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Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 14 Dual Nature of Radiation and Matter Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 14 Dual Nature of Radiation and Matter

Question 1.
Explain Planck’s idea of quantization of energy.
Answer:
Max Planck, in 1900, put forward the idea of quantization of energy to explain the blackbody radiation spectrum. He proposed that atoms behave as tiny oscillators and emit electromagnetic radiation, not continuously but as little packets of energy called quanta. He assumed that the energy associated with a quantum of radiation (now called a photon) is proportional to the frequency v of the oscillator. Thus, E = nhv, where n = 1, 2, 3, 4, … etc., and h is a universal constant, now called Planck’s constant. For n = 1, E = hv. A quantum of radiation is emitted when there is a transition from higher quantized level of energy of an oscillator to lower quantized level.

[Note : Historically, various terms have been used to denote a particle of light; quantum of electromagnetic radiation ≡ photon ≡ packet of energy ≡ atom of energy ≡ quantum of radiation ≡ bundle of energy. Interaction between two charged particles involves exchange of photons. The photon has zero rest mass, no charge, unit spin and travels in free space at a speed of 2.99792458 × 10⁸ m/s exact by definition. There is no conservation law for photons, i.e., they can be produced / absorbed.]

Question 2.
What was Hertz’s observation regarding emission of electrons from a metal surface?
Answer:
During his experiments on electromagnetic waves in 1887, Heinrich Rudolph Hertz (1857-94), Ger-man physicist, noticed that electric sparks occurred more readily when one of the electrodes of his spark-gap transmitter was exposed to ultraviolet radiation. This discovery was called the Hertz effect and is now known as the photoelectric effect.

Although Hertz did not follow up his discovery, others quickly established that the cause of the sparking ease was due to emission of negatively charged particles from the electrode irradiated. These particles were identified as electrons after the discovery of the electron in 1897.

Question 3.
Draw a neat labelled diagram to illustrate photoelectric effect.
Answer:

[Note : Positive metal ions and atoms are not shown in the figure.]

Question 4.
What were the investigations of Hallwachs and Lenard regarding photoelectric effect?
Answer:
Wilhelm Hallwachs (1859-1922), German physicist, found that a metal plate irradiated with ultraviolet radiation lost its charge more rapidly when the plate is negatively charged than when it is neutral or positive.

Investigations of photoelectric effect by Phillipp Lenard (1862-1947), German physicist, showed that

1. electron emission occurs only with radiations below a critical wavelength, i.e., above a critical frequency.
2. kinetic energy of the emitted electrons increases as wavelength decreases i.e., frequency increases but is independent of the intensity of radiation which determines the rate of emission of electrons (the number of electrons emitted per unit time).

Question 5.
What is a photosensitive surface?
Answer:
The surface which emits electrons when illuminated by electromagnetic radiation of appropriate frequency is called photosensitive surface.
[Note : The material that exhibits photoelectric effect is called photosensitive material.]

Question 6.
Why are alkali metals most suitable as photo-sensitive surfaces?
Answer:
The alkali metals e.g., caesium, potassium and sodium emit photoelectrons even when visible radiation (light) is incident on them. Hence, they are most suitable as photosensitive surfaces.

Question 7.
With a neat diagram, describe the apparatus to study the characteristics of photoelectric effect.
Answer:
Apparatus : A photoelectric cell G consists of the emitting electrode E (emitter) of the material being studied and the collecting electrode C (collector). The electrodes are sealed in an evacuated glass envelope provided with quartz window W that allows the passage of UV radiation and visible light. Monochromatic light of variable frequency from a suitable source S (such as a carbon arc) passes through a pair of polarizers P (permitting a change in the intensity of radiation) and falls on the emitter.

The electric circuit, as shown in below figure, allows the collector potential to be varied from positive through zero to negative with respect to the emitter, and permits the measurement of potential difference and current between the electrodes. When the collector is made negative, the voltmeter is connected in reverse.

[ Note : The radiation coming out of a filter is not truly monochromatic, it lies in the wavelength range between λ and λ + ∆λ that depends on the source and the filter. ]

Question 8.
In the experiment to study photoelectric effect, describe the effects of the frequency and intensity of the incident radiation on the photoelectric current, for a given emitter material and potential difference across the photoelectric cell.
Answer:
The emitter of the photoelectric cell is irradiated with monochromatic light whose frequency and intensity can be varied continuously and measured. Initially, the collector is made positive with respect to the emitter, so that photoelectrons ejected move quickly from the emitter to the collector. About 10 V is sufficient to do this. The photoelectric current as a function of intensity and frequency of incident radiation is studied.

(1) Effect of frequency : Keeping the light intensity and the accelerating potential difference V constant, the frequency of the incident radiation is varied from that of far-UV to red. It is found that for every material (usually, a metal) irradiated there is a limiting frequency below which no photoelectrons are emitted irrespective of the intensity of the radiation. This frequency, v₀, called the threshold frequency or cut-off frequency, is a characteristic of the material irradiated.

The graph of photoelectric current against frequency is shown in below figure; A and B represent two different metals. The photoelectric current is not the same in the two cases, because the intensity of light is different for different frequencies.

(2) Effect of intensity : With an emitter of a given material, the light intensity is varied by keeping the frequency v (≥ v₀) of the light and the accelerating potential difference V constant. It is found that the rate of electron emission, as indicated by the photoelectric current, is proportional to the light intensity. The graph of photoelectric current against light intensity is a straight line through (0, 0), below figure; if we vary either the frequency of the light or the material irradiated, only the slope of the line changes. No electrons are emitted in the absence of incident radiation.

[Note : The dark current, i.e., the current observed in the absence of light, is extremely low. Hence, it is ignored.]

Question 9.
In the experiment to study photoelectric effect, describe the variation of the photoelectric current as a function of the potential difference across the photoelectric cell, for incident radiation of (1) a given frequency above the threshold but different intensities (2) a given intensity but different frequencies above the threshold.
Answer:
(1) The potential difference (p.d.) across the photo-electric cell is varied keeping both the frequency v (≥ threshold frequency v₀) and the intensity of the light constant. Starting with the collector at about 10 V positive, we reduce this potential to zero and then run it negative.

When the p.d. across the tube is 10 V or more, all the emitted electrons are accelerated and travel across the tube, constituting the saturation current for a given light intensity; an increase in the potential of the collector does not cause an increase in current. As the collector potential is reduced from positive values through zero to negative values, the tube current reduces because of the applied retarding potential. In this case, some electrons stop and turn back before they can reach the collector. Eventually the potential difference is large enough to stop the current completely. This is called the stopping potential or cut-off potential VQ. The product of the stopping potential and electronic charge, V₀e, is equal to the maximum kinetic energy that an electron can have at the time of emission.
V₀e = KE_max \(\frac{1}{2}\)v²_max

In above figure, I₁ and I₂ are two intensities of the incident radiation for the same frequency v ( > v₀); I₂ = 2I₀. Doubling the intensity of light doubles the current at each potential, as in I₂, but V₀ is independent of I.

(2) The above experiment is repeated with different light frequencies for a given emitter material and light intensity. It is found that the stopping potential increases linearly with the frequency in below figure. Therefore, when photoejection occurs for frequencies above v₀, the maximum kinetic energy of the photoelectrons increases linearly with the frequency of the radiation.

[Note : It was shown by Hughes that the stopping potential depends linearly on the frequency of incident radiation. Lawrence and Beams established that the time interval between arrival of a photon on a metal surface and emission of an electron is less than 3 × 10^-9 sc.]

Question 10.
What is the effect of the intensity of incident radiation on the stopping potential in photo-electric emission?
Answer:
V₀ is independent of intensity.

Question 11.
In the experiment to study photoelectric effect, discuss the effect and significance of extremely weak radiation of frequency greater than the threshold frequency for the emitter material.
Answer:
The emitter of the photoelectric cell is irradiated with monochromatic light whose frequency is greater than the threshold frequency for the emitter material. The collector is kept at 10 V positive with respect to the emitter, so that photoelectrons ejected move quickly from the emitter to the collector.

The light is made extremely dim (i.e., the intensity is extremely weak). In this case, the number of ‘ photoelectrons emitted per unit time is very small (and special techniques are required to detect them); but, however few, they are emitted almost instantaneously and with the same maximum kinetic energy as for bright light of the same frequency.

According to the wave theory of light, wave trains of pulsating electromagnetic field spread out from the source. Dim light corresponds to waves of small amplitudes and small energy. If dim light spreads over a surface, conservation of energy requires that the electrons must store energy over long periods of time, which can be several hours, before gathering enough energy to become free of the metal. The fact that photoelectrons appear immediately, within about 10^-9 s, can be explained only by assuming that the light energy is not spread over the surface uniformly as required by the wave theory, but falls on the surface in concentrated bundles.

Question 12.
Define (1) threshold frequency (2) threshold wavelength (3) stopping potential.
Answer:
(1) The threshold frequency for a given metal surface is the characteristic minimum frequency of the incident radiation below which no photoelectrons are emitted from that metal surface (whatever may be the intensity of incident light).

(2) The threshold wavelength for a given metal surface is the characteristic maximum wavelength of the incident radiation above which no photoelectrons are emitted from that metal surface (whatever may be the intensity of incident light).

(3) The stopping potential is the value of the retarding potential difference that is just sufficient to stop the most energetic photoelectrons emitted from reaching the collector so that the photoelectric current in a photocell reduces to zero.

[Note : The threshold wavelength λ₀ = c/v₀, where c is the speed of light in free space and v₀ is the threshold frequency for the metal.]

Question 13.
State the characteristics of photoelectric effect.
Answer:
Characteristics of photoelectric effect:
(1) For every metal surface there is a limiting frequency of incident radiation below which no photoelectrons are emitted from that metal surface. This frequency, called the threshold frequency, is characteristic of the metal irradiated.

(2) The time rate of emission of photoelectrons in-creases in direct proportion to the intensity, of incident radiation.

(3) The photoelectrons have different speeds at the time of emission ranging from zero to a certain maximum value, which is characteristic for a given metal for a given frequency of the incident radiation. The maximum kinetic energy of the photoelectrons at the time of emission is independent of the intensity but increases linearly with the frequency of the incident radiation.

(4) For incident radiation of frequency greater than or equal to the threshold frequency for a given metal surface, photoelectric emission from the surface is almost instantaneous, even under extremely weak irradiation.

Question 14.
Can we get photoemission with an intense beam of radio waves ? Is photoemission possible at all frequencies ?
Answer:
The frequency of the incident radiation and not its intensity is the criterion for photoelectric effect. The lowest frequency of electromagnetic waves that can cause photoemission is about 4.6 × 10¹⁴ Hz (for the alkali metal caesium). Since radio waves have frequencies 1 GHz or lower, they cannot cause photoemission.

Only alkali metals are photosensitive to visible light; other metals are photosensitive only to far ultraviolet radiations.

Question 15.
Explain how wave theory of light fails to explain the characteristics of photoelectric effect.
OR
Explain the failure of wave theory of light to account for the observations from experiments on photoelectric effect.
Answer:
According to the wave theory of light, electromagnetic waves carry the energy stored in oscillating electric and magnetic fields. When enough energy is absorbed by an electron in a substance, it should be liberated as a photoelectron. Frequency of light does not come into picture in this case. Hence, there should not be any threshold frequency for emission of electrons. But it is found that there exists threshold frequency and it depends on the metal.

Experimentally, the maximum kinetic energy of photoelectrons increases linearly with the frequency of light. This cannot be accounted by the wave theory of light.

If a source of light is weak or far away from a metal surface, emission of an electron will not be almost instantaneous. The electron may have to wait for several hours/days for absorption of enough energy from the incident light as by the wave theory of light, energy is spread over the wavefront. But experimentally, for an appropriate frequency of incident light, photoelectric effect is almost instantaneous.

Only one observation, photoelectric current ∝ intensity of incident light can be accounted by the wave theory of light.

Question 16.
Give Einstein’s explanation of the photoelectric effect.
Answer:
Max Planck put forward the quantum theory in 1900 to explain blackbody spectrum. In the theory, he proposed that the electromagnetic radiation emitted by the body consists of discrete concentrated bundles of energy, each equal to hv, where h is a universal constant (now called Planck’s constant) and v is the frequency of the radiation.

Einstein put forth (1905) that these energy quanta, called light quanta/later called photons, interact with matter much like a particle. When a photon collides with an electron in an atom, the electron absorbs whole of the photon energy hv in a single collision or nothing. The electron uses this energy (1) to liberate itself from the atom, (2) to overcome the potential energy barrier at the surface thus liberating itself from the metal, and (3) retains the remaining part as its kinetic energy. Different electrons need different energies in the first two processes. There are some electrons which use minimum energy in the two processes, and hence come out of the metal with maximum kinetic energy. The minimum energy required, in the form of electromagnetic radiation, to free an electron from a metal is called the photoelectric work function Φ of that metal. Thus, for the most energetic photoelectrons at the time of emission,
maximum kinetic energy of the electron = photon energy – photoelectric work function
∴ \(\frac{1}{2}\)\(m v_{\max }^{2}\) = hv – Φ ∴ hv = Φ + \(\frac{1}{2}\)\(m v_{\max }^{2}\)
The above equation is called Einstein’s photo-electric equation.

Light interacts with matter as concentrated bundles of energy rather than energy spread over a Huygens type wavefront. Even under weak irradiation, an electron absorbs a photon’s energy in a single collision. But the rate of incident photons in dim light being less, the chances of such absorption diminish and consequently the photoelectric current diminishes. However, a photoelectron is emitted as soon as a photon is absorbed.

[Note : Albert Einstein (1879-1955), German-Swiss- US theoretical physicist, gave his photoelectric equation in 1905. In the period 1912-1916, Robert Andrews Millikan (1868 -1953), US physicist, was the first to obtain the precise experimental data from which the straight-line graphs, like the one shown in Fig. 14.6, were plotted for various metals. Einstein’s theoretically predicted equation-clearly having the right form for a straight-line graph-was thus verified.]

Question 17.
Define photoelectric work function of a metal.
Answer:
The photoelectric work function of a metal is defined as the minimum photon energy that ejects an electron from the metal.
It is equal to hv₀, where h is Planck’s constant and v0 is the threshold frequency for the metal.

Question 18.
Write Einstein’s photoelectric equation and explain its various tends. How does the equation explain the various features of the photoelectric effect?
Answer:
Einstein’s photoelectric equation :
hv = Φ + \(\frac{1}{2}\)\(m v_{\max }^{2}\) ………… (1)
where h ≡ Planck’s constant, v ≡ frequency of the electromagnetic radiation, hv ≡ energy of the photon incident on a metal surface, Φ ≡ photo-electric work function, i.e., the minimum energy of light quantum required to liberate an electron from the metal surface, v_max and – \(\frac{1}{2}\)\(m v_{\max }^{2}\) ≡ the maximum speed and maximum kinetic energy of the photoelectrons at the time of emission. Φ = hv₀, where v₀ is the threshold frequency for the metal.

Explanation of the characteristics of photoelectric effect:
(1) From the above equation we find that for photoejection, hv ≥ Φ. That is, hv_min = hv₀ must be equal to Φ. Hence, photoelectric effect is observed only if hv ≥ hv₀, i.e., v ≥ v₀. This shows the existence of a threshold frequency v₀ for which photoelectrons are just liberated from a metal surface (with zero kinetic energy). Since different metals differ in electronic configuration, the work function hv₀ and, therefore, frequency v₀ are different and characteristic of different metals.

(2) In this particle model of light,’ intensity of incident radiation’ stands for the number of photons incident on a metal per unit surface area per unit time. As the number of photons incident on a metal per unit surface area per unit time increases, there is a greater likelihood of a photon being absorbed by any electron. Therefore, the time rate of photoejection and hence photoelectric current increases linearly with the intensity of the incident radiation (v ≥ v₀).

(3) From Eq. (1), \(\frac{1}{2}\)\(m v_{\max }^{2}\), = hv – Φ= h(v – v₀)
This shows that the maximum kinetic energy in-creases linearly with the frequency v of the incident photon (v ≥ v₀) and does not depend on the time rate at which photons are incident on a metal surface.

(4) As the incident energy is concentrated in the form of a photon, and not spread over a wavefront, it is expected that an electron is emitted from the metal surface as soon as a photon (v ≥ v₀) is absorbed. This is in agreement with the experimental observation.

[ Note : The frequency v that appears in the formula E = hv is the frequency of the oscillating electric field / magnetic field in the electromagnetic wave. ]

Question 19.
Obtain the dimensions of Planck’s constant.
Answer:
The energy of a photon of frequency v is E = hv, where h is the Planck’s, constant.
∴ [h] = \(\frac{[E]}{[v]}=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}\) = ML²T^-1

Question 20.
Is the kinetic energy of all photoelectrons the same when emitted from a certain metal ? Explain.
Answer:
No. Explanation : Depending upon the position and state of an electron in a metal when it absorbs an incident photon, a photoelectron can have kinetic energy ranging from 0 to a certain maximum value equal to the photon energy minus the work function of the metal. Hence, the emitted photoelectrons have this range of kinetic energies.

Question 21.
In photoelectric effect, what does the stopping potential depend upon ?
Answer:
In photoelectric effect, the stopping potential depends upon the energy of the incident photon and the work function for the metal irradiated (or upon the frequency/wavelength of the incident radiation and the threshold frequency/wavelength for the metal irradiated).

Question 22.
What does the maximum kinetic energy (or the maximum speed) of a photoelectron depend on?
Answer:
The maximum kinetic energy (or the maximum speed) of a photoelectron depends upon the energy of the incident photon and the work function for the metal irradiated (or upon the frequency /wavelength of the incident radiation and the threshold frequency/wavelength for the metal irradiated).

Question 23.
In photoelectric effect, if a graph of stopping potential versus frequency of the incident radiation is plotted, what does the intercept on the frequency axis (v corresponding to V_o = 0) represent?
Answer:
The intercept on the frequency axis (v corresponding to V_o = 0) represents the threshold frequency for the metal.

