of an alternat­ing voltage is called a full-wave rectifier.
2. A half-wave rectifier cir­cuit uses a single diode which conducts for only one-half of each cycle.	2. A full-wave rectifier cir­cuit uses at least two diodes which conduct alternately for consecu­tive halves of each cycle.

Question 7.
State any two advantages of a full-wave rectifier.
Answer:

1. A full-wave rectifier rectifies both halves of each cycle of the ac input.
2. Efficiency of a full-wave rectifier is twice that of a half-wave rectifier.
3. The ripple in a full-wave rectifier is less than that in a half-wave rectifier.
   Ripple factors for a full-wave and half-wave rectifiers are respectively, 0.482 and 1.21.

Question 8.
Explain ripple in the output of a rectifier. What is ripple factor?
Answer:
The output of a rectifier is a pulsating dc. By the superposition theorem, this rectifier output can be looked upon as having two different components : a dc and an ac. The direct current is the average value of the pulsating current, averaged over each half cycle of the ac input. The ac component in the output is called the ripple. Ripple is undesirable in most electronic circuits and devices.

The ratio of the root-mean-square value of the ac component to the average value of the dc component in the filtered rectifier output is known as the ripple factor.
Ripple factor = \(\frac{(r m s \text { value of ac component })}{\text { (average value of dc component) }}\)
Percentage ripple = ripple factor × 100%
This factor mainly decides the effectiveness of a filter circuit in a power supply, i.e., smaller the value of this factor, lesser is the ac component in comparison to the dc component. Hence, more effective is the filter.

Question 10.
Explain the action of a capacitive filter with necessary diagrams.
Answer:
Consider a simple capacitive filter added to a full-wave rectifier circuit, Fig. 16.6(a). A capacitor is a charge storage device that it can deliver later to a load.

(a) Capacitive filter at the output of a full-wave rectifier (b) Waveform of filtered output
In the first quarter cycle, the capacitor charges as the rectifier output peaks. Later, as the rectifier output drops off during the second quarter cycle, the capacitor discharges and delivers the load current. The voltage across the capacitor, and the load, decreases up to a point B when the next voltage peak recharges the capacitor again. To be effective, a filter capacitor should be only slightly discharged between peaks. This will mean a small voltage change across the load and, thus, small ripple. As shown in Fig. (b), the capacitor supplies all the load current from A to B; from B to C, the rectifier supplies the current to the load and the capacitor.

The discharging time constant of a filter capacitor has to be long as compared to the time between the voltage peaks. For the same capacitor used with a half-wave rectifier, the capacitor will have twice the time to discharge, and the ripple will be greater. Thus, full-wave rectifiers are used when a low ripple factor is desired.

Question 11.
What is regulation in a dc power supply ? OR Explain unregulated power supply and regulated power supply.
Answer:
Voltage regulation is am important factor of a power supply. Regulation is its ability to hold the dc output steady under conditions of changing ac input or changing load. The output voltage under no-load condition (no current drawn from the supply) tends to drop to a lower value when load current is drawn from the supply (under load). The amount the dc voltage changes between the no-load and full-load conditions is described by a factor called voltage regulation.
Voltage regulation = \(\frac{\text { no-load voltage }-\text { full-load voltage }}{\text { full-load voltage }}\)

Question 12.
What is a regulated power supply?
Answer:
A dc power supply whose preset output voltage remains constant irrespective of variations in the line voltage or load current is called a regulated power supply.

Question 13.
What is a unregulated power supply ?
Answer:
A dc power supply whose output changes when a load is connected across it is called unregulated power supply.

Question 14.
Name any four common special-purpose diodes.
Answer:
Special-purpose diodes :

1. Zener diode
2. light emitting diode (LED)
3. photodiode
4. solar cell.

Question 15.
What is meant by breakdown of a pn-junction? Name two important mechanisms of junction breakdown.
Answer:
In a reverse-biased pn-junction, the depletion region is wider and the potential barrier is higher over equilibrium values. The majority charge carrier concentration in each region decreases from the equilibrium values and the diffusion current across the junction is zero. Only a very very small current flows due to the motion of minority charge carriers. Thus, the principal characteristic of a pn-junction diode is that it rectifies, i.e., it conducts significantly in one direction only.

When a sufficiently large reverse voltage is applied to a pn-junction, there is an abrupt strong increase in the reverse current and its rectifying properties are lost. This is known as junction breakdown. The absolute value V_B of the voltage at which the phenomenon occurs is called breakdown voltage. The breakdown process is not inherently destructive and is reversible.

Two important breakdown mechanisms are the Zener breakdown (due to tunneling effect) and avalanche breakdown (due to avalanche multipli-cation).

Question 16.
Explain Zener breakdown.
Answer:
In a reverse-biased pn-junction, the depletion region is wider and the potential barrier is higher over equilibrium values. The electric field in the depletion region is from the n- to the p-region. When a sufficiently large reverse voltage is applied to a pn-junction, the junction breaks down and conducts a very large current. Of the two important breakdown mechanisms, Zener breakdown takes place in heavily doped diodes.

Usually, the energy that an electron can gain from even a strong field is very small. However, the depletion region is very narrow in a heavily doped diode. Because of this, the electric field across the depletion region is intense enough to break the covalent bonds between neighbouring silicon atoms and pull electrons out of their orbits. This results in conduction electrons and holes. In the energy band diagram representation, this corresponds to the transition of an electron from the valence band to the conduction band and become available for conduction.

The current increases with increase in applied voltage, but without further increase in voltage across the diode. This process, in which an electron of energy less than the barrier height penetrates through the energy bandgap, is called tunneling (a quantum mechanical effect). The creation of electrons in the conduction band and holes in the valence band by tunneling effect in a reverse- biased pn-junction diode is called the Zener effect.

[Notes : (1) Tunneling occurs only if the electric field is very high. The typical field for silicon and gallium arsenide is > 10⁶ V / cm. To achieve such a high field, the doping concentrations for both p- and w-regions must be quite high (>10¹⁸ cm^-3). (2) Zener breakdown or Zener effect is named in honour of Clarence M. Zener (1905-93), US physicist, who explained the breakdown mechanism. (3) Avalanche breakdown occurs in diodes with a doping concentration of ≅ 10¹⁷ cm^-3 or less. The carriers gain enough kinetic energy to generate electron-hole pairs by the avalanche process when the value of reverse | V | becomes large. An electron in the conduction band can gain kinetic energy before it collides with a valence electron. The high-energy electron in the conduction band can transfer some of its kinetic energy to the valence electron to make an upward transition to the conduction band. An electron-hole pair is generated. All such electrons and holes accelerate in the high field of the depletion region and, in turn, generate other electron-hole pairs in a like manner. This process is called the avalanche process.]

Question 17. What is a Zener diode?
Answer:
A Zener diode is a heavily doped pn-junction diode operated in its breakdown region. Zener breakdown occurs when the breakdown voltage is less than about 6 V while avalanche breakdown occurs in lightly doped diodes and for breakdown voltage greater than 6 V. However, the Zener effect was discovered before the avalanche effect, so all diodes used in the breakdown region came to be known as Zener diodes.

Question 18.
Explain the use of a resistor in series with a Zener diode.
Answer:
The I-V characteristics in the breakdown region of a Zener diode is almost vertical. That is, the current I_Z can rapidly increase at constant V_Z. To prevent damage due to excessive heating, the Zener current should not exceed the rated maximum current, I_ZM. Hence, a current-limiting resistor Rs is connected in series with the diode.

I_Z and the power dissipated in the Zener diode will be large for I_L = 0 (no-load condition) or when I_L is less than the rated maximum (when Rs is small and RL is large). The current-limiting resistor R_s is so chosen that the Zener current does not exceed the rated maximum reverse current, I_ZM when there is no load or when the load is very high.

The rated maximum power of a Zener diode is
P_ZM = I_ZM = V_Z
At no-load condition, the current through R_s is Z = I_ZM and the voltage drop across it is V – V_Z, where V is the unregulated source voltage. The diode current will be maximum when V is maxi-mum at V_max and I = I_ZM. Then, the minimum value of the series resistance should be
R_{s, min} = \(\frac{V_{\max }-V_{Z}}{I_{Z M}}\)

A Zener diode is operated in the breakdown region. There is a minimum Zener current, I_Z(min), that places the desired operating point in the breakdown region. There is a maximum Zener current, I_ZM, at which the power dissipation drives the junction temperature to the maximum allowed. Beyond that current the diode can be damaged. Hence, the supply voltage must be greater than Vz and the current-limiting resistor must limit the diode current to less than the rated maximum, I_ZM.

Question 19.
State any two applications of a Zener diode.
Answer:
Applications of a Zener diode :

1. Voltage regulator
2. Fixed reference voltage in biasing transistors
3. Peak clipper in a wave shaping circuit
4. Meter protection from voltage fluctuations.

Question 20.
Solve the following :
(1) A Zener diode has a Zener voltage of 2.4 V and a 500 mW power rating. What should be the maxi-mum current through the diode if you design conservatively with a safety factor of 2?
Solution:
Data : V_z = 2.4 V, P_ZM = 500 mW A conservative design includes a safety factor to keep the power dissipation well below the rated maximum power. Thus, with a safety factor of 2, the operating power of the Zener diode is

(2) A 10 V regulated power supply uses a Zener diode of 500 mW power rating with an input voltage of 15 V dc and a current limiting resistor of 500 Ω. If a load of 1 kΩ is connected across the diode, is the diode in the breakdown region?
Solution:
Data: V = 15V, V_Z = 10V, R_s = 500 Ω, R_L =1000 Ω
The supply current,

∴ The diode current, I_Z = I – I_L = 0
Thus, the Zener diode is at the threshold of break down and there will not be any regulation.

(3) A 10 V regulated power supply is designed using a Zener diode of 500 mW power rating with an input voltage of 15 V dc. A load of 1 kΩ is to be connected across the diode. Calculate (a) the rated maximum current through the diode (b) the mini mum value of the series resistance.
Solution:
Data: V = 15V, V_Z = 10 V, P_ZM = 500 mW,
R_L = 1000 Ω, R_s = 200 Ω
The rated maximum Zener current,

(4) In the above problem, calculate the load current. If a series resistance of 200 Ω is used, what is the Zener current ?
Solution:
Data : V = 15 V, V_Z = 10 V, P_ZM = 500 mW, R_L = 1000 Ω, R_s = 200 Ω
The current through R_s is

∴ The zener current,
I_Z = I_s – I_L = 25 – 10 = 15 mA

(5) A Zener regulator has an input voltage that may vary from 15 V to 20 V while the load current may vary from 5 mA to 20 mA. If the Zener voltage is 12 V, calculate the maximum series resistance.
Solution:
Data : V_low = 15 V, V_high = 20 V, V_Z = 12 V,
I_{L, min} = 5mA, I_{L, max} = 20mA

Question 21.
What is a photodiode?
Answer:
A photodiode is a special purpose reverse-biased pn-junction diode that generates charge carriers in response to photons and high energy particles, and passes a photocurrent in the external circuit proportional to the intensity of the incident radiation. The term photodiode usually means a sensor that accurately detects changes in light level. Hence, it is sometimes called a photodetector or photosensor which operates as a photoelectric converter.

Question 22.
Explain the I-V characteristics of a photodiode.
Answer:
When a Si photodiode is operated in the dark (zero illumination), the current versus voltage characteristics observed are similar to the curve of a rectifier diode as shown by curve (1) in figure. This dark current in Si photodiodes range from 5 pA to 10 nA.

When light is incident on the photodiode, the curve shifts to (2) and increasing the incident illuminance (light level) shifts this characteristic curve still further to (3) in parallel. The magnitude of the reverse voltage has nearly no influence on the photocurrent and only a weak influence on the dark current. The normal reverse currents are in tens to hundreds of microampere range.

The I-V characteristics of a photodiode showing dark current and photocurrent for increasing illuminance

The almost equal spacing between the curves for the same increment in luminous flux reveals that the reverse current and luminous flux are almost linearly related. The photocurrent of the Si photo-diode is extremely linear with respect to the il-luminance. Since the total reverse current is the sum of the photocurrent and the dark current, the sensitivity of a photodiode is increased by minimizing the dark current.

Question 23.
Explain saturation current of a photodiode with a neat labelled diagram.
Answer:
When a reverse-biased photodiode is illuminated, the reverse current at a constant reverse voltage is directly proportional to the illuminance. The de-pendence of the photocurrent on the illuminance is very linear over six or more orders of illuminance, e.g., in the range from a few nanowatts to tens of milliwatts with an active area of a few mm2. But after a certain value of reverse current, the current does not increase further with increasing light intensity. This constant value is called the satura-tion current of the photodiode.