Question 24.
State the equation that relates the threshold wavelength (λ_o), the wavelength of incident radiation (λ) and the maximum speed of a photo-electron (v_max).
Answer:
\(\frac{h c}{\lambda}=\frac{h c}{\lambda_{0}}+\frac{1}{2} m v_{\max }^{2}\) is the required equation, where h is Planck’s constant, c is the speed of light in vacuum (free space) and m is the mass of the electron.

Question 25.
What is the energy of a photon (quantum of radiation) of frequency 6 × 10¹⁴ Hz?
[h = 6.63 × 10^-34 J∙s]
Answer:
hv = (6.63 × 10^-34)(6 × 10¹⁴)
= 3.978 × 10^-19 J is the energy of the photon.

Question 26.
If the total energy of a radiation of frequency 10¹⁴ Hz is 6.63 J, calculate the number of photons in the radiation.
Answer:
E = nhv, where hv is the energy of a photon in a radiation of frequency v and n is the number of photons in the radiation.
∴ n = \(\frac{E}{h v}=\frac{6.63}{\left(6.63 \times 10^{-34}\right)\left(10^{14}\right)}\) = 10²⁰

Question 27.
If in a photoelectric experiment, the stopping potential is 1.5 volts, what is the maximum kinetic energy of a photoelectron ? [e = 1.6 × 10^-19 C]
Answer:
\(\frac{1}{2}\)\(m v_{\max }^{2}\) = V_se = ( 1.5)(1.6 × 10^-19)
= 2.4 × 10^-19 J is the required kinetic energy.

Question 28.
What is the photoelectric work function for a metal if the threshold wavelength for the metal is 3.315 × 10^-7 m?
[h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s]
Answer:
Photoelectric work function for the metal =

Question 29.
Explain the utilization of energy absorbed by an electron in a metal during its collision with a photon.
Answer:
When a photon collides with an atomic electron inside an emitter metal, the electron absorbs whole of the photon energy in a single shot or nothing. The electron uses the absorbed energy (1) to liberate itself from the atom, (2) to overcome the potential energy barrier at the surface thus liberating itself from the metal, and (3) retains the remaining part as its kinetic energy.

30. Solve the following :
(h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, e = 1.6 × 10^-19 J, m (electron) = 9.1 × 10^-31 kg)

Question 1.
Find the energy of a photon if
(i) the frequency of radiation is 100 MHz
(ii) the wavelength of radiation is 10000 Å.
Solution:
Data : h = 6.63 × 10^-34 J∙s, v = 100 MHz = 100 × 10⁶ Hz, λ = 10000 Å = 10⁶ m, c = 3 × 10⁸ m/s
(i) The energy of a photon, E = hv
= (6.63 × 10^-34)(100 × 10⁶) = 6.63 × 10^-26 J

(ii) The energy of a photon, E = \(\frac{h c}{\lambda}\)
= \(\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{10^{-6}}\) = 1.989 × 10^-19 J

Question 2.
A monochromatic source emits light of wavelength 6000 Å. If the power of the source is 10 W, find the number of photons emitted by it per second assuming that 1% of electric energy is converted into light.
Solution:
Data : λ = 6000 Å = 6 × 10^-7 m ,h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, electric energy converted into light per second = \(\frac{1}{100}\) × 10W = 0.1J/s

Question 3.
Radiation of intensity 4 × 10^-5 W/m² is incident uniformly on a metal surface with work function 2.4 eV and area 1 cm². Assume that the radius of a metal atom is 2.4 Å and photoelectrons are ejected only from the surface of the metal. On the basis of the wave theory of light, how long will it take for an electron to be ejected from the metal surface ? (Assume one free electron/metal atom.)
Solution:
Data : Power/area = 4 × 10^-5 W/m², Φ = 2.4 eV = 2.4 × 1.6 × 10^-19 J = 3.84 × 10^-19 J, A = 1 cm² = 10^-4 m², r = 2.4 Å = 2.4 × 10^-10 m
Number of metal atoms on the surface = \(\frac{A}{\pi r^{2}}\)

For a single free electron, radiant energy incident per unit time

Ignoring reflection/scattering of light, time needed to absorb energy equal to 3.84 × 10^-19 J is
\(\frac{3.84 \times 10^{-19} \mathrm{~J}}{7.24 \times 10^{-24} \mathrm{~J} / \mathrm{s}}\) = 5.304 × 10⁴ s = 53040s
= 14 hours 44 minutes.

Question 4.
The energy of a photon is 2 eV. Find its frequency and wavelength.
Solution:
Data : E = 2 eV = 2 × 1.6 × 10^-19 J, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J∙s
(i) Frequency,

= 6.217 × 10^-7 = 6.217 × 10^-7 × 10¹⁰ Å
= 6217 Å = 621.7 nm

Question 5.
Find the wave number of a photon having an energy of 2.072 eV. [Given : e, c, h]
Solution:
Data : e = 1.6 × 10^-19 C, c = 3 × 10⁸ m / s, h = 6.63 × 10^-34 J∙s,
E = 2.072 eV = 2.072 × 1.6 × 10^-19 J
E = hv = \(\frac{h c}{\lambda}\)
Wave number, \(\frac{1}{\lambda}=\frac{E}{h c}\)
= \(\frac{2.072 \times 1.6 \times 10^{-19}}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}\) = 1.666 × 10⁶ m^-1

Question 6.
Calculate the energy of a photon, in joule and eV, in a light of wavelength 5000 Å.
Solution :
Data : λ = 5000 Å = 5000 × 10^-10 m = 5 × 10^-7 m, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J∙s
Energy of a photon, E = hv = \(\frac{h c}{\lambda}\)

Question 7.
The photoelectric work function for a metal surface is 2.3 eV. 1f the light of wavelength 6800 Å is incident on the surface of the metal, find the threshold frequency and the incident frequency. Will there be an emission of photoelectrons or not? [Given : c, h]
Solution:
Data: c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J.s, Φ = 2.3 eV = 2.3 × 1.6 × 10^-19 J, λ = 6800 Å = 6.8 × 10^-7 m
(i) Threshold frequency (v₀) : Φ = hv₀
∴ v₀ = \(\frac{\phi}{h}=\frac{2.3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}\) = 5.550 × 10¹⁴ Hz

(ii) Incident frequency (v) : c = vλ
∴ v = \(\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6.8 \times 10^{-7}}\) = 4.412 × 10¹⁴ Hz

(iii) Thus, v <v₀
As the frequency of the incident orange light is less than the threshold frequency there will be no emission of photoelectrons.

Question 8.
If the work function of a metal is 3 eV, calculate the threshold wavelength of that metal. [Given : c, h, 1 eV = 1.6 × 10^-19 J]
Solution:
Data : Φ = 3 eV, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J.s, 1 eV = 1.6 × 10^-19 J
∴ Φ = 3 × 1.6 × 10^-19 J

Question 9.
The photoelectric work function of copper is 4.7 eV. What are the threshold frequency and wavelength for photoemission from a copper surface? [1 eV = 1.6 × 10^-19 J]
Solution :
Data : Φ = 4.7 eV, 1 eV = 1.6 × 10^-19 J, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J.s

Question 10.
The work functions for potassium and caesium are 2.25 eV and 2.14 eV respectively. Will the photoelectric effect occur for either of these elements
(i) with incident light of wavelength 5650 Å
(ii) with light of wavelength 5180 Å?
Solution:
Data : Φ (potassium) = 2.25 eV,
Φ (caesium) = 2.14 eV, λ₁ = 5650 Å = 5.650 × 10^-7 m, λ₂ = 5180 Å = 5.180 × 10^-7 m, h = 6.63 × 10^-34 J.s, c = 3 × 10⁸ m/s
Φ (potassium) = 2.25 eV
= 2.25 × 1.6 × 10^-19 J =3.6 × 10^-19 J
Φ (caesium) = 2.14 eV = 2.14 × 1.6 × 10^-19 J
= 3.424 × 10^-19 J
Photon energy, E = \(\frac{h c}{\lambda}\)

(i) For λ₁ = 5650 Å
E₁ = \(\frac{h c}{\lambda_{1}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{5.650 \times 10^{-7}}\)
= 3.52 × 10^-19 J
This is greater than Φ (caesium), but less than Φ (potassium). Hence, photoelectric effect will occur in case of caesium, but not in case of potassium.

(ii) For λ₂ = 5180 Å
E₂ = \(\frac{h c}{\lambda_{2}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{5.180 \times 10^{-7}}\)
= 3.84 × 10^-19 J
This is greater than 0 for potassium and for caesium. Hence, photoelectric effect will occur in both the cases.

Question 11.
Photoemission just occurs from a lead surface when radiation of wavelength 3000 Å is incident on it. Find the maximum kinetic energy of the photoelectrons when the surface is irradiated by UV radiation of wavelength 2500 Å.
Solution:
Data : λ₀ = 3000 Å = 3 × 10^-7 m, λ = 2500 Å = 2.5 × 10^-7 m, h = 6.63 × 10^-34 J∙s,c = 3 × 10⁸ m/s
According to Einstein’s photoelectric equation, the maximum kinetic energy of the photoelectrons

Question 12.
The photoelectric work function for a metal is 4.2 eV. If the stopping potential is 3 V, find the threshold wavelength and the maximum kinetic energy of emitted electrons. [Given : c, h, e]
Solution :
Data : e = 1.6 × 10^-19 C, c = 3 × 10⁸ m/s, Φ = 4.2 eV = 4.2 × 1.6 × 10^-19 J = 6.72 × 10^-19 J, h = 6.63 × 10^-34 J∙s, v₀ = 3 V
(i) Threshold wavelength.

= 2.960 × 10^-7 m or 2960 Å

(ii) Maximum kinetic energy of emitted electrons,
KE_max = eV₀ = (1.6 × 10^-19)(3) = 4.8 × 10^-19 J

Question 13.
Radiation of wavelength 2 × 10^-7 m is incident on the cathode of a photocell. The current in the photocell is reduced to zero by a stopping potential of 2 V. Find the threshold wavelength for the cathode.
Solution:
Data : λ = 2 × 10^-19 m, V₀ = 2 V, e = 1.6 × 10^-19 C, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s
According to Einstein’s photoelectric equation,

∴ The threshold wavelength, λ₀ = 2.948 × 10^-7 m

Question 14.
The photoelectric current in a photoelectric cell can be reduced to zero by a stopping potential of 1.8 volts. Monochromatic light of wavelength 2200 Å is incident on the cathode. Find the maximum kinetic energy of the photoelectrons in joule. [Charge on the electron = 1.6 × 10^-19 C]
Solution:
Data: V₀ = 1.8 V, e = 1.6 × 10^-7 C .
The maximum kinetic energy of the photoelectrons,
KE_max = eV₀
= (1.6 × 10^-19) (1.8) = 2.88 × 10^-19 J

Question 15.
The photoelectric work function of a metal is 3 eV. Find the maximum kinetic energy and maximum speed of photoelectrons when radiation of wavelength 4000 Å is incident on the metal surface.
Solution:
Data : Φ = 3 eV = 3 × 1.6 × 10^-19 J, c = 3 × 10⁸ m / s, λ = 4000 Å = 4000 × 10^-10 m, h = 6.63 × 10^-34 J∙s, m = 9.1 × 10^-31 kg
(i) According to Einstein’s photoelectric equation, the maximum kinetic energy of photoelectrons,

= 1.725 × 10^-20 J

= 1.947 × 10⁵ m/s

Question 16.
The work function of tungsten is 4.50 eV. Calculate the speed of the fastest electron ejected from tungsten surface when light whose photon energy is 5.80eV shines on the surface.
Solution:
Data : Φ = 4.50 eV = 4.50 × 1.6 × 10^-19 J = 7.2 × 10^-19 J,
hv = 5.80eV = 5.80 × 1.6 × 10^-19 J = 9.28 × 10^-19 J,
m = 9.1 × 10^-31 kg

This is the speed of the fastest electron ejected.

Question 17.
If the work function for a certain metal is 1.8 eV,
(i) what is the stopping potential for electrons ejected from the metal when light of 4000 Å shines on the metal
(ii) what is the maximum speed of the ejected electrons?
Solution:
Data: Φ = 1.8eV, λ = 4000 Å = 4 × 10^-7 m,
h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s,
m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C

This is the maximum speed of the ejected electrons.

Question 18.
The work function of caesium is 2.14 eV. Find
(i) the threshold frequency for caesium
(ii) the wavelength of the incident light if photocurrent is brought to zero by a stopping potential of 0.60 V.
Solution:
Data : Φ = 2.14eV = 2.14 × 1.6 × 10^-19 J = 3.424 × 10^-19 J, V_O = 0.60 V,h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, e = 1.6 × 10^-19C
(i) Φ = hv₀
∴ The threshold frequency, v₀ = \(\frac{\phi}{h}\)
= \(\frac{3.424 \times 10^{-19}}{6.63 \times 10^{-34}}\) = 5.164 × 10¹⁴ Hz

(ii) V_Oe = 0.6 × 1.6 × 10^-19 = 0.96 × 10^-19 J
\(\frac{h c}{\lambda}\) – Φ = V_Oe ∴ \(\frac{h c}{\lambda}\) = Φ + V_Oe
∴ λ = \(\frac{h c}{\phi+V_{\mathrm{O}} e}\)

This is the required wavelength of the incident light.

Question 19.
The threshold wavelength for photoemission from silver is 3800 Å. Calculate the maximum kinetic energy in eV of photoelectrons emitted when ultraviolet radiation of wavelength 2600 Å falls on it. Also calculate the corresponding stop-ping potential. [1 eV = 1.6 × 10^-19 J]
Solution:
Data : λ₀ = 3800 A = 3.8 × 10^-7 m,
λ = 2600 Å = 2.6 × 10^-7 m, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J∙s, 1 eV = 1.6 × 10^-19 J
(i) According to Einstein’s photoelectric equation, the maximum kinetic energy of photoelectrons emitted,

Question 20.
When a surface is irradiated with light of wavelength 4950 Å, a photocurrent appears. The current vanishes if a retarding potential greater than 0.6 V is applied across the phototube. When a different source of light is used, it is found that the critical retarding potential is 1.1 V. Find the work function of the emitting surface and the wavelength of light from the second source.
Solution :
Data : λ₁ = 4.95 × 10^-7 m, V_O1 = 0.6 V, V_O2 = 1.1 V, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, e = 1.6 × 10^-19 C

∴ The wavelength of light from the second source,

Question 21.
The work function for the surface of aluminium is 4.2 eV. What potential difference will be required to stop the most energetic electrons emitted by light of wavelength 2000 Å? What should be the wavelength of the incident light for which the stopping potential is zero?
Solution:
Data: Φ = 4.2eV, λ₁ = 2 × 10^-7 m,V_O2 = 0,
h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, e = 1.6 × 10^-19 C

Question 22.
Radiation of wavelength 3000 Å falls on a metal surface having work function 2.3 eV. Calculate the maximum speed of ejected electrons.
Solution:
Data : λ = 3000 Å = 3 × 10^-7 m, h = 6.63 × 10^-34 J∙s, Φ = 2.3 eV = 2.3 × 1.6 × 10^-19 J, c = 3 × 10⁸ m/s, m = 9.1 × 10^-31 kg

Question 23.
If the total energy of radiation of frequency 10¹⁴ Hz is 6.63 J, calculate the number of photons in the radiation. [Planck’s constant = 6.63 × 10^-34 J∙s]
Solution:
Data: v = 10¹⁴ Hz, h = 6.63 × 10^-34 J∙s
The energy of a photon in a radiation of frequency v is hv.
∴ E = nhv,
so that the number of photons in the radiation is
n = \(\frac{E}{h v}\)
= \(\frac{6.63}{\left(6.63 \times 10^{-34}\right)\left(10^{14}\right)}\) = 10²⁰

Question 31.
Explain wave-particle duality of electromagnetic radiation.
Answer:
A particle is an object with a definite position in space at a given instant and having mass (or momentum), while a wave is a periodically repeated pattern in space and time, generally described by its velocity of propagation, wavelength and amplitude. It is a characteristic of a wave that it is not localized, i.e., it is spread over a region. Thus, these two concepts are contradictory and classical physics treats particles and waves as separate.

Under suitable circumstances, light and all other types of electromagnetic radiation exhibit typical ‘ wave phenomena like polarization, interference and diffraction. On the other hand, radiation exhibits a particle-like nature when it interacts with matter, as in the photoelectric effect and the Compton effect (scattering of X-rays by electrons in matter). It is emitted or absorbed only in terms of quanta of energy. This is the concept of photon : a particle with energy E = hv, where v is the frequency of the radiation and Planck’s constant h connects v and E, respectively the wave and particle aspects.

We see, therefore, that radiation exhibits a dual character. The synthesis of these two contradictory descriptions is called wave-particle duality of electromagnetic radiation.
[Notes : (1) Arthur Holly Compton (1892-1962), US physicist, discovered the effect, now known as the Compton effect, in 1923. (2) Planck’s constant h is also called the elementary quantum of action. Like e and c, it is one of the fundamental constants of nature.]

Question 32.
What is Compton effect? State the formula for the Compton shift and obtain its maximum value.
Answer:
When a high energy X-ray photon or γ-ray photon is scattered by an electron that is (almost) free, the photon loses energy and the electron gains energy shown in figure. This effect was discovered by A.H. Compton in 1923. It is now known as the Compton effect. This effect exhibits particle nature of electro-magnetic radiation.

If X is the wavelength of the incident photon, λ is the wavelength of the scattered photon, θ is the angle through which the photon is scattered, m₀ is the rest mass of an electron, c is the speed of light in free space and h is Plank’s constant, then, the wavelength shift, called the Compton shift is given by

[Note : In a collision between a low energy photon and a high energy electron, scattering results in loss in the energy of the electron and gain in the energy of the photon. This effect is known as the inverse Compton effect.]

Question 33.
What is the implication of Einstein’s interpretation of the photoelectric effect?
OR
What is the significance of the photoelectric effect?
Answer:
The phenomena of interference and polarization exhibit the wave nature of light, and James Clerk Maxwell (1831 – 79), British physicist, had established by 1865 that light is, and propagates as, an electromagnetic wave.