Question 24.
What is (i) dark current (ii) dark resistance of a photodiode?
Answer:
(i) Dark current: The current associated with a photo-diode with an applied reverse bias during operation in the dark (zero illumination) due to background radiation and thermally excited minority saturation current. It is of the order of picoamperes to nanoamperes. Larger active areas or increase in temperature and reverse bias result in higher dark current.

(ii) Dark resistance : The ratio of maximum withstandable reverse voltage to the dark current of a photo-diode is called dark resistance of that diode.
Dark resistance, R_d = \(\frac{\text { (maximum reverse voltage) }}{(\text { dark current })}\)

Question 25.
State any four advantages of a photodiode.
Answer:
Advantages of a photodiode :

1. Quick response to light.
2. High operational speed.
3. Excellent linear response over a wide dynamic range.
4. Low cost.
5. Wide spectral response.
6. Compact, lightweight, mechanically rugged and long life.

Question 26.
State any two disadvantages of a photodiode.
Answer:
Disadvantages of a photodiode :

1. Poor temperature stability : Reverse current is temperature dependent.
2. Reverse current for low illumination is small and requires amplification.

Question 27.
State any two applications of photodiodes.
Answer:
Applications of photodiodes :

1. A reverse-biased photodiode conducts only when illuminated, assuming that the dark current is essentially zero. Due to its quick response to radi-ation and high operational speed, photodiodes are used in high-speed counting or switching applications.
2. Extensively used in an fibreoptic communication system.
3. As photosensors/photodetectors for detection of UV radiations and accurate measurement of illumination. Avalanche photodiodes have increased responsivity and can be used as photomultipliers, especially for low illumination.
4. In burglar alarm systems as normally closed switch until exposure to radiation is interrupted. When interrupted, the reverse current drops to the dark current level and sounds the alarm.
5. In an optocoupler, a photodiode is combined with a light-emitting diode to couple an input signal to the output circuit. The key advantage of an optocoupler is the electrical isolation between the input and output circuits, especially in high voltage applications. With an optocoupler, the only contact between the input and the output is a beam of light.

Question 28.
Name two types of solar energy devices.
Answer:
Two major types of devices converting solar energy in usable form are

1. photothermal devices, which convert the solar energy into heat energy
2. photovoltaic devices, which convert solar energy into electrical energy.

Question 29.
What is a solar cell ? State the principle of its working.
Answer:
A solar cell is an unbiased pn-junction that converts the energy of sunlight directly into electricity with a high conversion efficiency.

Principle : A solar cell works on the photovoltaic effect in which an emf is produced between the two layers of a pn-junction as a result of irradiation.

Question 30.
Describe the output characteristics of a solar cell with a neat labelled graph.
Answer:
Output characteristic of a solar cell : The output characteristic, I-V curve, of a solar cell exposed to sunlight is plotted by varying the load resistance (Rr) from zero to infinity and measuring the corresponding current and voltage.

The short-circuit condition occurs when R_L = 0 so that V = 0. The current in this case is referred to as the short-circuit current I_SC. The open-circuit condition occurs when R_L = ∞. The net current in this case is zero and the voltage produced is called as the open-circuit voltage V_OC. Increasing the lightsensing area or light intensity per single solar cell produces a proportionate increase in I_SC. V_OC is independent of these parameters.

The operating point of a solar cell module is the point (V_p, I_p) on the I-V curve which delivers maximum power to the load. For this, V_p/ I_p must be equal to R_L.
[Note : The photocurrent in a solar cell is always in the reverse-bias direction so that the I-V graph is in the fourth quadrant.]

Question 31.
State the material selection criteria for solar cells.
Answer:
Criteria for materials to be used in solar cells :

1. Band gap energy must be between 1 eV and 1.8 eV. (The best band gap of a solar cell is in the region of 1.5 eV.)
2. It must have high optical absorption.
3. It must have high electrical conductivity.
4. The raw material must be available in abundance and the cost of the material must be low.

Question 32.
Name the common materials for solar cells.
Answer:
Optimized band gap for solar cells is close to 1.5 eV. Some of the common materials for solar cells are

1. silicon (Si), E_G = 1.12 eV – currently the most popular material but has low absorption coefficient and high temperature dependence,
2. gallium arsenide (GaAs), E_G = 1.42 eV -by far the most widely used, especially for high end applications like satellites. Its absorption coefficient is about ten times better than silicon and doesn’t have the same temperature dependence.
3. copper-indium diselenide (CIS), E_G = 1.01 eV – has the highest optical absorption, but gallium is introduced in the lattice to raise the band gap energy closer to the solar ideal. This resulted in the popular copper-indium-gallium diselenide (Culn- GaSe₂ or CIGS) material for photovoltaic cell. By variation of Ga fraction, a band gap of around 1.48 eV has been achieved.
4. cadmium telluride (CdTe), E_G = 1.44 eV-made from the II-VI group elements.

[Notes : (1) By far the most widely used III-V solar cell is gallium arsenide (GaAs). Other III-V semiconductors-indium phosphide (InP), gallium antimonide (GaSb), aluminium gallium arsenide (AlGaAs), indium gallium phosphide (InGaP), and indium gallium arsenide (InGaAs)-exchange group III elements to make different band gap energies. III-V semiconductors offer a great host of advantages over silicon as a material for photovoltaics. However, the biggest drawback is cost. (2) The theoretical limit on the thermodynamic efficiency of single-junction solar cells is ~ 30%. Hence, today’s most efficient technology for the generation of electricity from solar radiation is the use of multi-junction solar cells made of III-V compound semiconductors. Efficiencies up to 39% have al-ready been reported under concentrated sunlight. These solar cells have initially been developed for powering satellites in space and are now starting to explore the terrestrial energy market through the use of photovoltaic concentrator systems. A triple-junction solar cell, Ga_0.35In_0.65P/Ga_0.83In_0.17As/Ge, has been demonstrated by a conversion efficiency of 41.1% at 454 kW/m²]

Question 33.
State any four uses of solar cells.
Answer:
Uses of solar cells :

1. A solar cell array consisting of a set of solar cells is used during daylight hours to power an electrical equipment as well as to recharge batteries which can then be used during night.
2. Solar cell arrays provide electrical power to equipment on a satellite as well as at remote places on the Earth where electric power lines are absent.
3. Large-scale solar power generation systems linked with commercial power grid.
4. Independent power supply systems for radar detectors, monitoring systems, radio relay stations, roadlights and roadsigns.
5. Indoor uses include consumer products like, calculators, clocks, digital thermometers, etc. (They use very low levels of power and work under lowbrightness long-wavelength light from incandescent lamps, etc.)

Question 34.
What is a light-emitting diode (LED) ?
Answer:
A light-emitting diode (LED) is a forward-biased pn-junction diode formed from compound semiconductor materials such as gallium arsenide (GaAs) in which light emission can take place from direct radiative recombination of excess electron-hole pairs. A photon is emitted when an electron in the conduction band recombines with a hole in the valence band.

In infrared emitting LEDs, the encapsulating plastic lens may be impregnated or coated with phosphorus. Then, phosphorescence of the phos-phorus gives off visible light.

[Note: In an ordinary pn-junction diode, energy released in electron-hole recombination process is absorbed in the crystal structure as heat.]

Question 35.
Describe with a neat diagram the construction of an LED.
Answer:
Construction : A light-emitting diode is a forward- biased pn-junction diode formed from compound semiconductor materials. As shown in Fig. 16.14(a), the top metal contact to the n-layer (say) is provided with a window for the emitted light to escape. The diode chip is encapsulated in a transparent plastic lens. The cathode and anode leads from the metal contacts to the n-and p-layers, respectively, are provided for external connections, shown in figure.
The negative electrode (cathode) is identified by a notch or flat spot on the plastic body, or the cathode lead is shorter than the anode.

Question 36.
Draw the I-V forward characteristics of an LED and explain it.
Answer:
Explanation : The forward characteristic of an LED is similar to an ordinary junction diode. The diode starts conducting only after the forward-bias voltage overcomes the barrier potential. Thereafter the current increases exponentially beyond the knee region.

The threshold voltage is about 1.2 V for a standard red LED to about 3.6 V for a blue LED. However, these values depend on the manufacturer because of the different dopant concentrations used for the different wavelength ranges.

The intensity of the emitted light is directly propotional to the forward current. An LED is operated with a typical forward current of 20 mA- about 5 mA for a simple LED indicator to about 30 mA where a high intensity of light is needed. The LED forward current must be limited to a specified safe value using a series resistance.

The peak inverse voltage (PIV) or breakdown voltage of an LED is low, typically 5 V.

Question 37.
State any four advantages of an LED over common light sources.
Answer:
Advantages of an LED as a light source over common light sources :

1. Energy efficiency : An LED is small in size, requires low operating voltage and power and is extremely energy efficient, consuming up to 90% less power than incandescent bulbs. LED’s are now capable of outputting 135 lumens/watt.
2. Life and ruggedness : Being a solid state device, an LED is more rugged than bulbs with filament and has a typical life of 50000 hours or more.
3. Fast switching : An LED is very fast, i.e., its switching (on / off) time is less than 1 ns.
4. Brightness and colour control: The intensity of the emitted light can be varied continuously. The colour of the emitted light also can be controlled.
5. Small size : Because of their small size, they can be used to produce self-luminous, static or running, seven-segment alphanumeric displays.
6. Environmentally friendly : An LED being a semi-conductor device, does not contain hazardous substances like mercury (as in sodium and mercury vapour lamps).
7. Operationally cheap : Since LEDs use only a fraction of the energy of an incandescent light bulb there is a dramatic decrease in power costs.

Question 38.
State any four disadvantages of an LED.
Answer:
Disadvantages of an LED light source :
(1) Blue light hazard : There is a photobiological con-cern that bright blue LEDs and cool-white LEDs are capable of exceeding safe limits.

(2) Light quality : Most cool-white LEDs have spectral output significantly different from the Sun or an incandescent bulb-peak output being at 460 nm rather than peak retinal sensitivity of 550 nm. This can cause the colour of objects to be perceived differently under cool-white LED illumination than sunlight or incandescent bulbs.

(3) Temperature dependence : An LED luminaire can overheat in high ambient temperatures and effective cooling methods using heat sinks are essential for high-power LEDs. (Although the thermal power involved is not very large, it is released within a very small volume and area.) This is especially important for automotive, medical and military applications where the light unit must operate over a large range of temperatures and yet have a low failure rate.

(4) High initial cost: LEDs are currently more expen-sive-price per lumen-in initial capital cost, than most conventional lighting technologies.

(5) Voltage sensitivity : LEDs must be operated with the voltage above the threshold and the current below the rated maximum. This requires current- limiting resistors or current-regulated power supplies.

(6) Blue light pollution : Cool-white LEDs emit pro-portionally more blue light than conventional out-door light sources such as high-pressure sodium lamps. Due to Rayleigh scattering, these LEDs can cause more light pollution than other light sources.

Question 39.
State any four applications of LEDs.
Answer:
Applications of LEDs :

1. An LED is commonly used as an On/Off indicator lamp on electrical equipment.
2. LEDs are used in self-luminous seven-segment alphanumeric displays of calculators and digital clocks and meters, signages, etc.
3. Because of their low power consumption, LEDs are now commonly used in traffic signals, handheld torches, LED TV sets, domestic and decorative illumination and various indicator lamps in light motor vehicles and two-wheelers.
4. Under certain conditions, the essentially monochromatic light emitted by an LED is also coherent. Such diode lasers have found applications in optical fibre communications, CD players, CDROM drives, laser printers, bar code scanners, laser pointers, etc.

Question 40.
What is a junction transistor?
Answer:
A junction transistor consists of two back-to-back pn-junctions forming a sandwich structure in which a thin layer of n-type or p-type semiconductor is sandwiched between two layers of opposite type semiconductor.

The three terminals of a transistor connected to its three layers are known as the emitter (E), base (B) and collector (C). One pn-junction is between the emitter and the base while the other pn-junction is between the collector and the base.

The electric current is transported by both type of carriers, electrons and holes; for this reason the device is called a bipolar junction transistor (BJT).

There are two types of junction transistors : (i) pnp transistor (ii) npn transistor.

[Note : The point-contact transistor was invented in 1947 by US physicists John Bardeen (1908-91), Walter Brattain (1902-87) and William Shockley (1910-89). A month later Shockley invented the junction transistor.]

Question 41.
Draw the circuit symbols of (i) a pnp transistor (ii) an npn transistor.
Answer:

The arrow on the emitter shows the direction of current when the base-emitter junction is forward- biased. If the arrow points in (Points iN), it indicates the transistor is a pnp. On the other hand, if the arrow points out, the transistor is an npn (Not Pointing iN).