In his interpretation of the photoelectric effect in 1905, Einstein proposed that electromagnetic radiation behaves as a series of small packets or quanta of energy, later called photons. If the frequency of radiation is v, each photon has energy hv and momentum hv/c, where c is the speed of light in free space. Einstein’s photoelectric equation was verified experimentally by Robert Andrews Millikan (1868-1953), US physicist, in 1916.

A very strong additional evidence in support of the quantum theory of radiation was the discovery (in 1923) and explanation of the inelastic scattering of X-rays or γ-rays by electrons in matter by Arthur Holly Compton (1892-1962), US physicist. This inelastic scattering in which a photon transfers part of its energy to an electron is known as the Compton effect. It is similar to the Raman effect. The Compton effect shows particle nature of electro-magnetic radiation.

Since energy and momentum are considered in classical physics as characteristic properties of particles, the photoelectric effect and Compton effect exhibit the particle nature of radiation. But, to describe the photon energy, the quantum theory needs the frequency of the radiation, which is necessarily an attribute associated with a wave in classical physics. Thus, radiation exhibits the dual, seemingly contradictory, characters of particle and wave. In an experiment, we need to use only one of the descriptions, not both at the same time.

[Note : The momentum p and energy £ of a photon are related by the equation, p = E/c, where c is the speed of light in free space.]

Question 34.
Give a brief summary of the quantum theory of radiation.
OR
What is the photon picture of electromagnetic radiation?
Answer:
Quantum theory of radiation (The photon picture of electromagnetic radiation) :
(1) In its interaction with matter, electromagnetic radiation behaves as particles or quanta of energy. A quantum of energy is called a photon.

(2) If the frequency of radiation is v, irrespective of the intensity of radiation, each photon has energy hv and momentum hv/c, where c is the speed of light in free space.

(3) Intensity of radiation corresponds to the number of photons incident per unit time per unit surface area.

(4) Photons are electrically neutral and have zero rest mass.

(5) A photonRarticle collision (such as a photon-electron collision) obeys (he principles of conservation of energy arid momentum. However, in such a collision, an incident photon may be absorbed and/or a new photon may be created, so that the number of photons may not be conserved. For example, a γ-ray photon of energy greater than 1.02 MeV can produce an electron-positron pair in the presence of a heavy nucleus such as lead. In this case, the photon disappears and two particles (electron and positron) are produced. The total energy and momentum are conserved.

[Note: Photons have unit spin. Photons are influenced by gravitational field. A gravitational field can change the path and/or frequency/wavelength of a photon. Even after more than a century of its introduction, the concept of photon is not fully understood.]

Question 35.
What is the momentum of a photon of energy 3 × 10^-19 J? [c = 3 × 10⁸ m/s]
Answer:
Momentum of a photon = \(\frac{E}{c}=\frac{3 \times 10^{-19}}{3 \times 10^{8}}\)
= 10^-27 kg∙m/s .

Question 36.
What is the momentum of a photon of wave length 3.315 × 10^-7 m? [h = 6.63 × 10^-34J∙s]
Answer:
Momentum of a photon \(\frac{h}{\lambda}=\frac{6.63 \times 10^{-34}}{3.315 \times 10^{-7}}\)
= 2 × 10^-27 kg∙m/s .

37. Solve the following :

Question 1.
Find the momentum of a photon if the wavelength of the radiation is 6630 Å.
Solution:
Data : λ = 6.63 × 10^-7 m, h = 6.63 × 10^-34 J∙s
Energy of a photon, E = hv
c = λv
The momentum of a photon,

Question 2.
Find the momentum of a photon of energy 3 eV.
Solution :
Data : e = 1.602 × 10^-19 C,
E = 3 eV = 3 × 1.6 × 10^-19 J, c = 3 × 10⁸ m/s
The momentum of the photon,
p = \(\frac{E}{c}=\frac{3 \times 1.6 \times 10^{-19}}{3 \times 10^{8}}\) = 1.6 × 10^-27 kg∙m/s

Question 3.
Find the energy of a photon with momentum 2 × 10^-27 kg∙m/s.
Solution :
Data : p = 2 × 10^-27 kg∙m/s
The energy of the photon,
E = pc = (2 × 10^-27)(3 × 10⁸) = 6 × 10^-19 J
= \(\frac{6 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV = 3.75 eV

Question 38.
What is a photocell or photoelectric cell?
Describe its construction and working with a neat labelled diagram.
Answer:
A photocell or photoelectric cell is a device in which light energy is converted into electrical energy by photoelectric effect.

Construction : One form of the photoelectric cell shown in figure consists of a highly evacuated or gas-filled glass tube, an emitter (cathode) and a collector (anode). The light enters through a quartz window W and falls on the semicylindrical cathode C coated with a photosensitive metal. The anode is in the form of a straight wire of platinum or nickel, coaxial with the cathode.

If the cell is required to respond to the visible part of the spectrum, the cathode is coated with potassium or rubidium and the quartz window is replaced by glass. If the UV radiation only is to be used, cadmium is used as the sensitive surface. The cell is either highly evacuated (for accurate photometry) or filled with an inert gas at low pressure (if a larger current is desired).

Working : A photocell is connected in series with a battery and a variable resistance. The collector is kept at a positive potential with respect to the emitter. When UV radiation or visible light of frequency greater than the threshold frequency for the emitter surface is incident on the emitter, the ejected photoelectrons are focused by the cylindrical emitter (cathode) towards the collector (anode).

The photoelectrons collected by the collector constitute a photocurrent which may be measured by a microammeter in series with the photocell, as in an exposure meter or lux meter. Otherwise, the photocurrent is used to operate a relay circuit as in an alarm, or to drive the coils of a speaker as in reading an optical sound track in a cine film. The photocurrent becomes zero when the incident light is cut off.

Question 39.
State any four applications of a photoelectric cell.
OR
Explain any two applications of photoelectric effect.
Answer:
Applications of a photoelectric cell :
(1) In an exposure meter used for photography: A photographic film must be exposed to correct amount of light which, for a given film speed and lens aperture, depends on the exposure time. An exposure meter consists of a photocell, battery and microammeter connected in series. When the meter is directed towards an object, light reflected by the object enters the photocell and the photocurrent is directly proportional to the intensity of this light.
Usually, the microammeter scale is calibrated to read the exposure time directly.

(2) As a lux meter : A lux meter is used to measure the illumination and is similar in working to an exposure meter, except that the scale is calibrated to read the illumination in lux.

(3) In a burglar alarm as a ‘normally closed’ light- activated switch : It consists of a photocell, battery, relay system and a small directed light source. The radiation from the source falls on the photocell. If the light beam is interrupted by an intruder, the photoelectric current stops. This activates the relay system which sets off an alarm.

(4) In an optical reader of sound track in a cine film : The sound track of a cine film is recorded on one side of the positive film that is run in a cinema hall. The track consists of a dark wavy patch modulated by the recorded sound. Light from the projector lamp also passes through the sound track and falls on a photocell behind. The photocurrent is proportional to the transmitted light intensity and changes according to the recorded sound wave. The photocurrent is amplified and is used to drive the loudspeaker.

(5) A photocell can be used to switch on or off street lights.

Question 40.
Name any two instruments in which photo-electric effect is used.
Answer:
Exposure meter used in photography and lux meter.

Question 41.
State the de Broglie hypothesis and the de Broglie equation.
Answer:
De Broglie hypothesis : Louis de Broglie (1892-1987), French physicist, proposed (in 1924) that the wave-particle duality may not be unique to light but a universal characteristic of nature, so that a particle of matter in motion also has a wave or periodicity associated with it which becomes evident when the magnitude of Planck’s constant h cannot be ignored.

De Broglie equation : A particle of mass m moving with a speed v should under suitable experimental conditions exhibit the characteristics of a wave of wavelength
λ = \(\frac{h}{m v}=\frac{h}{p}\)
where p = mv = momentum of the particle.
The relation λ = h/p is called the de Broglie equation, and the wavelength λ associated with a particle momentum is called its de Broglie wavelength. The corresponding waves are termed as matter waves or de Broglie waves or Schrodinger waves.

[Note: This hypothesis was revolutionary at that time and accepted by others because Einstein supported it. Erwin Schrodinger (1887-1961), Austrian physicist, formulated a wave equation for matter waves.]

Question 42.
Explain the concept of de Broglie waves or matter waves.
Answer:
According to de Broglie, a particle of mass m moving with a speed v should, under suitable experimental conditions, exhibit the characteristics of a wave of wavelength
λ = \(\frac{h}{m v}=\frac{h}{p}\) … ………. (1)
where p = mv ≡ momentum of the particle and h is Planck’s constant.

This dual character of matter contained in Eq. (1) is usually referred to as the wave nature of matter or matter waves. They are a set of waves that represent the behaviour of particles under appropriate conditions. It does not, however, mean that the particles themselves are oscillating in space.

Interpretation of matter waves by Max Born (1882-1970), German bom British physicist, is that they are waves of probability, since the square of their amplitude at a given point is linked to the likelihood of finding the particle there. Hence, the wavelength λ may be regarded as a measure of the degree to which the energy is localized. If λ is exceedingly small, the energy is very localized and the particle character of the object is dominant. On the other hand, if λ is very large, the energy is distributed over a large volume; under these circumstances, the wave behaviour is dominant.

The wave nature of material particles such as the electron, neutron and helium atom has been established experimentally beyond doubt.

Question 43.
Derive an expression for the de Broglie wavelength associated with an electron accelerated from rest through a potential difference V. Consider the nonrelativistic case.
Answer:
Consider an electron accelerated from rest through a potential difference V. Let v be the final speed of the electron. We consider the nonrelativistic case, v << c, where c is the speed of light in free space. The kinetic energy acquired by the electron is
\(\frac{1}{2}\) mv² = \(\frac{1}{2m}\) (mv)² = eV ………. (1)
where e and m are the electronic charge and mass (nonrelativistic).

Therefore, the electron momentum,
p = mv = \(\sqrt{2 m e V}\) ………… (2)
The de Broglie wavelength associated with the electron is
λ = \(\frac{h}{p}\) ………….. (3)
where h is Planck’s constant.
From Eqs. (2) and (3),
λ = \(\frac{h}{\sqrt{2 m e V}}\)
Equation (4) gives the required expression.
[Note : Substituting the values of h = 6.63 × 10^-34 J-s, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C in Eq. (4), we obtain λ = \(\sqrt{150 / V}\) × 10^-10 m = \(\sqrt{150 / V}\) Å = \(12.25 / \sqrt{V}\) Å, where V is in volt. Therefore, electrons accelerated from rest through 150 volts have a de Broglie wavelength of 1 Å. This corresponds to the X-ray region of the electromagnetic spectrum.]

Question 44.
Derive an expression for the de Broglie wavelength.
Answer:
For the particle-like aspects of electromagnetic radiation, we consider radiation to consist of particles whose motion is governed by the wave propagation properties of certain associated waves.

To determine the wavelength of such waves, consider a beam of electromagnetic radiation of frequency v whose quanta have energy E.
E = hv
where h is the Planck constant.
For a quantum of radiation of momentum p, by Einstein’s theory,
E = pc
where c is the speed of propagation of the radiation in free space.
∴ pc = hv
∴ p\(\frac{c}{v}\) = h
The wavelength X of the associated wave governing the motion of the quanta is given by the relation
λ = c/v.
∴ pλ = h ∴ λ = \(\frac{h}{p}\)
V ’
This is the required expression.

Question 45.
What is the de Broglie wavelength associated with a particle having momentum 10^-26 kg∙m/s? [h = 6.63 × 10^-34 J∙s]
Answer:
The de Broglie wavelength associated with the particle,
λ = \(\frac{h}{p}=\frac{6.63 \times 10^{-34}}{10^{-26}}\) = 6.63 × 10^-8 m

Question 46.
With a neat labelled diagram, describe the Davisson and Germer experiment in support of the concept of matter waves.
Answer:
Davisson and Germer experiment (1927) :
The experimental arrangement, as shown in below figure, consists of an electron gun, a crystal holder and an electron detector enclosed in a vacuum chamber. In the electron gun, electrons emitted by a heated metallic filament (cathode) are accelerated by a potential difference V between the cathode and the anode, and emerge through a small hole in the anode. The electron gun directs a narrow collimated beam of electrons at a nickel crystal. Scattered electrons are detected by a movable detector.

The angle Φ between the incident and scattered beams is the scattering angle. Polar graphs of the number of scattered electrons as a function of angle Φ are plotted for different values of the accelerating voltage.

It is found that the electrons are scattered at a certain angle more than at others. Also, the number of scattered electrons in this direction is maximum for a certain kinetic energy of the incident electrons.

The detector registered a maximum at a scattering angle Φ = 50° for V = 54 V from figure. This electron diffraction can be understood only on the basis of de Broglie’s matter wave model. The de Broglie wavelength of the electrons accelerated from rest through a p.d. of 54 V is λ = \(\sqrt{150 / 54}\) Å = 1.67 Å
The wavelength calculated from the diffraction effect is 1.65 Å, nearly 1.67 Å.
[ Note : Clinton Joseph Davisson (1881 -1958), US physicist. Lester Halbert Germer (1896-1971), US physicist.]

47. Solve the following

Question 1.
(2) Find the momentum of the electron having de Brogue wavelength of 0.5 Å.
Solution:
Data: λ = 0.5Å = 5 × 10^-11 m, h = 6.63 × 10^-34 J∙s
The momentum of the electron, .
p = \(\frac{h}{\lambda}=\frac{6.63 \times 10^{-34}}{5 \times 10^{-11}}\) = 1.326 × 10^-23 kg∙m/s

Question 2.
A cracker of mass M at rest explodes in two parts of masses m₁ and m₂ with non-zero velocities. Find the ratio of the de Broglie wavelengths of the two particles.
Solution:
The cracker has zero momentum before explosion. By the principle of conservation of momentum, after the explosion,

Question 3.
Calculate the de Brogue wavelength of a proton if it is moving with the speed of 2 × 10⁵ m/s. [m^p = 1.673 × 10^-27 kg]
Solution:
Data: m^p = 1673 × 10^-27 kg, v = 2 × 10⁵ m/s, h = 6.63 × 10^-34 J∙s
De Brogue wavelength, λ = \(\frac{h}{p}=\frac{h}{m v}\)
∴ λ = \(\frac{6.63 \times 10^{-34}}{\left(1.673 \times 10^{-27}\right)\left(2 \times 10^{5}\right)}\)
= 1.981 × 10^-12 m

Question 4.
Calculate the de Brogue wavelength of an electron moving with \(\frac{1}{300}\) of the speed of light in vacuum. [Take m (electron) = 9.11 × 10^-28 g]
Solution:

Question 5.
Find the de Broglie wavelength of a dust particle of radius 1 μm and density 2.5 g/cm³ drifting at 2.2 m/s. (Take π = 3.14)
Solution:
Data : r = 1 μm = 10^-6 m, h = 6.63 × 10^-34 J∙s, ρ = 2.5 g/cm³ = 2.5 × 10³ kg/m³, v = 2.2 m/s,

Question 6.
Find the de Broglie wavelength associated with a car (mass = 1000 kg) moving at 20 m/s.
Solution:
Data : m = 1000 kg, v = 20 m/s,h = 6.63 × 10^-34 J∙s
The de Broglie wavelength,
λ = \(\frac{h}{p}=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{(1000)(20)}\) = 3.315 × 10^-38 m

Question 7.
What is the de Broglie wavelength of an electron accelerated from rest through 25000 volts ?
Solution:
Data: V = 25 × 10³ V, e = 1.6 × 10^-19 C, m_e = 9.11 × 10^-31 kg, h = 6.63 × 10^-34 J∙s
Kinetic energy of the electron,
E = eV
=(1.6 × 10^-19)(25 × 10³ V)
=4 × 10^-15 j
The momentum of the electron,

I Note : Here, the kinetic energy of the electron, 25 keV, is far less than the electron’s rest mass energy (m0c²) which is about 0.51 MeV. Hence, it is a nonrelativistic case.]

Question 8.
Find the de Broglie wavelength of a proton accelerated from rest by a potential difference of 50 V. [m_p = 1.673 × 10^-27 kg]
Solution:
Data : m_p = 1.673 × 10^-27 kg, h = 6.63 × 10^-34 J∙s, KE = 50 eV = 50 × 1.6 × 10^-19 J = 8 × 10^-18 J
The kinetic energy of the proton,

= 4.053 × 10^-12 m = 0.04053 A

Question 9.
A moving electron and a photon have the same de Brogue wavelength. Show that the electron possesses more energy than that carried by the photon.
Solution:
The de Brogue wavelength, λ = \(\frac{h}{p}\)
If an electron and a photon have the same de Brogue wavelength, they must have the same momentum, p.
For the photon, E_p = hv = \(\frac{h c}{\lambda}=\left(\frac{h}{\lambda}\right) c\) = pc … (1)
For the electron, mass m = \(\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}}\)
where m₀ is the rest mass of the electron and y is its speed.
∴ \(m^{2}\left(\frac{c^{2}-v^{2}}{c^{2}}\right)=m_{0}^{2}\)
∴ m²c⁴ – m²v²c² = \(m_{0}^{2} c^{4}\)
∴ (m²c⁴ = (m₀c²)² + p²c² (where p = mv)
∴ \(E_{\mathrm{e}}^{2}\) = (m₀c²)² + p²c²
where E_e = mc² = m₀c² + K is the total energy of the electron, m₀c² being he rest mass energy and K, the kinetic energy.
∴ E_e = \(\sqrt{\left(m_{0} c^{2}\right)^{2}+p^{2} c^{2}}\) …………. (2)
From Eqs. (1) and (2), we have E_e > E_p.
[Note : The result m = \(\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}}\) was obtained by Einstein in 1905.]

Multiple Choice Questions

Question 1.
The energy of a photon of wavelength λ is

Answer:
(D) \(\frac{h c}{\lambda}\)

Question 2.
The number of photoelectrons emitted
(A) varies inversely with the frequency of radiation
(B) varies directly with the frequency of radiation
(C) varies inversely with the intensity of radiation
(D) varies directly with the intensity of radiation.
Answer:
(D) varies directly with the intensity of radiation.

Question 3.
A metal emits no electrons if the incident light energy falls below certain threshold. For photo-emission, you would decrease
(A) the intensity of light
(B) the frequency of light
(C) the wavelength of light
(D) the collector potential.
Answer:
(C) the wavelength of light

Question 4.
When light of wavelength 5000 Å falls on a metal surface whose photoelectric work function is 1.9 eV, the kinetic energy of the most energetic photoelectrons is
(A) 0.59 eV
(B) 1.39 eV
(C)1.59eV
(D)2.59eV.
Answer:
(A) 0.59 eV

Question 5.
The threshold wavelengths for photoemission of two metals A and B are 300 nm and 600 nm, respectively. The ratio Φ_A/Φ_B of their photoelectric work functions is
(A) \(\frac{1}{4}\)
(B) \(\frac{1}{2}\)
(C) 2
(D) 4.
Answer:
(C) 2

Question 6.
The photoelectric threshold wavelength of a certain metal is 3315 Å. Its work function is
(A) 6 × 10^-19 J
(B) 7.286 × 10^-19 J
(C) 9 × 10^-19 J
(D) 9.945 × 10^-19 J.
Answer:
(A) 6 × 10^-19 J

Question 7.
The photoelectric work function of a certain metal is 2.5 eV. If the metal is separately irradiated with photons of energy 3 eV and 4.5 eV, the ratio of the respective stopping potentials is
(A) 1
(B) \(\frac{2}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{1}{5}\)
Answer:
(B) \(\frac{2}{3}\)

Question 8.
Sodium and copper have photoelectric work functions 2.3 eV and 4.7 eV, respectively. The ratio λ_0Na/λ_oCu 0f the threshold wavelengths for photoemission is about
(A) 1 : 4
(B) 1 : 2
(C) 2 : 1
(D) 4 : 1.
Answer:
(C) 2 : 1

Question 9.
When light of wavelength A falls on the cathode of a photocell, the kinetic energy of the most energetic photoelectrons emitted is £. If light of wavelength λ/2 is used, what can be said about the new value E’?
(A) E’ = E/2
(B) E’ = E
(C) E’ = 2E
(D) E’ > 2E.
Answer:
(D) E’ > 2E.

Question 10.
Electrons are ejected from a metallic surface when light with a wavelength of 6250 Å is used. If light of wavelength 4500 Å is used instead,
(A) there may not be any photoemission
(B) the photoelectric current will increase
(C) the stopping potential will increase
(D) the stopping potential will decrease.
Answer:
(C) the stopping potential will increase

Question 11.
UV radiation of energy 6.2 eV falls on molybdenum surface whose photoelectric work function is 4.2 eV. The kinetic energy of the fastest photoelectrons is
(A) 3.2 × 10^-19 J
(B) 3.52 × 10^-19 J
(C) 6.72 × 10^-19 J
(D) 9.92 × 10^-19 J.
Answer:
(A) 3.2 × 10^-19 J

Question 12.
In a photocell, increasing the intensity of light increases
(A) the stopping potential
(B) the photoelectric current
(C) the energy of the incident photons
(D) the maximum kinetic energy of the photo-electrons.
Answer:
(B) the photoelectric current

Question 13.
In a photocell, doubling the intensity of the incident light (v > v₀) doubles the
(A) stopping potential
(B) threshold frequency
(C) saturation current
(D) threshold wavelength.
Answer:
(C) saturation current

Question 14.
In the usual notation, the momentum of a photon is
(A) hvc
(B) \(\frac{h v}{c}\)
(C) \(\frac{h \lambda}{c}\)
(D) hλc.
Answer:
(B) \(\frac{h v}{c}\)

Question 15.
The momentum of a photon with λ = 3315 Å is
(A) 2 × 10^-27 kg∙m/s
(B) 5 × 10^-27 kg∙m/s
(C) 2 × 10^-41 kg∙m/s
(D) 5 × 10^-41 kg∙m/s.
Answer:
(A) 2 × 10^-27 kg∙m/s

Question 16.
Let p and E denote the linear momentum and energy of emitted photon, respectively. If the wavelength of incident radiation is increased,
(A) both p and E decrease
(B) p increases and E decreases
(C) p decreases and E increases
(D) both p and E decrease.
Answer:
(C) p decreases and E increases

Question 17.
When radiations of wavelength λ₁ and λ₂ are incident on a certain photosensitive material, the energies of electron ejected are E₁ and E₂ respectively, such that E₁ > E₂. Then, Planck’s constant h is [c = speed of light]

Answer:
(C) \(\frac{\left(E_{1}-E_{2}\right) \lambda_{1} \cdot \lambda_{2}}{c\left(\lambda_{2}-\lambda_{1}\right)}\)

Question 18.
If the frequency of incident light falling on a photosensitive material is doubled, then the kinetic energy of the emitted photoelectron will be
(A) the same as its initial value
(B) two times its initial value
(C) more than two times its initial value
(D) less than two times its initial value.
Answer:
(C) more than two times its initial value

Question 19.
The kinetic energy of emitted photoelectrons is independent of
(A) the frequency of incident radiation
(B) the intensity of incident radiation
(C) the wavelength of incident radiation
(D) the collector plate potential.
Answer:
(B) the intensity of incident radiation

Question 20.
In a photon-electron collision
(A) only total energy is conserved
(B) only total momentum is conserved
(C) both total energy and total momentum are conserved
(D) both total momentum and total energy are not conserved.
Answer:
(C) both total energy and total momentum are conserved

Question 21.
The de Broglie equation for the wavelength of matter waves is

Answer:
(A) λ = \(\frac{h}{p}\)

Question 22.
The momentum associated with a photon is given by
(A) hv
(B) \(\frac{h v}{c}\)
(C) hE
(D) hλ
Answer:
(B) \(\frac{h v}{c}\)

Question 23.
The momentum of a photon of de Broglie wave-length 5000 Å is [h = 6.63 × 10^-34 J∙s]
(A) 1.326 × 10^-28 kg∙m/s
(B) 7.54 × 10^-28 kg∙m/s
(C) 1.326 × 10^-27 kg∙m/s
(D) 7.54 × 10^-27 kg∙m/s.
Answer:
(C) 1.326 × 10^-27 kg∙m/s

Question 24.
The de Broglie wavelength of a 100-m sprinter of mass 66 kg running at a speed of 10 m/s is about
[h = 6.63 × 10^-34 J∙s]
(A) 10^-34 m
(B) 10^-33 m
(C) 10^-32 m
(D) 10^-31 m.
Answer:
(C) 10^-32 m

Question 25.
Which of the following particles moving with the same speed has the longest de Broglie wavelength?
(A) Proton
(B) Neutron
(C) α-particle
(D) β-particle
Answer:
(D) β-particle

Question 26.
If p and E are respectively the momentum and energy of a photon, the speed of the photon is given by
(A) p∙E
(B) E/p
(C) (E/p)²
(D) \(\sqrt{E / p}\)
Answer:
(B) E/p

Question 27.
If the kinetic energy of a free electron is doubled, its de Broglie wavelength
(A) decreases by a factor of 2
(B) increases by a factor of 2
(C) decreases by a factor of \(\sqrt {2}\)
(D) increases by a factor of \(\sqrt {2}\).
Answer:
(C) decreases by a factor of \(\sqrt {2}\)

Question 28.
The de Broglie wavelength of an a-particle accelerated from rest through a potential difference V is λ. In order to have the same de Broglie wavelength, a proton must be accelerated from rest through a potential difference of .
(A) V
(B) 2V
(C) 4V
(D) 8V.
Answer:
(D) 8V.

Question 29.
If a photon has the same wavelength as the de Broglie wavelength of an electron, they have the same
(A) velocity
(B) energy
(C) momentum
(D) angular momentum.
Answer:
(C) momentum

Question 35.
The de Broglie wavelength of a grain of sand, of mass 1 mg, blown by a wind at the speed of 20 m/s is [h = 6.63 × 10^-34 J∙s]
(A) 33.15 × 10^-36m
(B) 33.15 × 10^-33 m
(C) 33.15 × 10^-30 m
(D) 33.15 × 10^-30 m.
Answer:
(C) 33.15 × 10^-30 m

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 6 Superposition of Waves Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 6 Superposition of Waves

Question 1.
What is a progressive wave?
Answer:
A progressive wave or wave motion is a periodic or oscillatory disturbance in a medium or in a vacuum which is propagated without any damping and obstruction from one place to another at a finite speed.
[Note: A progressive wave is also called a traveling wave.]

Question 2.
Define :

1. transverse
2. longitudinal progressive wave.

Answer:

1. A progressive wave in which the vibration of the individual particles of the medium is perpendicular to the direction of propagation of the wave is called a transverse progressive wave.
2. A progressive wave in which the vibration of the individual particles of the medium is along the line of propagation of the wave is called a longitudinal progressive wave.

Question 3.
What is a mechanical wave ? Explain.
Answer:
A mechanical wave is a wave motion in a material medium.

Such a wave originates in the displacement of some portion of an elastic medium from its normal position. This causes the layers of matter to oscillate about their equilibrium positions. Because of the elastic properties of the material, the disturbance is transmitted from one layer to the next and so the waveform progresses through the medium.

Question 4.
What is a simple harmonic progressive wave ?
Answer:
A simple harmonic progressive wave is a periodic disturbance in a medium or in vacuum which propagates at a finite speed and in which the vibrations of the particles of the medium, as in a mechanical wave, or the oscillations of the electric and magnetic fields as in an electromagnetic wave are simple harmonic.

Question 5.
Define the following physical quantities related to a progressive wave :
(1) wave speed
(2) frequency (n)
(3) wavelength
(4) amplitude
(5) period
(6) wave number.
Answer:
(1) Wave speed : The distance covered by a progressive wave per unit time is called the wave speed.
(2) Frequency : The number of waves that pass per unit time across a given point of the medium is called the frequency of the wave.
[Note : It is equal to the number of vibrations per unit time made by a particle of the medium.]
(3) Wavelength : Wavelength is the distance between consecutive particles of the medium which are moving in exactly the same way at the same time and have the same displacement from their equilibrium positions.
[Note : Such particles are said to be in the same phase (the same state of vibration).]
(4) Amplitude : The magnitude of the maximum displacement of a particle of the medium from its equilibrium position is called the amplitude of the wave.
(5) Period : The time taken for a complete wave (one wavelength long) to pass a given point in the medium is called the period of the wave.
[Note : It is equal to the periodic time of the vibrational motion of a particle of the medium.]
(6) Wave number : The number of waves present per unit distance is called the wave number.

Question 6.
Write the equation of a progressive wave travelling along the positive x-direction.
Answer:
A progressive wave travelling along the positive x-direction is given by
y(x, t) = A sin (kx – ωt)
where A is the amplitude of the wave, k is the wave number and co is the angular frequency.
[Note : y(x, t) = A sin (kx – ωt), y(x, t) = A sin (ωt – kx), y(x, t) = A cos (kx – ωt), y(x, t) = A cos (ωt – kx) also represent a progressive wave travelling in the positive x-direction. Hence, any one of them can be used. y(x, t) can be written simply as y.]

Question 7.
Write the equation of a progressive wave travelling along the negative x-direction.
Answer:
A progressive wave travelling along the negative x-direction is given by y(x, t) = A sin (kx + ωt)
where A is the amplitude of the wave, k is the wave number and ω is the angular frequency.

Question 8.
Express the equation of a simple harmonic progressive wave in different forms.
Answer:
A simple progressive wave travelling along the positive x-direction is given by y = A sin (ωt – kx) … (1)
where A is the amplitude of the wave, k is the wave number and co is the angular frequency.

Frequency of vibrations, n = \(\frac{1}{T}\), Eq. (2) can be written as
y = A sin 2π(\(\overline{\mathrm{T}}\) – \(\frac{x}{\lambda}\))
Equations (1), (2), (3), (4), (5) and (6) are the different forms of the equation of a simple harmonic progressive wave.

Question 9.
A simple harmonic progressive wave is given by y = A sin (ωt – kx), where the symbols have their usual meaning. What is

1. the particle velocity at a point x and time t
2. the wave speed ?

Answer:

1. Particle velocity, \(\frac{d y}{d t}\) = ωA cos (ωt – kx)
2. Wave speed, v = \(\frac{\omega}{k}\).

Question 10.
A simple harmonic progressive wave has frequency 25 Hz and wavelength 4 m. If the phase difference between motions of two particles is (π/10) rad, what is the corresponding path difference?
Answer:
Path difference = \(\frac{\lambda}{2 \pi}\) × phase difference
= \(\frac{4 \mathrm{~m}}{2 \pi \mathrm{rad}}\) × \(\frac{\pi}{10}\) rad = 0.2 m

Question 11.
A simple harmonic progressive of frequency 100 Hz and wavelength 0.5 m travels through a medium. If the path difference between two points in the path of the wave is 0.1 m, what is the corresponding phase difference ?
Answer:
Phase difference = \(\frac{2 \pi}{\lambda}\) × path difference
= \(\frac{2 \pi}{0.5 \mathrm{~m}}\) × 0.1 m = 0.4 rad

Question 12.
The displacement of a particle of a medium when sound wave propagates is represented by y = A cos (ax + bt) where A, a and b are positive constants. The wave is reflected by an obstacle situated at x = 0.

1. What is the wavelength and frequency of the incident wave ?
2. Write the equation of the reflected wave.

Answer:
The incident sound wave is represented by y = A cos (ax + bt) where A, a and b are positive constants. The equation of a progressive wave of amplitude A, wavelength λ and frequency n = ω/2π, travelling along the negative direction of the x-axis is
y = A cos (ωt + kx),
where k = 2π/λ is the propagation constant. Comparing the two equations, ω = b and k = a.

1. Therefore, the wavelength of the incident wave,
   λ = \(\frac{2 \pi}{k}\) = \(\frac{2 \pi}{a}\) and its frequency, n = \(\frac{\omega}{2 \pi}\) = \(\frac{b}{2 \pi}\)
2. The equation of a progressive wave travelling along the positive direction of the x-axis is
   y = A cos (ωt – kx)
   ∴ The equation of the reflected wave is
   y = A cos (bt – ax).

Question 13.
Solve the following :

Question 1.
A simple harmonic progressive wave travels along a string. The time for a particle of the string to move from maximum displacement to zero is 0.004 s. What are the period and frequency of the wave ? If the wavelength is 1.2 m, what is the wave speed?
Solution :
Data : t = 0.004 s, λ = 1.2 m
The period T of the wave = the periodic time of the vibrational motion of the particle of the string = 4t = 4 × 0.004 = 0.016
∴ The frequency of the wave,
n = \(\frac{1}{T}\) = \(\frac{1}{0.016}\) = 62.5 Hz
The wave speed, v = nλ = 1.2 × 62.5 = 75 m/s

Question 2.
Write the equation of a simple harmonic progressive wave of amplitude 0.05 m and period 0.04 s travelling along the positive x-axis with a velocity of 12.5 m/s.
Solution:
Data : A = 0.05 m, T = 0.04 s, v = 12.5 m/s
∴ Equation of the wave travelling in the positive direction of the x-axis is

Question 3.
A simple harmonic progressive wave is given by the equation y = 0.1 sin 4π (50t – 0.1 x), in SI units. Find the amplitude, frequency, wavelength and speed of the wave.
Solution :
Data : y = 0.1 sin 4π (50t – 0.1 x)
= 0.1 sin 2π (100t – 0.2 x)
= 0.1 sin 2π( 100t – \(\frac{x}{5}\))
Let us compare this equation with that of a simple harmonic progressive wave.
∴ y = A sin 2π(nt – \(\frac{x}{2}\)) = 0.1 sin2π(100t – \(\frac{x}{5}\))
Comparing the quantities on both sides, we get,

1. amplitude (A) = 0.1 m
2. frequency (n) = 100 Hz
3. wavelength (λ) = 5 m
4. speed (v) = nλ = 100 × 5 = 500 m/s

Question 4.
The equation of a transverse wave on a stretched string is y = 0.2 sin 2π(\(\frac{t}{0.02}\) – \(\frac{x}{20}\)) where distances are in metre and time in second.
Find the

1. amplitude
2. frequency
3. speed of the wave.

Solution:
Let us compare the given equation with the equation of a simple harmonic progressive wave :

Comparing the quantities on both sides, we get,
A = 0.2 m, T = 0.02 s, λ = 20 m
∴

1. Amplitude (A) = 0.2 m
2. Frequency (n) = \(\frac{1}{T}\) = \(\frac{1}{0.02}\) = 50 Hz
3. Speed (v) = nλ – 50 × 20 = 1000 m/s

Question 5.
The equation of a simple harmonic progressive wave is y = 0.4 sin 100π (t – \(\frac{x}{5}\)) where all quantities are in SI units. Calculate the

1. wavelength
2. speed of the wave.