Question 42.
How is a junction transistor formed? Draw schematic diagrams showing the structure of the two types of BJTs.
Answer:
A bipolar junction transistor has three separately doped regions and two pn-junctions. A pnp transistor is formed by starting with a p-type substrate. An zz-type region is grown by thermally diffusing do-nor impurities into this substrate. A very heavily doped p+ region is then diffused into the n-type region. The heavily doped p + -region is called the emitter, symbol E in below figure.The narrow central n-region, with lightly doped concentration, is called the base (symbol B). The width of the base is small compared with the minority carrier diffusion length. The moderately doped p-region is called the collector (symbol C). The doping concentration in each region is assumed to be uniform.

The npn transistor is the complementary structure to the pnp transistor : A narrow p region grown into an n type substrate, by thermally diffusing acceptor impurities, forms the base. The heavily doped n + region diffused into the base forms the emitter.

Question 43.
Draw diagrams showing the two-diode analogues of npn and pnp transistors.
Answer:

[Important note : The diode equivalents are pretty much useless except for biasing a transistor or testing with a ‘digital multimeter (DMM). The transistor action of a BJT, as explained in Question 55 is substantially different from that of two independent back-to-back pn-junctions.]

Question 46.
Explain the working of an npn transistor with a neatly labelled circuit diagram.
OR
Explain the action of a junction transistor with a neatly labelled circuit diagram.
Answer:
For normal operation of a junction transistor, the emitter-base junction is always forward biased and the collector-base junction is always reverse biased. Below figure shows the biasing of the junctions for an npn transistor connected as an amplifier with the common-base configuration, that is, the base lead is common to the input and output circuits. The emitter-base junction is forward biased by the battery V_BB while the collector-base junction is reverse biased by the battery V_CC. V_BB should be greater than the emitter-base barrier potential (the threshold voltage). The arrows of the various currents indicate the direction of current under normal operating conditions (also called the active mode).

Since the emitter-base junction is forward biased, majority carriers electrons in the n⁺ emitter are injected into the base and holes (majority carriers in the p-type base) are injected from the base into the emitter. Under the ideal-diode condition, these two current components constitute the total emitter current IE.

The emitter is a very heavily doped n-type region. Hence, the current between emitter E and base B is almost entirely electron current from E into B across the forward-biased emitter junction.

The p-type base is narrow and the hole density in the base is very low. Therefore, virtually all the injected electrons (more than 95%) diffuse right across the base to the collector junction without recombining with holes. Since the collector junction is reverse biased, the electrons on reaching the collector junction are quickly swept by the strong electric field there into the n-type collector region, where they constitute the collector current I_C.

In practice, about 1% to 5% of the holes from the emitter recombine with holes in the base layer and cause a small current I_B in the base lead. Therefore,
I_E = I_B + I_C ≈ I_C
Therefore, carriers injected from a nearby emitter junction can result in a large current flow in a reverse-biased collector junction. This is the transistor action, and it can be realized only when the two junctions are physically close enough to interact as described.

If a pnp transistor is used, the battery connections must be reversed to give the correct bias. The conduction process is similar but takes place instead by migration of holes from emitter to collector. A few of these holes recombine with electrons in the base.

[Notes : (1) If, the two junctions are so far apart that all the injected electrons are recombined in the base before reaching the base-collector junction, then the transistor action is lost and the p-n-p structure becomes merely two diodes connected back to back. (2) Use of double-subscripted voltage notation in transistor circuits : same subscripts (viz., Vm and Vcc) represent the voltage of a biasing battery; different subscripts (viz, V_BB and V_CC) are used to indicate voltage between two points. Single subscripts (as in Fig. 16.19) are used for a node voltage, that is, the voltage between the subscripted point and ground.]

Question 47.
What are the different transistor configurations in a circuit? Show them schematically.
Answer:
There are three configurations in which a transistor may be connected in a circuit:
(a) Common-emitter (CE) : The emitter terminal is common to the input and output circuits.
(b) Common-base (CB) : The base terminal is common to the input and output circuits.
(c) Common-collector (CC) : The collector terminal is common to the input and output circuits.

[Notes : (1) CE: Produces the highest current and power gain of all the three bipolar transistor configurations the reason why it is the most commonly used configuration for transistor based amplifiers. The emitter is grounded and the input signal applied between the base and emitter. The input impedance is small due to the forward biasing of the EB-junction. The output taken from between the collector and emitter, as well as the outputimpedance, are large due to the reverse biased CB-junction. However, its voltage gain is much lower. – (2) CB : The base terminal is grounded, the input signal is applied between the base and emitter terminals while the output signal is taken from between the base and collector terminals. Though its high frequency response is good for single stage amplifier circuits, it is not very common due to its low current gain characteristics and low input impedance. (3) CC : Very useful for impedance matching applications because of the very large ratio of input impedance to output impedance. The collector is grounded and the input signal is directly given to the base. The output is taken across the load resistor in series with the emitter. Hence, the current through the load resistor is the emitter current and the current gain of the configuration is approximately equal to the β value of the transistor.]

Question 48.
State the relation between the dc common-base current ratio (α_dc) and the dc common-emitter current ratio (β_dc).
Answer:

Question 49.
Draw a neat labelled circuit diagram to study the characteristics of a transistor in common- emitter configuration.
Answer:

Question 50.
Explain with necessary graphs (1) input (base) (2) output (collector) characteristics of a transistor in common-emitter configuration.
Answer:
Common-emitter characteristics :
(1) Input or base characteristics : This is a set of curves of base current IB against base to emitter voltage VBE for different constant collector to emitter voltages V_CE shown in figure.

Keeping collector voltage V_CE constant, base voltage is gradually increased from zero. Initially, I_B is zero till V_BE is less than the threshold voltage for the forward-biased base-emitter junction, 0.7 V for silicon and 0.3 V for germanium. For V_BE greater than the threshold voltage, I_B increases steeply.

For a constant collector voltage V_CE, the dynamic input resistance (r_i) is defined as the ratio of the differential change in the base-to-emitter voltage (∆V_BE) to the corresponding change in the base current (∆I_CE).
r_i = (\(\frac{\Delta V_{\mathrm{BE}}}{\Delta I_{\mathrm{B}}}\))_VCE = constant

(2) Output or collector characteristics : This is a set of curves of collector current I_C against collector-to- emitter voltage V_CE for different constant base currents I_B, shown in figure.