Solution:

Comparing the quantities on both sides, we get,
A = 0.4 m, n = 50Hz and v = 40m/s
∴

1. Speed (v) =40 m/s
2. Wavelength (λ) = \(\frac{v}{n}\) = \(\frac{40}{50}\) = 0.8 m

Question 6.
The equation of a simple harmonic progressive wave is given by y = 0.05 sin π (20t – \(\frac{x}{6}\)), where all quantities are in SI units. Calculate the displacement of a particle at 5 m from the origin and at the instant 0.1 second. Also find the phase difference between two particles separated by 5 m.
Solution :
Data : x = 5 m, t = 0.1 s
(i) The displacement of the particle,

Comparing the two sides, we get, λ = 12 m
The phase difference between two points separated by x = 5 m is given by

Question 7.
A sound wave of amplitude 0.2 cm, frequency 1000 Hz and wavelength 0.31 m is travelling in air. Calculate the displacement of the particle at 3.1 m from the origin after 1.004 s. What would be the phase difference for two positions of the vibrating particle after an interval of 0.001s?
Solution :
Data : A = 0.2 cm = 0.002 m, n = 1000 Hz, λ = 0.31 m, x = 3.1 m, t = 1.004 s, t₂ – t₁ = 0.001 s
(i) The displacement of the particle,
y = A sin 2π(nt – \(\frac{x}{\lambda}\))
= 0.002 sin 2π(1000 × 1.004 – \(\frac{3.1}{0.31}\))
= 0.002 sin 2π (1004 – 10)
= 0.002 sin 2π(994) = 0 [or y = 0 metre]

(ii) Phase difference = 2πn (t₂ – t₁)
= 2π × 1000 × 0.001 = 2π radians

Question 8.
The equation of a simple harmonic progressive wave is given by y = 4 sin π(\(\frac{t}{0.02}\) – \(\frac{x}{75}\)). Find the displacement and velocity of a particle at 50 cm from the origin and at 0.1 second. (All quantities are expressed in CGS units.)
Solution :

The displacement of the particle, y = 3.464 cm = 3.464 × 10^-2 m
The velocity of the particle,

Question 14.
What is meant by reflection of a wave ?
Answer:
When a wave travelling in a medium is incident on a boundary with another medium, a part of it returns into the original medium with a change in its direction of propagation while a part of it is transmitted into the second medium. The phenomenon in which a part of the wave is returned into the original medium with reduction in its intensity and energy is called reflection.

Question 15.
Explain the reflection of transverse waves at the surface of
(1) a denser medium
(2) a rarer medium.
OR
Explain the effect on the phase of transverse waves reflected from
(1) a denser medium
(2) a rarer medium.
Answer:
(1) Reflection of transverse waves from a denser medium : Suppose that a crest of a transverse wave travels along a string and is incident on the surface of a denser medium such as a rigid wall at point B, as shown in below figure. As the crest cannot travel further, it is reflected.

Since point B is fixed, its displacement is always zero. Therefore, the crest must be reflected in such a way, that the displacement at B due to the reflected

wave is exactly equal in magnitude and opposite in direction to that due to the incident wave. Therefore, the crest is reflected as a trough. Hence, there is a phase change of 180° or π radians when transverse waves are reflected from a denser medium.

(2) Reflection of transverse waves from a rarer medium : In this case, the particles of the medium are free to vibrate. Hence,

1. there is no change of phase
2. a crest is reflected as a crest and a trough is reflected as a trough.

Question 16.
Consider a heavy string X and a light string Y joined together at point O. Explain what happens when a wave pulse
(1) travelling from the string X reaches the junction O
(2) travelling from the string Y reaches the junction O.
Answer:
The tension in both strings is the same. Hence, the junction O is a discontinuity between string X of greater linear density than string Y because the wave speed is less on X than on Y.

(1) When a pulse travelling on the heavy string X reaches O, the light string Y gets pulled upwards. Thus the pulse, gets partially transmitted and partially reflected as a crest, as shown in below figure. However, the amplitude of transmitted pulse is greater than that of the reflected pulse.

(2) When a pulse travelling on a light string Y reaches O, the heavier string X pulled slightly upwards. Thus, the pulse is partly transmitted as a crest but the reflected part is inverted as a trough, as shown in below figure. Here, the amplitude of transmitted pulse is smaller than that of the reflected pulse.

[Note : In either case, the relative heights of the reflected and transmitted pulses depend on the relative densities of the two strings. Obviously, if the strings are identical, there is no discontinuity at the boundary and no reflection takes place.]

Question 17.
A heavy string X is joined to a light string Y at point O. How will a pulse get reflected
(1) travelling on the string X towards O
(2) travelling on the string Y towards O ?
Answer:
(1) A pulse travelling on a heavy string will reflect without inversion at its boundary with a lighter string. Thus, a crest will reflect as a crest and a trough will reflect as a trough.

(2) When a pulse travelling on a light string encounters a boundary with a heavier string, the reflected pulse is inverted. Thus, a crest will reflect as a trough and vice versa.

Question 18.
Explain the reflection of sound waves (i.e., longitudinal waves) at the surface of
(1) a denser medium
(2) a rarer medium.
OR
Explain the effect on the phase of longitudinal waves reflected from
(1) a denser medium
(2) a rarer medium.
Answer:
(1) Reflection of a longitudinal wave from a denser medium : Consider a sound wave incident on a denser medium such as a rigid wall. When a compression is incident on the wall, the particles of air close to the wall are in a compressed state. To return to their normal condition, the particles begin to press in the opposite direction and therefore a compression gets reflected as a compression and a rarefaction is reflected as a rarefaction. However, the displacements of the particles in the reflected wave are opposite to their displacements in the incident wave, so that there is a change of phase of 180° or π radians.

(2) Reflection of a longitudinal wave from a rarer medium : When sound waves are reflected from

the surface of a rarer medium, there is no change of phase. Therefore, a compression is reflected as a rarefaction and vice versa. The reason is as follows :
When a compression is incident on the surface of a rarer medium, it can pass into that medium. This is because the particles of the rarer medium are free to move and they get compressed, leaving a rarefaction behind, which travels in the opposite direction. In a similar manner an incident rarefaction gets reflected as a compression.

Question 19.
State the principle of superposition of waves.
Answer:
Principle of superposition of waves : The displacement of a particle at a given point in space and time due to the simultaneous influence of two or more waves is the vector sum of the displacements due to each wave acting independently.

Notes :

1. The principle of superposition is applicable to all types of waves.
2. The phenomena of interference, beats, formation of stationary waves, etc. are based on the principle of superposition of waves.

Question 20.
Explain the superposition of two wave pulses of equal amplitude and same phase moving towards each other.
OR
Explain constructive interference when two wave pulses of equal amplitude and same phase are superimposed.
Answer:
Consider two wave pulses of the same amplitude and phase moving towards each other, as shown in below figure.

When the two pulses cross each other, stages (b) to (d), the resultant displacement is equal to the vector sum of the displacements due to the individual pulses (shown by dashed lutes). Therefore, the resultant displacement at that point becomes maximum. This phenomenon is called constructive interference. After crossing each other, both the pulses continue to propagate with their initial amplitude.

Question 21.
Explain the superposition of two wave pulses of equal amplitude and opposite phase moving towards each other.
OR
Explain destructive interference when two wave pulses of equal amplitude and opposite phase are superimposed.
Answer:
Consider two wave pulses of the same amplitude but opposite phase moving towards each other, as shown in below figure.

When the two pulses cross each other, stages (b) to (d), the resultant displacement is equal to the vector sum of the displacements due to the individual pulses (shown by dashed lines). Therefore, the resultant displacement at that point becomes minimum, equal to zero, because the individual pulses are exactly 180° out of phase.

[Notes: Music lovers often find many types of ambient sounds that interfere with the sounds coming through their headphones. Active noise-cancelling headphones not only block out some high frequency sound waves but also actively cancel out lower-frequency sound waves by destructive interference. They actually create sound waves with the same amplitude that mimic the incoming noise but with inverted phase (also known as antiphase), i.e., exactly 180° out of phase to the original noise. This inverted signal (in antiphase) is then amplified and a transducer creates a sound wave directly proportional to the amplitude of the original waveform, creating destructive interference. This effectively reduces the volume
of the perceivable noise.]

Question 22.
Derive an equation for the resultant wave produced due to superposition of two waves. Hence, state the expression for the amplitude of the resultant wave when two waves are
(1) in phase
(2) out of phase.
Answer:
Consider two waves of the same frequency, different amplitudes A₁ and A₂ and differing in phase by φ. Let these two waves interfere at x = 0.
The displacement of each wave at x = 0 are
y₁ = A₁ sin ωt
y₂ = A₂ sin (ωt + φ)

According to the principle of superposition of waves, the resultant displacement at that point is
y₁ = y₁ + y₂
= A₁ sin ωt + A₂ sin (ωt + φ)
Using the trigonometrical identity,
sin (C + D) = sin C cos D + cos C sin D,
y = A₁ sin ωt + A₂ ωt cos φ + A₂ cos ωt sin φ
y = (A₁ + A₂ cos φ) sin ωt + A₂ sin φ cos ωt … (1)
Let (A₁ + A₂ cos φ) = A cos θ … (2)
and A₂ sin φ = A sin θ … (3)
Substituting Eqs. (2) and (3) in EQ. (1), we get the equation of the resultant wave as
y = A cos θ sin ωt + A sin θ cos ωt = A sin (ωt + θ) … (4)
It has the same frequency as that of the interfering waves. The amplitude A of the resultant wave is given by squaring and adding Eqs. (2) and (3).

Thus, the amplitude of the resultant wave is maximum when the two interfering waves are in phase.

Case (2) : When the two interfering waves are out of phase, ivarphi = ipi. Then, the amplitude of the resultant wave is,

Thus, the amplitude of the resultant wave is minimum when the two interfering waves are in opposite phase.

Question 23.
What is the relation between the amplitude of a wave and its intensity?
Answer:
The intensity of a wave is proportional to the square of its amplitude.

Question 24.
Two interfering waves of the same frequency are out of phase but have different amplitudes A₁ and A₂. What can you say about the intensity of the resultant wave ?
Answer:
The two interfering waves are out of phase. Thus, the amplitude and hence the intensity of the resultant wave is minimum, I_min ∝ (A_min)² where (A_min)² = (A₁ – A₂)².

Question 25.
Two interfering waves of the same frequency are in phase but have different amplitudes A₁ and A₂. What can you say about the intensity of the resultant wave ?
Answer:
The two waves interfere in phase. Thus, the amplitude and hence the intensity of the resultant wave is maximum, I_max ∝ (A_min)² where (A_max)² = (A₁ + A₂)².

Question 26.
What is a stationary wave? Why is it called stationary?
Answer:
When two progressive waves having the same amplitude, wavelength and speed, travel through the same region of a medium in opposite directions, their super-position under certain conditions creates a stationary interference pattern called as a stationary or standing wave.

It is called stationary because the resultant harmonic disturbance of the particles does not travel in any direction and there is no transport of energy in the medium.

Question 27.
Define :

1. transverse stationary wave
2. longitudinal stationary wave.

Answer:

1. When two identical transverse progressive waves travelling in opposite directions along the same line superimpose, the resultant wave produced is called a transverse stationary wave.
2. When two identical longitudinal progressive waves superimpose, the resultant wave produced is called a longitudinal stationary wave.

Question 28.
When stationary waves of wavelength 40 cm are formed in a medium, what is the distance between

1. successive nodes
2. a node and the next antinode?

Answer:

1. 20 cm
2. 10 cm.

Question 29.
The equation of a stationary wave is y = 0.04 cos \(\frac{2 \pi x}{0.6}\) sin 2π (100t) with all quantities in SI units. What is the length of one loop ?
Answer:
Comparing the given equation with

∴ Length of one loop = \(\frac{\lambda}{2}\) = 0.3 m.

Question 30.
What is the speed of the waves superposed ? For data, see Question 29.
Answer:
n = 100 Hz. v = nλ = 100 × 0.6 = 60 m/s.

Question 31.
What is the maximum speed of a particle at an antinode ? For data, see Question 29.
Answer:
v_max = 2A(2πn) = 0.04 × 2π × 100 = 8π m/s.

Question 32.
Distinguish between progressive waves and stationary waves.
Answer:
Progressive and stationary waves :

Question 33.
Solve the following :

Question 1.
A sound wave of frequency 1000 Hz and travelling with speed 340 m/s is reflected from the closed end of the tube. At what distance from that end will the adjacent node occur?
Solution :
Data : n = 1000 Hz, v = 340 m/s
The wavelength of the stationary wave set up in the tube, λ = \(\frac{v}{n}\).
The distance between successive nodes

Question 2.
Two simple harmonic progressive waves are represented by y₁ = 2 sin 2π(100 t – \(\frac{x}{60}\)) cm and y₂ = 2 sin 2π(100t + \(\frac{x}{60}\)) cm. The waves combine to form a stationary wave. Find

1. the amplitude at an antinode
2. the distance between adjacent node and antinode
3. the loop length
4. the wave speed.

Solution :

we get, λ = 60 cm and n = 100 Hz. Therefore,

1. the amplitude at an antinode, | 2A | = 4 cm
2. the distance between adjacent node and antinode \(\frac{\lambda}{4}\) = \(\frac{60}{4}\) = 15 cm
3. the loop length = \(\frac{\lambda}{2}\) = \(\frac{60}{2}\) cm = 30 cm
4. the wave speed = nλ= 100 × 60 = 6000 cm/s

Question 3.
The equation of a standing wave is given by y = 0.02 cos (πx) sin (100 πt) m. Find the amplitude of either wave interfering, wavelength, time period, frequency and wave speed of interfering waves.
Solution :
Data : y = 0.02 cos (πx) sin (100 πt) m
Comparing this equation with
y = 2A cos\(\left(\frac{2 \pi x}{\lambda}\right)\) sin (2π nt)
we get for either interfering waves,

1. the amplitude, | A | = \(\frac{0.02}{2}\) = 0.01 m
2. the wavelength, λ = 2 m
3. the time period, T = \(\frac{1}{n}\) = \(\frac{1}{50}\)s = 0.02 s
4. the frequency, n = 50 Hz
5. the wave speed = nλ = 50 × 2 = 100 m/s

Question 34.
Explain
(1) free vibrations
(2) forced vibrations.
Answer:
(1) Free vibrations : A body capable of vibrations is said to perform free vibrations when it is disturbed from its equilibrium position and left to itself.

In the absence of dissipative forces such as friction due to surrounding air and internal forces, the total energy and hence the amplitude of vibrations of the body remains constant. The frequencies of the free vibrations of a body are called its natural frequencies and depend on the body itself.

In the absence of a maintaining force, in practice, the total energy and hence the amplitude decreases due to dissipative forces and the vibration is said to be damped. The frequency of damped vibrations is less than the natural frequency.

(2) Forced vibrations : The vibrations of a body in response to an external periodic force are called forced vibrations.

The external force supplies the necessary energy to make up for the dissipative losses. The frequency of the forced vibrations is equal to the frequency of the external periodic force.

The amplitude of the forced vibrations depends upon the mass of the vibrating body, the amplitude of the external force, the difference between the natural frequency and the frequency of the periodic force, and the extent of damping.

Question 35.
Define resonance.
Answer:
Resonance : If a body is made to vibrate by an external periodic force, whose frequency is equal to the natural frequency (or nearly so) of the body, the body vibrates with maximum amplitude. This phenomenon is called resonance.
The corresponding frequency is called the resonant frequency.

For low damping, the amplitude of vibrations has a sharp maximum at resonance, as shown in above figure. The flatter curve without a pronounced maximum is for high damping.

Question 36.
Distinguish between
(1) free vibrations and resonance
(2) forced vibrations and resonance (Two points of distinction).
Answer:
(1) Free vibrations and resonance:

(2) Forced vibrations and resonance :

Question 37.
Give any two applications of resonance.
Answer:

1. A radio or TV receiver set is tuned to the frequency of the desired broadcast station by adjusting the resonant frequency of its electrical oscillator circuit.
2. The speed of sound at room temperature can be determined by making the air column of a resonance tube resonate with a vibrating tuning fork of known frequency.
3. The frequency of a tuning fork can be determined by making a sonometer wire, of known mass per unit length and under known tension, resonate with the vibrating fork.
4. The amplitude of the oscillations of a child on a swing is increased by pushing with a frequency equal to the natural frequency of the swing.

Question 38.
Give any two disadvantages of resonance.
Answer:

1. A column of soldiers marching in regular step on a narrow and structurally flexible bridge can set it into dangerously large amplitude oscillations. The bridge may even collapse at the resonance.
2. Structural resonance of a suspension bridge induced by the winds can lead to its catastrophic collapse. Several early suspension bridges were destroyed by structural resonance induced by modest winds.
3. Vibrations of a motor or engine can induce resonant vibrations in its supporting structures if their natural frequency is close to that of the vibrations of the engine. A common example is the rattling sound of a bus body when the engine is left idling. Vibrations in an aircraft are caused by the engine and the aerodynamic effects. The vibrations cause metal fatigue, especially in the fuselage, wings and tail, and eventually lead to metal fracture.
4. Every ship has a natural period of rolling (side to side oscillation about an axis along its length). If the ship encounters a series of waves such that the wave period matches the rolling, it will have no time righting itself before the next wave strikes. Resonant conditions can occur when the combination of wave period, vessel speed and heading with respect to the waves lead to an encounter close to the natural roll period of the vessel. This situation, if not corrected, can lead to severe rolling, with roll angle exceeding 15°.
   Large containerships are particularly vulnerable to rolling. Possible consequences are loss of containers, machinery failure, structural damage and even capsizing of the ship. The speed and direction of the ship can be changed to avoid the consequences of synchronous rolling.

Question 39.
What are overtones? What is the meaning of first overtone ?
Answer:
The higher allowed harmonics above the first harmonic or fundamental are called overtones.
The first overtone is the higher allowed harmonic immediately above the first harmonic.

Question 40.
Distinguish between harmonics and overtones.
Answer:

Question 41.
What is end correction ? State the cause of end correction. How is it estimated ?
Answer:
When sound waves are sent down the air column in a narrow closed or open pipe, they are reflected at the ends-without phase reversal at an open end and with a phase reversal at a closed end. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column. Thus, the stationary waves have an antinode at an open end.

However, because air molecules in the plane of an open end are not free to move in all directions, reflection of the longitudinal waves takes place slightly beyond the rim of the pipe at an open end. The distance of the antinode from the open end of the pipe is called end correction. According to Reynolds, the distance of the antinode from the rim is approximately 30% of the inner diameter of a cylindrical pipe. This distance must be taken into account in accurate determination of the wavelength of sound. Hence, this distance is called the end correction.

Therefore, if d is the inner diameter of a cylindrical pipe, an end correction e = 0.3 d for each open end must be added to the measured length of the pipe. If l is the measured length, the effective length of the air column in the case of a pipe closed at one end is l + 0.3d, while that for a pipe open at both ends is l + 0.6 d.