(i) For V_BE less than the threshold voltage for the junction, I_B = 0 and I_C = 0, i.e., there is no current through the transistor. Both the junctions are reverse biased and the transistor acts like an open switch. Then the transistor is said to be switched Fully-OFF or in the cut-off region or OFF-mode.

(ii) The base current is set to a suitable value by varying the base-to-emitter voltage V_BE to a value greater than the threshold voltage. Then, the collector-to-emitter voltage V_CE is varied and the variation of I_C is plotted against V_CE. For very small values of V_CE, ideally zero, I_C is maximum, equal to V_CC/ R_L. Both the junctions are forward biased and the transistor acts like a closed switch. The transistor is said to be switched Fully-ON, or in the saturation region or ON-mode.

(iii) But for values of V_CE above about the threshold voltage, I_C is constant and V_CE has relatively little effect on it. In this region, called as the active region, I_C is determined almost entirely by I_B. In this region, the common-emitter current gain β is large.

For a constant base current IB, the dynamic output resistance (r₀) is defined as the ratio of the differential change in the collector-to-emitter voltage (∆V_CE) to the corresponding change in the collector current (∆I_C).
r₀ = \(\left(\frac{\Delta V_{\mathrm{CE}}}{\Delta I_{\mathrm{C}}}\right)_{I_{\mathrm{B}}=\text { constant }}\)

Question 51.
What is an amplifier? Explain the use of a transistor as an amplifier.
OR
Draw a neat circuit diagram of a transistor CE- amplifier and explain its working.
Answer:
A device that increases the amplitude of voltage, current or power of a weak alternating signal, by drawing energy from a separate source other than the signal, is called an amplifier.

Principle : The collector current can be controlled by a small change in the base current.

Electric circuit: Consider the use of a npn transistor as an amplifier in the widely used common- emitter (CE) configuration in which the emitter is common to the input and output circuits, shown in figure. The emitter-base junction is forward biased by the battery V_BB while the collector-base junction is reverse biased by the battery V_CC.

The voltage V_i to be amplified, called the signal voltage, is applied between the base and the emitter.

Working : The collector characteristics shows that in the active region, the collector current Ic is determined almost entirely by the base current IB, and collector potential Vc has relatively little effect on it.

An npn transistor amplifier in CE configuration

For V_i = 0, applying Kirchhoff’s loop law to the output and input loops, we get respectively,
V_CC – I_CR_L – V_CE = 0 ……………. (1)
and P_BB – I_BR_B – V_BE = 0 …………… (2)
The applied signal voltage causes small changes ∆V_BE in the emitter-base p.d. thereby producing variations ∆V_BE in the base current.
∆V_BE = r_i∆I_B …………. (3)
where r_i is the dynamic input resistance.
B_BB + V_i = I_BR_B + V_BE + ∆I_B(R_B + r_i)
∴ V_i = ∆I_B(R_B + r_i) = r_i∆I_B

The variations in ∆I_B cause proportionately larger variations ∆I_C in the collector current because ∆I_C = β_ac ∆I_B, where the ac common-emitter current gain (β_ac) is always greater than 50. For normal operating voltages, β_ac is almost the same as β_dc. From EQuestion(l), since V_CC is constant,
∆V_CC = ∆I_CR_L + ∆V_CE = 0
The time-varying collector current produces a time-varying output voltage P0 across the load resistance R_L.
V₀ = ∆V_CE = – ∆I_CR_L = – [β_ac∆I_BR_L

Thus, V₀ > V_i, so that the circuit produces a voltage gain. The amplifier’s voltage gain (A_v) is defined as the ratio of the output voltage to the input voltage.
A_v = \(\frac{V_{\mathrm{o}}}{V_{\mathrm{i}}}=-\frac{\beta_{\mathrm{ac}} R_{\mathrm{L}}}{r_{\mathrm{i}}}\)

The minus sign indicates that the output voltage is 180° out of phase with the input voltage.
[ Note : Amplifiers use emitter bias by moving the resistor from the base circuit to the emitter circuit. This important change keeps the operating point of the transistor fixed and immune to changes in current gain. The base supply voltage is now applied directly to the base.]

Question 52.
Solve the following :
(1) The diagram shows a CE circuit using a silicon transistor. Calculate the (a) base current (b) collector current. [V_BB = 2V, V_CC = 10V, R_B = 100 kΩ, R_L = 1 kΩ, [β_dc = 200]

Solution:
Data : V_BB = 2 V, V_CC = 10 V, R_B = 100 kΩ, R_L = 1 kΩ,
β_dc = 200.
Since it is a silicon transistor, the emitter-base
barrier potential, V_BE = 0.7 V.
The voltage across the base resistor is
V_BB – V_BE = 2 – 0.7 = 1.3 V
Therefore, the base current,
I_B = \(\frac{V_{\mathrm{BB}}-V_{\mathrm{BE}}}{R_{\mathrm{B}}}=\frac{1.3}{10^{5}}\) = 1.3 × 10^-5 = 13 μA
The collector current,
I_C = βI_B = 200 × 1.3 × 10^-5 = 2.6 × 10^-3 A = 2.6mA

(2) In the above problem, calculate the collector- emitter voltage (V_CE).
Solution:
The collector-emitter voltage,
V_CE = V_CC – I_CR_L = 10 – (2.6 × 10^-3)(10³) = 2.6V

Question 53.
What is meant by an analog signal and an analog electronic circuit?
Answer:
An analog signal consists of a continuously varying voltage or current. An analog electronic circuit takes an analog signal as input and outputs a signal that varies continuously according to the input signal.

Question 54.
What is meant by a digital signal?
Answer:
A digital signal consists of a sequence of electrical pulses whose waveform is approximately regular, with the potential switching alternately between two values. The lower value of the potential is labelled as LOW or 0 and the higher value as HIGH or 1.

With just two bits of information, 1 or 0 (HIGH or LOW), digital circuits use binary system. A sequence of 1s and 0s is encoded to represent numerals, letters of the alphabet, punctuation marks and instructions.

[ Note : The meaning of a voltage being high or low at a particular location within a circuit can signify a number of things. For example, it may represent the on or off state of a switch or saturated transistor. It may represent one bit of a number, or whether an event has occurred, or whether some action should be taken. The high and low states can be represented as true and false statements, which are used in Boolean logic. When positive-true logic is used high = true, while high = false when negative-true logic is used.]