Question 42.
What are the frequencies of the notes produced in an open and closed pipes in terms of the length of pipe L and velocity of waves v?.
Answer:
The frequencies of all the harmonics present in an open pipe are

The frequencies of the odd harmonics present in a closed pipe are

where p = 0,1, 2, 3, … .

Question 43.
State the factors on which the fundamental frequency of air column in a pipe depends.
Answer:

1. Speed of sound in air
2. length of the pipe
3. diameter of the pipe.

Question 44.
The fundamental frequency of air column in a pipe closed at one end is 300 Hz. What is the frequency of the

1. second overtone
2. third harmonic ? (Ignore the end correction.)

Answer:
Closed pipe.

1. Second harmonic = fifth harmonic = 5 × 300 = 1500 Hz
2. Third harmonic = 3 × 300 = 900 Hz.

Question 45.
The fundamental frequency of air column in a pipe open at both ends is 200 Hz. What is the frequency of the

1. second harmonic
2. third overtone ? (Ignore the end correction.)

Answer:
Open pipe.

1. Second harmonic = 2 × 200 = 400 Hz
2. Third overtone = fourth harmonic
   = 4 × 200 = 800 Hz.

Question 46.
Stationary waves in the air column inside a pipe of length 50 cm and closed at one end have three nodes and three antinodes. What is the wavelength ?
Answer:
Here, L = 5\(\frac{\lambda}{4}\)
∴ Wavelength λ = \(\frac{4 L}{5}\) = \(\frac{4 \times 50 \mathrm{~cm}}{5}\) = 40 cm

Question 47.
Show that the fundamental frequency of vibration of the air column in a pipe open at both ends is double that of a pipe of the same length and closed at one end.
Answer:
Let L_O and L_C be the lengths of a pipe open at both ends and a pipe closed at one end, respectively. Let n_O and n_C be their corresponding fundamental frequencies. Then, ignoring the end corrections,
n_O = \(\frac{v}{2 L_{\mathrm{O}}}\) and n_C = \(\frac{v}{4 L_{\mathrm{C}}}\)
where v is the speed of the sound in air.
Given that L_O = L_C = L (say),
n_O = \(\frac{v}{2 L}\) and n_O = \(\frac{v}{4 L}\)
∴ n_O = 2\(\left(\frac{v}{4 L}\right)\) = \(2 n_{\mathrm{C}}\)

Question 48.
Prove that a pipe of length 2L open at both ends has the same fundamental frequency as a pipe of length L closed at one end.
Answer:
Let L_O and L_C be the lengths of a pipe open at both ends and a pipe closed at one end, respectively. Let n_O and n_C be their corresponding fundamental frequencies. Then, ignoring the end corrections,
n_O = \(\frac{v}{2 L_{\mathrm{O}}}\) and n_C = \(\frac{v}{4 L_{\mathrm{C}}}\)
where v is the speed of the sound in air.
Given that L_C = L and L_O = 2L,
n_O = \(\frac{v}{4 L}\) and n_C = \(\frac{v}{4 L}\) ∴ n_O = n_C

Question 49.
A pipe open at both ends has the fundamental frequency n. If the pipe is immersed vertically in water up to half its length, what would be the fundamental frequency of the resulting air column?
Answer:
Let L be the length of the pipe open at both ends whose fundamental frequency is n. Then, ignoring the end correction, n = \(\frac{v}{2 L}\)
where v is the speed of sound in air.

When the pipe is immersed vertically in water up to half its length, it becomes a pipe closed at one end with an air column of length L’ = L / 2. Then, its fundamental frequency n’ is

which is equal to n, the fundamental frequency of the open pipe.

Question 50.
The pth overtone of an organ pipe open at both ends has a frequency n. When one end of the pipe is closed, the qth overtone has a frequency N. Show that N = \(\frac{(2 q+1) n}{2(p+1)}\).
Answer:
Let L be the length of an organ pipe and v be the speed of sound in air.
When the pipe has both its ends open, the frequency of the pth overtone, ignoring the end correction, is (p + 1)\(\frac{v}{2 L}\).

When one end of the pipe is closed, the frequency of the qth overtone of a pipe of length L and closed at one end is (2q + 1)\(\frac{v}{4 L}\)

which is the required expression.

Question 51.
Two organ pipes open at both ends and of same length but different radii (or diameters) produce sounds of different frequencies. Why?
Answer:
Stationary waves formed in the air column of a pipe open at both ends have an antinode at each end. These antinodes are slightly beyond the rim of the pipe and an end correction of approximately 30% of the inner diameter must be added to the measured length of the air column for each open end.

Suppose two organ pipes, open at both ends and of same length 1, have inner diameters d₁, and d₂. Then, the effective lengths of the air columns are respectively L₁ = l + 0.6dt and L₂ = l + 0.6d₂. The fundamental frequencies of the corresponding air columns are

where v is the speed of sound in air. Thus, if d₁ and d₂ are different, n₁ and n₂ will also be different.

Question 52.
Two organ pipes closed at one end have the same diameters but different lengths. Show that the end correction at each end is e = \(\frac{n_{1} l_{1}-n_{2} l_{2}}{n_{2}-n_{1}}\), where the symbols have their usual meanings.
Answer:
Suppose two organ pipes, closed at one end and of the same inner diameter d, have lengths l₁ and l₂.
Then, the effective lengths of the air columns are respectively
L₁ = l₁ + 0.3d and L₂ = l₂ + e = l₂ + 0.3d
where e = 0.3d is the end correction for the open end.
The fundamental frequencies of the corresponding air columns are

where v j the speed of sound in air.
∴ v = 4n₁(l₁ + e) = 4n₁(l₂ + e)
∴ n₁I₁ + n₁e = n₂I₂ + n₂e
∴ n₁I₁ – n₂I₂ = (n₂ – n₁)e
∴ e = \(\frac{n_{1} l_{1}-n_{2} l_{2}}{n_{2}-n_{1}}\)
which is the required expression.

Question 53.
State any two limitations of end correction.
Answer:
Limitations of end correction :

1. Inner diameter of the tube must be uniform.
2. Effects of air flow and temperature outside the tube are ignored.
3. The prongs of the tuning fork should be perpendicular to the air column in the tube, with their tips at the centre of the tube and a small distance above the rim of the tube.

Question 54.
A tuning fork is in resonance with a closed pipe. But the same tuning fork cannot be in resonance with an open pipe of the same length. Why ?
Answer:
For the same length of air column, and the same speed of sound, the fundamental frequency of the air column in a closed pipe is half that in an open pipe. Hence, a tuning fork in unison with the air column in a closed pipe cannot be in unison with the air column of the same length in an open pipe.

Question 55.
Solve the following :

Question 1.
Calculate the fundamental frequency of an air column in a tube of length 25 cm closed at one end, if the speed of sound in air is 350 m/s.
Answer:
v = 350 m/s, L = 25 cm = 0.25 m
∴ The fundamental frequency of the air column is
n = \(\frac{v}{4 L}\) = \([\frac{350}{4 \times 0.25}]\) = 350 Hz

Question 2.
A pipe which is open at both ends is 47 cm long and has an inner diameter of 5 cm. If the speed of sound in air is 348 m/s, calculate the fundamental frequency of the air column in that pipe.
Solution :
Data : l = 47 cm = 0.47 m, d = 5 cm = 0.05 m, v = 348 m/s
e = 0.3 d = 0.3 × 0.05 = 0.015 m
As the tube is open at both ends, the corrected length (L) is
L = l + 2e = 0.47 + (2 × 0.015) = 0.5 m
∴ The fundamental frequency of the air column is
n = \(\frac{v}{2 L}\) = \(\frac{348}{2 \times 0.5}\) = 348 Hz

Question 3.
A tube open at both ends is 47 cm long. Calculate the fundamental frequency of the air column. (Ignore the end correction. Speed of sound in air is 3.3 × 10² m/s.)
Solution :
Data : L = 47 cm = 0.47 m, v = 330 m/s.
The fundamental frequency of the air column,
n = \(\frac{v}{2 L}\) = \(\frac{330}{2 \times 0.47}\) = \(\frac{165}{0.47}\) = 351.1 Hz

Question 4.
The speed of sound in air at room temperature is 350 m/s. A pipe is 35 cm in length. Find the frequency of the third overtone in the pipe when it is
(i) closed at one end
(ii) open at both ends. Ignore the end correction.
Solution :
Data : v = 350 m/s, L = 35 cm = 35 × 10^-2 m
(i) For a pipe closed at one end, the fundamental frequency is

As only odd harmonics are present in this case, the frequency of pth overtone is n_p = (p × 2 + 1) n_C
∴ The frequency of the 3rd overtone is
n₃ = (3 × 2 + 1)n_C = 7n_C = 7 × 250 = 1750 Hz

(ii) For a pipe open at both ends, the fundamental frequency is

In this case, all harmonics are present.
∴ The frequency of the pth overtone is
n_p = (p + 1) n_O
∴ The frequency of the 3rd overtone is n₃ = (3 + 1) n_O = 4n_O = 4 × 500 = 2000 Hz

Question 5.
Find the frequency of the fifth overtone of an air column vibrating in a pipe closed at one end. The length of the pipe is 42.10 cm and the speed of sound in air at room temperature is 350 m/s. The inner diameter of the pipe is 3.5 cm.
Solution :
Data : L = 42.10 cm = 0.4210 m, v = 350 m/s, d = 3.5 cm = 3.5 × 10^-2 m, pipe closed at one end

Question 6.
Determine the length of a pipe open at both ends which is in unison with a pipe of length 20 cm closed at one end, in the fundamental mode. Ignore the end correction.
Solution :
Let n_O and L_O be the fundamental frequency and length respectively, of the pipe open at both ends and let n_C and L_C be the corresponding values for a pipe closed at one end. If v is the speed of sound in air,
n_O = \(\frac{v}{2 L_{\mathrm{O}}}\) and n_C = \(\frac{v}{4 L_{\mathrm{C}}}\)
As the air columns in the two pipes vibrate in unison,
n_O = n_C
∴ 2L_O = 4L_C ∴ L_O = 2L_C
But L_C = 20cm, ∴ L_O = 2 × 20 = 40 cm

Question 7.
The fundamental frequency of an air column in a pipe closed at one end is in unison with the third overtone of an open pipe. Calculate the ratio of the lengths of their air columns.
Solution :
Pipe closed at one end : fundamental frequency,
n_C = \(\frac{v}{4 L_{C}}\)
column is 51.8 cm.
n_O = \(\frac{v}{2 L_{\mathrm{O}}}\)
In this case, the frequency of the third over-tone = \(\frac{4 v}{2 L_{\mathrm{O}}}\) = \(\frac{2 v}{L_{\mathrm{O}}}\)
By the data, \(\frac{v}{4 L_{C}}\) = \(\frac{2 v}{L_{\mathrm{O}}}\)
∴ \(\frac{L_{\mathrm{O}}}{L_{\mathrm{C}}}\) = 8 or \(\frac{L_{\mathrm{C}}}{L_{\mathrm{O}}}\) = \(\frac{1}{8}\)

Question 8.
The consecutive overtones of an air column closed at one end are 405 Hz and 675 Hz respectively. Find the fundamental frequency of a similar air column but open at both ends.
Solution:
For the air column closed at one end, let
L = the length of the air column,
n_C = the fundamental frequency,
n_q, n_{q + 1} = the frequencies of the qth and (q + 1)th overtones, where q = 1, 2, 3, …
Since only odd harmonics are present as overtones, n_q = (2q + 1)n_C and n_{q + 1} = [2(q + 1) + n]n_C
= (2q + 3)n_C
Data : n_q = 405 Hz, n_q+1 = 675 Hz

Solving for q, q = 1
Therefore, the two given frequencies correspond to the first and second overtones, i.e., the third and fifth harmonics.
∴ 3n_C = 405 Hz
∴ n_C = 135 Hz
This gives the fundamental frequency of the air column closed at one end.
The fundamental frequency (n_O) of an air column of same length but open at both ends is double that of the air column closed at one end (ignoring the end correction).
∴ n_O = 2n_C = 2 × 135 = 270 Hz
This gives the fundamental frequency of a similar air column but open at both ends.
Solution :
Data : n = 480 Hz, l₁ = 16.8 cm, l₂ = 51.8 cm
(1) The speed of sound in air is v = 2n (l₂ – l₁)
= 2 × 480 × (51.8 – 16.8) = 33600 cm/s = 336 m/s

(2) Let λ be the wavelength of sound waves and e be the end correction.
For the first resonance, l₁ + e = \(\frac{\lambda}{4}\) … (1)
For the second resonance, l₂ + e = \(\frac{3 \lambda}{4}\) … (2)
From Eq. (1), λ = 4(l₁ + e).
Substituting this value in Eq. (2), we get,

Question 9.
In a resonance tube experiment, a tuning fork v resonates with an air column 10 cm long and again resonates when it is 32.2 cm long. Calculate the wavelength of the wave and the end correction.
Solution:

This gives the wavelength of the wave. We have,

This gives the end correction.

Question 10.
The length of air column in a resonance tube for fundamental mode is 16 cm and that for second resonance is 50.25 cm. Find the end correction.
Solution:
Data: l₁ = 16 cm, l₂ = 50.25 cm
End correction,

Question 56.
State the formula for the speed of transverse waves on a stretched string (or wire). Hence obtain an expression for the fundamental frequency of the vibrating string (or wire).
Answer:

1. If a string (or a wire) stretched between two rigid supports is plucked at some point, the disturbance produced travels along the string in the form of transverse waves. If T is the tension applied to the string and m is the mass per unit length (i.e., linear density) of the string, the speed of the transverse waves is
   v = \(\sqrt{\frac{T}{m}}\)
2. The transverse waves moving along the string are reflected from the supports. The reflected waves interfere and under certain conditions set up stationary waves in the string. At each support, a node is formed.
3. The possible or allowed stationary waves are subject to the two boundary conditions that there must be a node at each fixed end of the string. The different ways in which the string can then vibrate are called its modes of vibration.
4. In the simplest mode of vibration, there are only two nodes (N), one at each end and an antinode (A) is formed midway between them, as shown in
   
   In this case, the distance between successive nodes is equal to the length of the string (L) and is equal to λ/2, where λ is the wavelength.
   ∴ L = \(\frac{\lambda}{2}\) or λ = 2L
   The frequency of vibrations is n = \(\frac{v}{\lambda}\)
   Substituting v = \(\sqrt{\frac{T}{m}}\) and λ = 2L in this relation, we get,
   n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
   This is the lowest frequency of the stationary waves on a stretched string and is called the fundamental frequency.

Question 57.
What is the minimum frequency with which a stretched string of length L, linear density m can vibrate under tension T?
Answer:
The minimum frequency with which a stretched string of length L, linear density m can vibrate under tension T is the fundamental frequency given by n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\).

Question 58.
The mass per unit length of a wire is 1 × 10^-4 kg/m and the tension in the wire is 25 N. What is the speed of the transverse waves on the wire?
Answer:

Question 59.
With neat labelled diagrams, explain the three lowest modes of vibration of a string stretched between rigid supports.
Answer:
Consider a string of linear density m stretched between two rigid supports a distance L apart. Let T be the tension in the string.

Stationary waves set up on the string are subjected to two boundary conditions : the displacement y = 0 at x = 0 and at x = L at all times. That is, there must be a node at each fixed end. These conditions limit the possible modes of vibration to only a discrete set of frequencies such that there are an integral number of loops p between the two fixed ends.

Since, the length of one loop (the distance between consecutive nodes) corresponds to half a wavelength (λ),

In the simplest mode of vibration, only one loop (p = 1) is formed. The corresponding lowest allowed frequency, n, given by
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\) … (4)
is called the fundamental frequency or the first harmonic. The possible modes of vibration with frequencies higher than the fundamental are called the overtones.

In the first overtone, two loops are formed (p = 2). Its frequency,
n₁ = \(\frac{2}{L} \sqrt{\frac{T}{m}}\) = 2n … (5)
is twice the fundamental and is, therefore, the second harmonic.

In the second overtone, three loops are formed (p = 3). Its frequency,
n₂ = \(\frac{3}{2 L} \sqrt{\frac{T}{m}}\) = 3n … (6)
is the third harmonic.

Question 60.
The speed of transverse waves on a vibrating string is 50 m/s. If the length of the string is 0.25 m, what is the fundamental frequency of vibration?
Answer:
Fundamental frequency, n = \(\frac{v}{(2 L)}\) = \(\frac{50}{(2 \times 0.25)}\)
= 100 Hz

Question 61.
State and explain the laws of vibrating strings.
Answer:
The fundamental frequency of vibration of a stretched string or wire of uniform cross section is 1
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
where L is the vibrating length, m the mass per unit length (linear density) of the string and T the tension in the string. From the above expression, we can state the following three laws of vibrating strings.

(1) Law of length : The fundamental frequency of vibrations of a stretched string is inversely proportional to its vibrating length, if the tension and mass per unit length are kept constant. If T and m are constant,
n ∝ \(\frac{1}{L}\) or nL = constant.

(2) Law of tension : The fundamental frequency of vibrations of a stretched string is directly proportional to the square root of the applied tension, if the length and mass per unit length are kept constant. If L and m are constant,
n ∝ or \(\sqrt{T}\) or n²/T = constant.

(3) Law of mass (or law of linear density) : The fundamental frequency of vibrations of a stretched string is inversely proportional to the square root of its mass per unit length, if the length and tension are kept constant. If L and T are constant,
n ∝ \(\frac{1}{\sqrt{m}}\) or n²m = constant.

Question 62.
How does the fundamental frequency of a vibrating string depend on the radius of cross section of the string and the mass density of the material of the string ?
Answer:
Consider a string stretched between two rigid supports a distance L apart. Let T be the tension in the string, r be its radius of cross section and p be the mass density of its material. Then, the mass of the string M = (πr²L)p, so that its linear density, i.e., mass per unit length, m = M/L = πr²p.

According to the law of mass of a vibrating string, the fundamental frequency (n) is inversely proportional to the square root of its linear density, when T and L are constant.

Question 63.
A string/wire is stretched between two rigid supports. State any two factors on which the fundamental frequency of the string/wire depends.
Answer:

1. Tension in the string/wire
2. length of the string/wire (or radius or mass per unit length or mass density of the material of the string/wire)

Question 64.
Why are strings of different thicknesses and materials used in a sitar or some other such instruments?
Answer:
The linear density of a string, m = πr²p, where r is the radius of cross section of the string and p is the mass density of its material. By the law of mass of a vibrating string, the frequency of vibrations n of the string is inversely proportional to \(\sqrt{m}\). Therefore, n ∝ \(\frac{1}{r}\) and n ∝\(\frac{1}{\sqrt{\rho}}\). Hence, the strings of different thicknesses and materials in a stringed musical instrument like sitar can be set to different scales.

Question 65.
If Y and ρ are Young’s modulus and mass density of the material of a stretched wire of length L, show that the fundamental frequency of vibration of the wire is n = \(\frac{1}{2 L} \sqrt{\frac{Y \Delta L}{\rho L}}\), where ∆L is the elastic extension of the wire.
Answer:
Consider a wire stretched between two rigid supports a distance L apart. Let T ≡ the tension in the wire, r ≡ the radius of cross section of the wire,
Y, ρ ≡ Young’s modulus and mass density of the material of the wire,
M,m ≡ the mass and linear density of the wire.
Then, M = (πr²L)ρ and m = \(\frac{M}{L}\) = πr²ρ … (1)
The stress in the wire = \(\frac{T}{\pi r^{2}}\)
∴ \(\frac{T}{m}\) = \(\frac{T}{\pi r^{2} \rho}\) = \(\frac{\text { stress }}{\rho}\) … (2)
The The fundamental frequency of vibration of the wire,

if ∆L is the elastic extension of the wire under tension T, strain = ∆L/L.

which is the required expression.

Question 66.
What is the linear density of a wire of mass density 8 g/cm³ and cross-sectional radius 0.05 mm ?
Answer:
Linear density of the wire = nr2p
= π(5 × 10^-3)² (8) = 2π × 10^-4 g/cm.

Question 67.
Stationary waves on a vibrating string of length 30 cm has three loops. What is the wavelength ?
Answer:

Question 68.
Write a short note on sonometer.
Answer:
A sonometer consists of a uniform wire stretched over a rectangular sounding box, and passes over

two movable bridges (or knife edges) and a pulley, in above figure. It works on the phenomenon of resonance. The tension in the wire is adjusted by adding weights to the hanger attached to the free end of the wire. The length of the wire between the movable bridges, L, is adjusted to vibrate in unison with a given timing fork either by beats method or by paper-rider method. L is called the vibrating length. First, the vibrating length is set to minimum and then gradually increased in small steps. In the beats method, the wire and the tuning fork are simultaneously set into vibrations for each vibrating length. Beats can be heard when the two frequencies are very close. Then, a finer adjustment of the wire is needed so that no beats are heard. This is when the two are in unison.

For the paper-rider method, a small light paper in the form of Λ is placed on the wire at its centre. The stem of the vibrating timing fork is gently pressed on the sonometer box. The vibrating length is gradually increased from minimum till the paper rider vibrates and thrown off. Because, when the wire resonates with the tuning fork at its lowest fundamental mode, the wire vibrates with maximum amplitude and the centre of the wire is an antinode. Hence, the paper rider is thrown off.

A sonometer is used to determine the frequency of a tuning fork and to verify the laws of vibrating strings.

Question 69.
Explain the use of a sonometer to verify
(i) the law of length
(ii) the law of tension
(iii) the law of linear density.
Answer:
(i) Verification of law of length : According to this law, n ∝ \(\frac{1}{L}\), if T and m are constant. To verify this
law, the sonometer wire of given linear density m is kept under constant tension T. The length of the wire is adjusted for the wire to vibrate in unison with tuning forks of different frequencies n₁, n₂, n₃,… . Let L₁, L₂, L₃, … be the corresponding resonating lengths of the wire. It is found that, within experimental errors, n₁L₁ = n₂L₂ = n₃L₃ = …. This implies that the product, nL = constant, which vertifies the law of length.

(ii) Verification of law of tension : According to this law, n ∝ \(\sqrt{T}\), if L and m are constant. To verify this law, the vibrating length L of the sonometer wire of given linear density m is kept constant.

A set of tuning forks of different frequencies is used. The tension in the wire is adjusted for the wire to vibrate in unison with tuning forks of different frequencies n₁, n₂, n₃,….. . T₁, T₂, T₃, ….. corresponding tensions. It is found that, within experimental errors, \(\frac{n_{1}}{\sqrt{T_{1}}}\) = \(\frac{n_{2}}{\sqrt{T_{2}}}\) = \(\frac{n_{3}}{\sqrt{T_{3}}}\) = ….. This implies \(\frac{n_{1}}{\sqrt{T}}\) = constant which verifies the law of tension.

(iii) Verification of linear density : According to this law, n ∝ \(\frac{1}{\sqrt{m}}\), if T and L are constant. To verify this law, two wires having different linear densities m₁ and m₂ are kept under constant tension T.
A tuning fork of frequency n is used. The lengths of the wires are adjusted for the wires to vibrate in unison with the tuning fork. Let L₁ and L₂ be the corresponding resonating lengths of the wires. It is found that, within experimental errors, \(L_{1} \sqrt{m_{1}}\) = \(L_{2} \sqrt{m_{2}}\). This implies \(L \sqrt{m}\) = constant. According to the law of length of a vibrating string, n ∝ \(\frac{1}{L}\).
∴ n ∝ \(\frac{1}{\sqrt{m}}\) which verifies the law of linear density.

Question 70.
A stretched sonometer wire vibrates at 256 Hz. If its length is increased by 10%, without changing the tension in the wire, what will be the frequency of the wire ?
Answer:
L₂ = 1.1 L₁ = \(\frac{n_{2}}{n_{1}}\) = \(\frac{L_{2}}{L_{1}}\) ∴ \(\frac{n_{2}}{256}\) = 1.1
∴ n₂ = 256 × 1.1 = 281.6 Hz

Question 71.
Solve the following :

Question 1.
Find the speed of a transverse wave along a string of linear density 3.6 × 10^-3 kg/m, when it is under a tension of 1.8 kg wt.
Solution :
Data : m = 3.6 × 10^-3 kg/m, g = 9.8 m/s²
∴ T = 1.8 kg wt = 1.8 × 9.8N
The speed of transverse waves along the string is

Question 2.
The speed of a transverse wave along a uniform metal wire, when it is under a tension of 1000 g wt, is 68 m/s. If the density of the metal is 7900 kg/m³, find the area of cross section of the wire.
Solution :
Data : g = 9.8 m/s², T = 1000 g wt = 1 kg wt = 9.8 N,
V = 68 m/s, ρ = 7900 kg/m³

This gives the area of cross section of the wire.

Question 3.
A transverse wave is produced on a string 0.7 m long and fixed at its ends. Find the speed of the wave when it vibrates emitting the second overtone of frequency 300 Hz.
Solution :
Data : L = 0.7 m, n (second overtone) = 300 Hz
In this case, three loops are formed on the string.
∴ L = 3\(\frac{\lambda}{2}\)
∴ λ = \(\frac{2 L}{3}\) ∴ v = nλ = \(\frac{2}{3}\)nL
∴ v = \(\frac{2}{3}\) × 300 × 0.7 = 140 m/s
This gives the speed of the wave.

Question 4.
A uniform wire under tension is fixed at its ends. If the ratio of the tension in the wire to the square of its length is 360 dyn/cm² and the fundamental frequency of vibration of the wire is 300 Hz, find its linear density.
Solution :

Question 5.
A metal wire of length 20 cm and diameter 0.2 mm is stretched by a load of 2 kg wt. If the density of the material of the wire is 7.8 g/cm³, find the fundamental frequency of vibration of the wire.
Solution :
Data : L = 20 cm = 0.2 m, d = 0.2 mm, g = 9.8 m/s², T = 2 kg wt = 2 × 9.8 N = 19.6 N, ρ = 7.8 g/cm³ = 7.8 × 10³ kg/m³
∴ r = \(\frac{d}{2}\) = 0.1 mm = 0.1 × 10^-3 m
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\) and m = πr²ρ

The fundamental frequency of vibration of the wire is 707 Hz.

Question 6.
Two wires, each 1 m long and of the same diameter, have densities 8 × 10³ kg/m³ and 2 × 10³ kg/m³ and are stretched by tensions 196 N and 49 N, respectively. Compare their fundamental frequencies.
Solution :
Data : L₁ = L₂ = 1 m, d₁ = d₂ (∴ r₁ = r₂), ρ₁ = 8 × 10³ kg/m³, ρ₂ = 2 × 10³ kg/m³, T₁ = 196 N, T₂ = 49N

∴ Their fundamental frequencies are the same.

Question 7.
A uniform wire of length 49 cm and linear density 4 × 10^-4 kg/m is subjected to a tension of 28 N. Determine its frequency for

1. the fundamental mode
2. the second harmonic
3. the third overtone.

Solution :
Data : L = 49 cm = 0.49 m, T = 28 N, m = 4 × 10^-4 kg/m

1. Fundamental frequency :
   
2. Second harmonic : 2n = 2 × 270 = 540 Hz
3. Third overtone : 4n = 4 × 270 = 1080 Hz

Question 8.
Two wires of the same material, having lengths in the ratio 2 : 1 and diameters in the ratio 3 :1 are subjected to tensions in the ratio 1 : 4. Find the ratio of their fundamental frequencies.
Solution :
Let n₁, L₁, T₁, m₁, r₁ and ρ₁ be the fundamental frequency, vibrating length, tension, mass per unit length, radius and density of the first wire respectively and let n₂, L₂, T₂, m₂, r₂ and ρ₂ be the corresponding quantities of the second wire.

Substituting these values in the above relation,
\(\frac{n_{1}}{n_{2}}\) = \(\frac{1}{2}\) × \(\frac{1}{3}\) × \(\sqrt{\frac{1}{4} \times 1}\) = \(\frac{1}{12}\)

Question 9.
A wire under a certain tension, gives a note of fundamental frequency 320 Hz. When the tension is changed, the frequency of the fundamental note rises to 480 Hz. Compare the tensions in the wire.
Solution :
Data : n₁ = 320 Hz, n₂ = 480 Hz Fundamental frequency is n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
As the length (L) of the wire and its mass per unit length (m) are kept constant,
n ∝ \(\sqrt{T}\)
∴\(\frac{n_{1}}{n_{2}}\) = \(\sqrt{\frac{T_{1}}{T_{2}}}\)
∴ The ratio of the tensions in the wire,
\(\frac{T_{1}}{T_{2}}\) = \(\left(\frac{n_{1}}{n_{2}}\right)^{2}\) = \(\left(\frac{320}{480}\right)^{2}\) = \(\left(\frac{2}{3}\right)^{2}\) = \(\frac{4}{9}\)

Question 10.
A transverse wave is produced on a stretched string 0.9 m long and fixed at its ends. Find the speed of the transverse wave, when the string vibrates while emitting second overtone of frequency 324 Hz.
Solution :
Data : L = 0.9 m, n (second overtone) = 324 Hz
For the second overtone of a vibrating string, λ = \(\frac{2}{3}\)L
The speed of the transverse wave formed on the string, v = nλ
∴ v = n × \(\frac{2}{3}\)L = 324 × \(\frac{2}{3}\) × 0.9
= 324 × 0.6 = 194.4 m/s

Question 11.
The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency of 160 Hz, where should the person press the string ?
Solution :
Data : L₁ = 80 cm n₁ = 112 Hz, n₂ = 160 Hz
According to the law of length, n₁L₁ = n₂L₂.
∴ The vibrating length to produce the fundamental frequency of 160 Hz,
L₂ = \(\frac{n_{1} L_{1}}{n_{2}}\) = \(\frac{112(80)}{160}\) = 56 cm

Question 12.
What should be the tension applied to a wire of length 1 m and mass 10 grams, if it has to vibrate with the fundamental frequency of 50 Hz ?
Solution :
Data : L = 1 m, mass of the wire = 10 g = 0.01 kg, n = 50 Hz
∴ m = mass per unit length of the wire = \(\frac{0.01}{1}\)
= 0.01 kg/m
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
∴ Tension = T = 4n²L²m
= 40(50)²(1)²(0.01)
= 4 × 2500 × 0.01 = 100 N

Question 13.
A sonometer wire of length 1 m weighing 2 g is subjected to a suitable tension. The vibrating length of the wire in unison with a tuning fork of frequency 512 Hz is 12 cm. If the vibrating length of the wire in unison with another fork under the same conditions is 12.8 cm, find the frequency of this fork.
Solution :
Data : L₁ = 12 cm, n₁ = 512 Hz, L₂ = 12.8 cm n₁L₁ = n₂L₂
∴ The frequency of the second fork,
n₂ = \(\frac{n_{1} L_{1}}{L_{2}}\) = \(\frac{512(12)}{12.8}\) = 480 Hz

Question 14.
A stretched sonometer wire emits a fundamental note of frequency 256 Hz. Keeping the stretching force constant and reducing the length of the wire by 10 cm, the frequency becomes 320 Hz. Calculate the original length of the wire.
Solution:
Data : n₁ = 256 Hz, T and m constant, L₂ = L₁ – 10 cm, n₂ = 320 Hz

∴ 5L₁ – 50 = 4L₁
∴ L₁ = 50 cm = 0.5 m

Alternative method :
Since T and m are constant, nL = constant.
∴ n₁L₁ = n₂L₂ ∴ \(\frac{L_{1}}{L_{2}}\) = \(\frac{n_{2}}{n_{1}}\)
∴ \(\frac{L_{1}}{L_{1}-10}\) = \(\frac{320}{256}\) = \(\frac{20}{16}\) = \(\frac{5}{4}\)
∴ 4L₁ = 5L₁ – 50
∴ 5L₁ – 4L₁ = 50
∴L₁ = 50cm = 0.5 m

Question 15.
A sonometer wire, 36 cm long, vibrates with a frequency of 280 Hz in the fundamental mode when it is under a tension of 24.5 N. Calculate the linear density of the material of the wire.
Solution :
Data: L = 36 cm = 0.36 m, n = 280 Hz, T = 24.5 N,

Question 16.
A sonometer wire 36 cm long, vibrates with a fundamental frequency of 280 Hz, when it is under tension of 24.5 N. Calculate mass per unit length of wire.
Solution :
Data : L = 36 cm = 0.36 m, n = 280 Hz, T = 24.5 N
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
∴ Linear density, m = \(\frac{T}{4 L^{2} n^{2}}\)
∴ m = \(\frac{24.5}{4(0.36)^{2}(280)^{2}}\)
= 6.0 × 10^-4 kg/m

Question 17.
The length of a sonometer wire between two fixed ends is 110 cm. Where should be the two bridges be placed so as to divide the wire into three segments, whose fundamental frequencies are in the ratio of 1: 2 : 3 ?
Solution :
Data : L₁ + L₂ + L₃ = 110 cm, n₁ : n₂ : n₃ = 1 : 2 : 3
According to the first law of length, n ∝ \(\frac{1}{L}\) if T and m are constant.
By given data, n₁ = 2n₂ = 3n₃
∴ \(\frac{1}{L_{1}}\) = \(\frac{2}{L_{2}}\) = \(\frac{3}{L_{3}}\)
∴ L₂ = 2L₁ and L₃ = 3L₁
L₁ + 2L₁ + 3L₁ = 110
∴ 6L₁ = 110
∴ L₁ = 18.3 cm
∴ L₂ = 2 × 18.3 = 36.6 cm
∴ L₃ = 3 × 18.3 = 54.9 cm
Therefore, the two bridge should be kept in such a way that the distance between them in 36.6 cm and distance of 1st bridge from the fixed end of the wire is 18.3 cm.

Question 72.
What are beats? Define
(1) the period of beats
(2) beat frequency (1 mark each)
Answer:
A periodic variation in loudness (or intensity) when two sound notes of slightly different frequencies are sounded
at the same time is called beats.

If two notes of slightly different frequencies n₁ and n₂ are played simultaneously, the resulting
note from their interference has a frequency of (n₁ + n₂)/2. However, the amplitude of this resulting note varies from the sum to the difference of the amplitudes of the two notes n₁ and n₂. An intensity maximum and an intensity minimum are respectively called waxing and waning. Thus, the resulting note will be heard as one of periodic loud (waxing) and faint (waning) sound. One waxing and one waning form one beat. Formation of beats is an example of interference in time.

The time interval between successive maxima or minima of sound at a given place is called the period of beats.
The number of beats produced per unit time is called the beat frequency.

Question 73.
Distinguish between stationary waves and beats. (Two points of distinction)
Answer:

Question 74.
Discuss analytically the formation of beats and show that
(1) the beat frequency equals the difference in frequencies of two interfering waves
(2) the waxing and waning occur alternately and with the same period.
OR
Explain the production of beats and deduce analytically the expression for beat frequency.
Answer:
Consider two sound waves of equal amplitude (A) and slightly different frequencies n₁ and n₂ (with n₁ > n₂) propagating through the medium in the same direction and along the same line. These waves can be represented by the equations y₁ = A sin 2πn₁t and y₂ = A sin 2πn₂t at x = 0, where y denotes the displacement of the particle of the medium from its mean position.

By the principle of superposition of waves, the resultant displacement of the particle of the medium at the point at which the two waves arrive simultaneously is the algebraic sum
y = y₁ + y₂ = A sin 2πn₁t + A sin 2πn₂t
Now, sin C + sin D

∴y = R sin 2πnt

The above equation shows that the resultant motion has amplitude IRI which changes periodically with time. The period of beats is the period of waxing (maximum intensity of sound) or the period of waning (minimum intensity of sound). The intensity of sound is directly proportional to the square of the amplitude of the wave. It is maximum (waxing) when R becomes maximum;

The intensity of sound is minimum (waning) when R = 0

From Eqs. (1) and (2), it can be seen that the waxing and waning occur alternately and with the same period.