Question 55.
What is a digital circuit?
Answer:
An electronic circuit that processes only digital signals is called a digital circuit. There are only two voltage states present at any point within a digital circuit. These voltage states are either high or low.

The branch of electronics which deals with digital circuits is called digital electronics.

[Note : Digital circuits can store and process bits of information needed to make complex logical decisions. Digital electronics incorporate logical decision-making processes into a circuit.]

Question 56.
What are the three ways of representing a logic gate?
Answer:
A logic gate can be represented by its logic symbol, Boolean expression and the truth table.

Question 57.
How many rows are there in the truth table of a 3-input gate?
Answer:
Each of the 3 inputs can take 2 values (0 and 1).
Hence, the number of rows in the truth table of a 3-input gate = 2 × 2 × 2 = 2³ = 8.

Question 58.
Name the common logic gates.
Answer:
The five common logic gates are the AND, OR, NOT, NAND and NOR gates.

Of these the AND, OR and NOT gates which respectively perform the logical AND, logical OR and logical NOT operations are called the basic logic gates. These three gates form the basis for other types of logical gates. The NAND and NOR are called the universal logic gates because any gate can be implemented by the combination of NAND and NOR gates.

Question 59.
Define the following logic gates :
(1) AND
(2) OR
(3) NOT.
Give the logic symbol, Boolean expression and truth table of each. (1 mark each )
Answer:
(1) The AND gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if and only if all the inputs are HIGH simultaneously.
The AND operation represents a logical multiplication.

Below figure shows the 2-input AND gate logic symbol and the Boolean expression and the truth table for the AND function.

(2) The OR gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if any one or more of the inputs is HIGH.
The OR operation represents a logical addition.

Below figure shows the 2-input OR gate logic symbol, and the Boolean expression and the truth table for the OR function.

(3) The NOT gate or INVERTER : It is a circuit with one input whose output is HIGH if the input is LOW and vice versa.
The NOT operation outputs an inverted version of the input. Hence, a NOT gate is also known as an INVERTER.

The small invert bubble on the output side of the inverter logic symbol, Fig. 16.30 and the over bar (^__) in the Boolean expression represent the invert function.

Question 60.
A gate generates a HIGH output when at least one of its inputs is HIGH. Which is this gate?
Answer:
It is an OR gate.

Question 61.
How will a 2-input AND gate work when both its input terminals are shorted? Give the circuit symbol and truth table.
Answer:
When both the input terminals of a 2-input AND gate are shorted, i.e., the same signal X goes to both inputs, the output is Y = X ∙ X which will give 1 if X = 1, and 0 if X = 0; hence Y = X ∙ X = X.
Circuit symbol:

Truth table :

[Note : The OR operation gives exactly the same result. The laws Y = X ∙ X = X and Y = X + X = X are called idem potent laws.]

Question 62.
Write the Boolean expression and give the circuit symbol for a 3-input AND gate.
Answer:
Consider an AND gate with 3-inputs, A, B and C. Boolean expression : Y = A ∙ B ∙ C
Circuit symbol:

Question 63.
If only 2-input OR gates are available, draw the circuit to implement the Boolean expression Y = A + B + C.
Answer:
Implementation of Y = A + B + C using two-input OR gates :

Question 64.
Define the logic gates (1) NAND (2) NOR.
Give the logic symbol, Boolean expression and truth table of each.
How are the above gates realized from the basic gates?
Answer:
(1) The NAND gate : It is a circuit with two or more inputs and one output, whose output is HIGH if any one or more of the inputs is LOW; the output is LOW if all the inputs are HIGH.

The NAND gate is a combination of an AND gate followed by a NOT gate so that the truth table of the NAND function is obtained by inverting the outputs of the AND gate.

(2) The NOR gate: It is a circuit with two or more inputs and one output, in which the output is HIGH if and only if all the inputs are LOW.

The NOR gate is realized by connecting the output of OR gate to the input of a NOT gate, so that the truth table of the NOR function is obtained by inverting the outputs of the OR gate.

Question 65.
How will a NAND gate work when all its input terminals are shorted?
Answer:
A NOT gate.

Question 66.
Define the XOR (Exclusive OR) logic gate. Give its logic symbol, Boolean expression and truth table. How is the XOR gate realized from the basic gates?
Answer:
The XOR (Exclusive OR) gate : It is a circuit with only two inputs and one output in which the output signal is HIGH if and only if the inputs are different from each other.

Question 67.
Prepare the truth tables for the following logic circuits. Write the Boolean expression for the output and name (or identify) the output function in each case.

OR
Draw the logic diagram and write the truth table for the Boolean equations
(1) Y = \(\overline{\mathrm{A}+\mathrm{B}}\)
(2) Y = \(\overline{\mathrm{A} \cdot \mathrm{B}}\)
(3) \(\overline{\mathrm{A}}+\overline{\mathrm{B}}=\overline{\mathrm{A} \cdot \mathrm{B}}\)
(4) \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}=\overline{\mathrm{A}+\mathrm{B}}\)
Answer:
(1)

The same signal X is fed to both the inputs of the 2-input NOR gate.

The truth table shows that the output is HIGH if the input is LOW and vice versa. Therefore, the circuit functions as a NOT gate or INVERTER.
∴ Boolean expression for the output logic is Y = \(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{X}}\)

(2)

The same signal X is fed to both the inputs of the 2-input NAND gate.
Truth table :

The truth table shows that the output is HIGH if the input is LOW and vice versa. Therefore, the circuit functions as a NOT gate or INVERTER.
∴ Boolean expression for the output logic is Y = \(\overline{\mathrm{A} \cdot \mathrm{B}}=\overline{\mathrm{X}}\)

(3)

Each input of the 2-input OR gate is fed through a NOT gate, i.e., the inputs to the OR gate are \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\).
Truth table:

The truth table shows that the output is HIGH if any one of the inputs is LOW; the output is LOW if all the inputs are HIGH. Therefore, the circuit functions as a NAND gate.
∴ Boolean expression for the output logic is Y = latex]\overline{\mathrm{A}}+\overline{\mathrm{B}}=\overline{\mathrm{A} \cdot \mathrm{B}}[/latex]

(4)

Inverted inputs \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\) are fed to the 2-input AND gate.
Truth table :

The truth table shows that the output is HIGH if and only if all the inputs are LOW. Therefore, the circuit functions as a NOR gate.
∴ Boolean expression for the output logic is Y = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}=\overline{\mathrm{A}+\mathrm{B}}\)

Question 68.
What is the equivalent logic gate for the following logic circuit?

Answer:
The given logic circuit (combination of logic gates) is

i. e., the output is HIGH if and only if both the inputs are LOW, which is obtained by a NOR gate.

Question 69.
Write the truth table for the Boolean equation Y = \(\overline{\mathrm{A}}\) – B + A – \(\overline{\mathrm{B}}\).
Answer:

Multiple Choice Questions

Question 1.
An electronic circuit which converts alternating voltage into unidirectional pulsating voltage is called
(A) a transistor
(B) a rectifier
(C) an oscillator
(D) a transformer.
Answer:
(B) a rectifier

Question 2.
A pn junction exhibits rectifying property because of
(A) the potential barrier across the pn junction
(B) the difference in the doping concentrations in the p- and n-layers
(C) the avalanche breakdown when reverse biased
(D) the Zener breakdown when reverse biased.
Answer:
(A) the potential barrier across the pn junction

Question 3.
The stepped down output of a transformer, with turns ratio 5 : 1, is fed to an ideal diode D and load resistance R_L. The peak load voltage is

(A) 22 V
(B) 31 V
(C) 44 V
(D) 62 V
Answer:
(D) 62 V

Question 4.
In the given circuit, the peak value of the ac source voltage is 10 V, and the diode has a negligible forward resistance and infinite reverse resistance. During the negative half cycle of the source voltage, the peak voltage across the diode is

(A) – 10 V
(B) 0
(C) 5 V
(D) 10 V
Answer:
(A) – 10 V

Question 5.
A full-wave rectifier uses a grounded centre tap on the secondary winding of a transformer and a 60 Hz source voltage across the primary winding. The output frequency of the full-wave rectifier is
(A) 30 Hz
(B) 60 Hz
(C) 120 Hz
(D) 240 Hz.
Answer:
(C) 120 Hz

Question 6.
Avalanche breakdown in a Zener diode takes place due to
(A) thermal energy
(B) light energy
(C) magnetic field
(D) accelerated minority charge carriers.
Answer:
(D) accelerated minority charge carriers.

Question 7.
A Zener diode is used as a
(A) half wave regulator
(B) half wave rectifier
(C) simple voltage regulator
(D) voltage amplifier.
Answer:
(C) simple voltage regulator

Question 8.
The current through the Zener diode in the following circuit is

(A) 5 mA
(B) 10 mA
(C) 15 mA
(D) 30 mA.
Answer:
(A) 5 mA

Question 9.
In operation, a photodiode is
(A) unbiased
(B) always forward-biased
(C) always reverse-biased
(D) either forward-or reverse-biased.
Answer:
(C) always reverse-biased

Question 10.
The photocurrent in a photodiode is a few
(A) nanoamperes
(B) microamperes
(C) milliamperes
(D) amperes.
Answer:
(B) microamperes

Question 11.
A photodiode is used in
(A) a brake indicator
(B) an optocoupler
(C) a regulated power supply
(D) a logic gate.
Answer:
(B) an optocoupler

Question 12.
When the load resistance across a solar cell is zero, the current in the external circuit passed by the solar cell is called
(A) the open-circuit current
(B) the reverse saturation current
(C) the short-circuit current
(D) the photocurrent.
Answer:
(C) the short-circuit current

Question 13.
Which of the following is the correct circuit symbol for an LED ?

Answer:

Question 14.
The colour of light emitted by an LED depends upon
(A) its forward bias
(B) its reverse bias
(C) the band gap of the material of the semiconductor
(D) its size.
Answer:
(C) the band gap of the material of the semiconductor

Question 15.
The centre terminal of a junction transistor is called
(A) the emitter
(B) the opposite semiconductor
(C) the collector
(D) the base.
Answer:
(D) the base.

Question 16.
A transistor acts as a ‘closed switch’ when it is in
(A) the cutoff region
(B) the active region
(C) the breakdown region
(D) the saturation region.
Answer:
(D) the saturation region.

Question 17.
A junction transistor acts as
(A) a rectifier
(B) an amplifier
(C) the oscillator
(D) a voltage regulator.
Answer:
(B) an amplifier

Question 18.
When an npn junction transistor is used as an amplifier in CE-mode,
(A) the central p-type region is common to both input and output circuits
(B) the emitter terminal is common to both input and output circuits
(C) the emitter junction is reverse biased while the collector junction is forward biased
(D) the signal voltage is applied between the two n regions.
Answer:
(B) the emitter terminal is common to both input and output circuits

Question 19.
When a pnp transistor is operated in saturation region, then its
(A) the base-emitter junction is forward biased and base-collector junction is reverse biased
(B) both the base-emitter and base-collector junctions are reverse biased
(C) both the base-emitter and base-collector junctions are forward biased
(D) the base-emitter junction is reverse biased and base-collector junction is forward biased.
Answer:
(C) both the base-emitter and base-collector junctions are forward biased

Question 20.
Which logic gate corresponds to the logical equation, Y = \(\overline{\mathrm{A}+\mathrm{B}}\) ?
(A) NAND
(B) NOR
(C) AND
(D) OR
Answer:
(B) NOR

Question 21.
The output of a NAND gate is HIGH if
(A) any one or more of the inputs is LOW
(B) all the inputs are HIGH
(C) only all the inputs are simultaneously LOW
(D) only if an inverter is connected at its output.
Answer:
(A) any one or more of the inputs is LOW

Question 22.
The output of NOR gate is HIGH, when
(A) all inputs are HIGH
(B) all inputs are LOW
(C) only one of its inputs is HIGH
(D) only one of its inputs is LOW.
Answer:
(B) all inputs are LOW

Question 23.
Which logic gate corresponds to the truth table

(A) AND
(B) NOR
(C) OR
(D) NAND
Answer:
(B) NOR

Question 24.
The logic gate which produces LOW output when any one of the input is HIGH and produces HIGH output only when all of its inputs are LOW is called
(A) an AND gate
(B) an OR gate
(C) a NOR gate
(D) a NAND gate.
Answer:
(C) a NOR gate

Question 25.
The Boolean expression for an Exclusive OR gate is
(A) A + B
(B) A ⊕ B
(C) \(\overline{\mathrm{A}+\mathrm{B}}\)
(D) A ∙ B
Answer:
(B) A ⊕ B