Question 75.
The speed of sound in air under certain conditions is 350 m/s. If two sound waves of wavelengths \(\) m and \(\) m arrive at a point at the same time, what will be the beat frequency?
Answer:
Beat frequency = |n₁ – n₂|

Question 76.
State the conditions for hearing beats.
Answer:
Conditions for hearing beats : For two sound waves to interfere and give rise to beats,

1. they should travel in the same medium and arrive at the listener at the same time
2. their frequencies should not differ by more than about 7 Hz for distinct beats
3. their amplitudes should be equal or nearly so.

Question 77.
When two tuning forks of slightly different frequencies are sounded together to produce beats, what is the effect on the beat frequency if the prongs of the tuning fork with higher frequency are waxed a little or filed a little?
Answer:
Filing the prongs of a tuning fork reduces its mass and thereby raises its frequency. Applying a little wax to the prongs increases its mass and thereby reduces its frequency. Therefore, filing the prongs of the tuning fork of the higher frequency increases the beat frequency while applying a little wax to its prongs decreases the beat frequency.

Question 78.
When two tuning forks of slightly different frequencies are sounded together to produce beats, what is the effect on beat frequency if the prongs of the tuning fork with lower frequency is waxed a little or filed a little?
Answer:
Filing the prongs of a tuning fork reduces its mass and thereby raises its frequency. Applying a little wax to the prongs increases its mass and thereby reduces its frequency. Therefore, filing the prongs of the tuning fork of the lower frequency decreases the beat frequency while applying a little wax to its prongs increases the beat frequency.

Question 79.
A tuning fork has frequency 512 Hz. What can you say about its frequency when

1. its prongs are filed
2. some wax is applied to its prongs ?

Answer:

1. Its frequency will be more than 512 Hz.
2. Its frequency will be less than 512 Hz.

Question 80.
Explain any two applications of beats.
Answer:
Applications of beats :

(1) Listening for beats – or rather, their absence-is the usual method of tuning musical instruments and in the determination of the frequency of a musical note.

(2) Ultrasonic vocal sounds made by bats and dolphins may be detected by superimposing a sound of different frequency to produce audible beats.

(3) In music, beats are used to produce a low frequency sound (a grave tone). Two notes whose difference in frequency is equal to the desired low frequency are used for this purpose. When two notes are nearly in tune, the beats are slow. But as the beat frequency increases to 20 Hz or more, the beats may ultimately merge into a continuous tone known as a difference tone.

(4) (i) Speed of a moving object can be determined using a Doppler RADAR. Radio waves from the RADAR are reflected off a moving object, such as an aeroplane. The superposition of the incident and reflected waves produces beats. The frequency of beats helps to determine the speed of the aeroplane.
The same principle is used in speed guns used by traffic police to determine the speed of cars on a highway.

(ii) In medicine, a Doppler ultrasound test (sonography) uses reflected sound waves to evaluate blood flow through the major arteries and veins of the arms, legs and neck. It can show blocked or reduced blood flow because of narrowing of the major arteries. Duplex (or 2D) Doppler, Colour Doppler and Power Doppler are different techniques of the same test.

Notes: Some other applications of beats are as follows :

1. Detection of toxic gases inside mines, especially collieries : Air from inside a mine and pure air are blown through two separate identical organ pipes. If beats are heard it would indicate that the composition of air inside the mines is different from that outside. This can serve as an early warning system.
2. In music, consonance and dissonance depend upon the beats produced when two notes are sounded simultaneously. A beat frequency between 10 Hz and 50 Hz (between the fundamental notes being played as well as any of their overtones) is unpleasant and results in dissonance.
3. Superheterodyne reception of radio waves in most radio, television and radar receivers : A low-frequency signal produced in the receiver is beat against an incoming high-frequency radio signal to produce an intermediate (beat) frequency (IF). This IF signal retains the information of the incoming signal. The receiver can be tuned to different broadcast frequencies by adjusting the frequency of the low-frequency signal. The IF signal though can be kept the same in every case and can therefore be amplified with higher gain.]

Question 81.
If beat frequency is 10 Hz, what is the time interval between
(i) successive waxings
(ii) a waxing and subsequent waning of sound.
Answer:
Beat period = \(\frac{1}{\text { beat frequency }}\) = \(\frac{1}{10 \mathrm{~Hz}}\) = 0.1 s
Hence, the time interval will be 0.1 s in case (i) and 0.05 s in case (ii).

Question 82.
A sonometer wire of length L₁ is in unison with a tuning fork of frequency n. When the vibrating length of the wire is reduced to L₂, it produces x beats per second with the fork. Show that n = x₂.\(\frac{L_{2}}{L_{1}-L_{2}}\).
Answer:
The fundamental frequency of vibration of a wire of length L₁, mass per unit length m and under tension T is
n₁ = \(\frac{1}{2 L_{1}} \sqrt{\frac{T}{m}}\) = n … (1)
since it is in unison with a tuning fork of frequency n. When the vibrating length of the wire is L₂, its fundamental frequency is
n₂ = \(\frac{1}{2 L_{2}} \sqrt{\frac{T}{m}}\) … (2)
T and m remaining constant.
∴ \(\frac{n_{2}}{n_{1}}\) = \(\frac{L_{1}}{L_{2}}\) … (3)
Since L₂ < L₁, n₂ > n₁ so that n₂ – n₁ = x
∴ n₂ = n₁ + x …. (4)
Substituting for n₂ in Eq. (3),
\(\frac{n_{1}+x}{n_{1}}\) = \(\frac{L_{1}}{L_{2}}\) or \(\frac{x}{n_{1}}\) = \(\frac{L_{1}-L_{2}}{L_{2}}\)
∴ n₁ = n = x.\(\frac{L_{2}}{L_{1}-L_{2}}\)
which is the required expression.

Question 83.
Solve the following :

Question 1.
A tuning fork C produces 6 beats/second- with another tuning fork D of frequency 320 Hz. When a little wax is put on the prongs of D, the number of beats reduces to 4 per second. Find the frequency of C.
Solution :

1. Initially 6 beats per second are heard. Hence, the difference between the frequencies of the tuning forks is 6 Hz. As the frequency of fork D is 320 Hz, the frequency of fork C = 320 + 6 Hz
   = 326 Hz or 314 Hz
2. When the prongs of fork D are loaded with a little wax, the frequency of fork D decreases and becomes less than 320 Hz.
3. If the frequency of fork C is 326 Hz, the number of beats heard per second must increase.
4. However, as the number of beats heard per second has decreased from 6 to 4, the frequency of fork C must be 314 Hz.

Question 2.
A tuning fork C produces 8 beats per second with another tuning fork D of frequency 340 Hz. When the prongs of tuning fork C are filed a little, the number of beats produced per second decreases to 4. Find the frequency of tuning fork C before filing its prongs.
Solution :
The frequency of timing fork D is 340 Hz. Let n be the frequency of tuning fork C. Since tuning forks C and D produce 8 beats per second when sounded together,
n – 340 = 8 or 340 – n = 8
∴ n = 348 Hz or 332 Hz
When the prongs of a tuning fork are filed a little, its frequency increases. Let n’ be its frequency after filing : n’ > n.

It is given that the beat frequency is reduced from 8 Hz to 4 Hz.
If n was 348 Hz, n’ will be more than 348 Hz. Hence, the beat frequency should increase. Hence, n ≠ 348 Hz.
∴ n = 332 Hz

Question 3.
A sonometer wire 100 cm long produces a resonance with a tuning fork. When its length is decreased by 10 cm, 8 beats per second are heard. Find the frequency of the tuning fork.
Solution :
Data : L₁ = 100 cm, L₂ = 100 – 10 = 90 cm, n₂ – n₁ = 8 beats per second

Now, n₂ – n₁ = 8 ∴ \(\frac{10}{9}\)n₁ – n₁ = 8
∴ 10n₁ – 9n₁ = 9 × 8 = 72
∴ n₁ = 72 Hz
This gives the frequency of the tuning fork as the wire of length 100 cm is in unison with the fork.

Question 4.
A stretched sonometer wire is in unison with a tuning fork. When its length is increased by 4 %, the number of beats heard per second is 6. Find the frequency of the fork.
Solution :
Data : \(\frac{L_{2}}{L_{1}}\) = 1.04, n₁ – n₂ = 6Hz
n₁ = \(\frac{1}{2 L_{1}} \sqrt{\frac{T}{m}}\) and n₂ = \(\frac{1}{2 L_{2}} \sqrt{\frac{T}{m}}\)
∴ \(\frac{n_{1}}{n_{2}}\) = \(\frac{L_{2}}{L_{1}}\) = 1.04
∴ n₁ = 1.04n₂
Now, n₁ – n₂ = 6
∴ 1.04 n₂ – n₂ = 6
∴ n₂ = \(\frac{6}{0.04}\) = 150 Hz
∴ n₁ = n₂ + 6 = 150 + 6 = 156 Hz
This gives the frequency of the tuning fork as initially the wire and the fork vibrate in unison.

Question 5.
The wavelengths of two notes in air are \(\frac{83}{170}\) m and \(\frac{83}{172}\) m. Each of these notes produces 4 beats per second with a third note of a fixed frequency. Find the speed of sound in air.
Solution :
Data : λ₁ = 83/170 m, λ₂ = 83/172 m
Let n₁ and n₂ be the corresponding frequencies.
∴ v = n₁λ₁ = n₂λ₂
where v is the speed of sound in air.
But λ₁ > λ₂
∴ n₁ < n₂
If n is the third frequency, n₁ < n < n₂
∴ n – n₁ = 4 and n₂ – n = 4
Method 1 :
∴ n₂ – n₁ = 8 ∴ \(\frac{v}{\lambda_{2}}\) – \(\frac{v}{\lambda_{1}}\) = 8
∴ v\(\left[\frac{172}{83}-\frac{170}{183}\right]\) = 8 ∴ v × \(\frac{2}{83}\) = 8
∴ v = 4 × 83 = 332 m/s
Method 2 :
∴ n₁ = n – 4 and n₂ = n + 4
∴ (n – 4) × \(\frac{83}{170}\) = (n + 4) × \(\frac{83}{172}\)
Simplifying, we get, 2n = 1368
∴ n = 684 Hz
∴ n₁ = 684 – 4 = 680 Hz
∴ v = n₁λ₁ = 680 × \(\frac{83}{170}\) = 4 × 83
∴ v = 332 m/s

Question 6.
Two sound notes have wavelengths \(\frac{83}{170}\) m and \(\frac{83}{172}\) m in air. These notes, when sounded together, produce 8 beats per second. Calculate the velocity of sound in air and frequencies of the two notes.
Solution :
Data : λ₁ = 83/170 m, λ₂ = 83/172 m
Let nλ₁ and nλ₂ be the corresponding frequencies.
∴ v = n₁λ₁ = n₂λ₂
where v is the velocity of sound in air.
∴ n₁ = \(\frac{v}{\lambda_{1}}\) and n₂ = \(\frac{v}{\lambda_{2}}\) … (1)
But λ₁ > λ₂ ∴ n₁ < n₂
∴ n₂ – n₁ = 8
∴ v\(\left(\frac{1}{\lambda_{2}}-\frac{1}{\lambda_{1}}\right)\) = 8 [From Eq. (1)] …(2)

Question 7.
32 tuning forks are arranged in descending order of frequencies. If any two consecutive tuning forks are sounded together, the number of beats heard is eight per second. The frequency of the first tuning fork is octave of the last fork. Calculate the frequency of the first, last and the 21st fork.
Solution :
Data : n₁ = 2n₃₂, (n₁ is octave of n₃₂) beat frequency = 8 Hz
The set of tuning forks is arranged in descending order of their frequencies.
∴ n₂ = n₁ – 8
n₃ = n₂ – 8 = n₁ – 2 × 8
n₄ = n₃ – 8 = n₁ – 3 × 8
∴ n₃₂ = n₁ – 31 × 8 = n₁ – 248
Since n₁ = 2n₃₁, n₃₂ = 2n₃₂ – 248
∴ The frequency of the last fork, n₃₂ = 248 Hz
The frequency of the first fork, n₁ = 2n₃₂ = 2 × 248 = 496 Hz
∴ The frequency of the 21st fork, n₂₁ = n₁ – 20 × 8 = 496 -160 = 336 Hz

Question 8.
A set of 12 tuning forks is arranged in order of increasing frequencies. Each fork produces V beats per second with the previous one. The last is an octave of the first. The fifth fork has a frequency of 90 Hz. Find V and the frequency of the first and the last tuning forks.
Solution :
Data : n_i+1 – n_i = Y, n₁₂ = 2n₁, n₅ = 90 Hz n₂ – n₁ = Y beats/s
∴ n₂ = n₁ + Y beats/s
Similarly, n₃ = n₂ + Y = n₁ + Y + Y
∴ n₃ = n₁ + 2Y = n₁ + (3 – 1) Y
∴ n_x = n₁ + (x – 1) Y
Similarly, n₁₂ = n₁ + (12 – 1) Y = n₁ + 11Y
∴ n₁₂ = 2n₁ = n₁ + 11Y
∴ n₁ = 11Y
Also, n₅ = n₁ + (5 – 1) Y = n₁ + 4Y
∴ n₅ = 11Y + 4Y = 15Y
∵ n₅ = 90 Hz ∴ 15Y = 90 ∴ Y = 6
∴ n₁ = 11Y beats/s = 11 × 6 beats/s = 66 Hz and n₁₂ = 2n₁ = 2 (66) = 132 Hz

Question 9.
Two tuning forks when sounded together produce 5 beats per second. A sonometer wire of length 0.24 m is in unison with one of the forks. When the length of wire is increased by 1 cm, it is in unison with the other fork. Find the frequencies of the tuning forks.
Solution :
Data :L₁ = 0.24 m = 24 cm, L₂ = 24 + 1 = 25 cm, beat frequency = 5 Hz

∴ n₁ = n₂ + 5 = 125 Hz
∴ The frequencies of the two tuning forks are 125 Hz and 120 Hz.

Question 10.
A closed pipe and an open pipe sounded together produce 5 beats/s. If the length of the open pipe is 30 cm, find by how much should the length of the closed pipe be changed to make the air columns in the two pipes vibrate in unison. [Speed of sound in air = 330 m/s]
Solution :
Data : Beat frequency = 5 s^-1, L_O = 0.3 m, v = 330 m/s
The fundamental frequencies of a closed pipe and open pipe are respectively
n_C = \(\frac{v}{4 L_{\mathrm{C}}}\) and n₀ = \(\frac{v}{2 L_{O}}\)
Let L’_C and n’_C be the changed length and frequency of the closed pipe,
n’_C = n_O

Method 1 :
Since the beat frequency = 5 Hz,
|n_C – n_O| = |n_C – n’_C| = 5

∴ The length of the given closed pipe should be changed by \(\frac{10}{11}\) % to bring it unison with the open pipe.

Method 2 :
Since the beat frequency = 5 Hz,
n_C – n_O = 5 or n_O – n_C = 5
i. e., n_C = 555 Hz or 545 Hz
∴ L_C = \(\frac{v}{4 n_{\mathrm{C}}}\) = \(\frac{330}{4(555)}\) or \(\frac{330}{4(545)}\)
= 0.1486 m or 0.1514 m
= 14.86 cm or 15.14 cm
∴ The length of the given closed pipe should be changed by 0.14 cm.

Question 11.
The forks, A and B, produce 4 beats/s when sounded together. Fork A is in unison with 30 cm length of a sonometer wire and fork B is in unison with 25 cm length of the same wire under the same tension. Calculate the frequencies of the forks.
Solution :
Data : L_A = 30 cm, L_B = 25 cm, beat frequency = 4 s^-1
n ∝ \(\frac{1}{L}\)
Since L_A > L_B, n_A < n_B
∴ n_B – n_A = 4 Hz
and, for the same tension and linear density,
\(\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}}\) = \(\frac{L_{\mathrm{A}}}{L_{\mathrm{B}}}\) = \(\frac{30}{25}\) = \(\frac{6}{5}\)
∴ \(\left(\frac{6}{5}-1\right)\)n_A = 4
∴ \(\frac{1}{5}\)n_A = 4
∴ n_A = 20 Hz
∴ n_B = 24 Hz

Question 84.
Explain the following characteristics of sound :
(1) loudness
(2) pitch
(3) quality or timbre.
Ans.
(1) Loudness : The loudness of a note is the magnitude of the sensation produced by the sound waves on the ear. It depends upon
(a) the energy of the vibration
(b) the sensitiveness of the individual ear
(c) the pitch of the sound.

The loudness of a sound depends on the intensity of the sound wave, which is in turn proportional to the square of the amplitude of the wave itself. Loudness is a physiological (subjective) sensation, while intensity is an objectively measurable physical property of the wave. There is no direct relation between loudness and intensity. Near the middle of the audible range of frequencies, the ear is very sensitive to changes in intensity, which it interprets as changes in loudness.

The unit of loudness is the phon. It is equal to the loudness in decibel of any equally loud pure tone of frequency 1000 Hz.

(2) Pitch : By pitch we mean whether the note is high or low. The pitch of a note depends upon the frequency of the sound. But pitch is not determined by frequency alone. A physiological factor is involved and the sense of pitch is modified by the loudness and quality of the sound.

The average range of frequencies that the human ear detects as sound is approximately 20 Hz to 20000 Hz (the audible range). The human ear is capable of detecting a difference in pitch between two notes. The smallest difference in frequency that the ear can detect as a difference in pitch is approximately proportional to the frequency of one of the notes. That is, a given change in frequency of a low note will produce a greater change in pitch than it will in a high note.

(3) Quality : By quality or timbre is meant that characteristic of a sound by which it is possible to distinguish it from all other sounds of the same pitch and loudness. The same note played at the same loudness on two different musical instruments are easily distinguished from each other by their timbre.

[Note : A pure note, consisting of only one frequency, is different from a musical note, which may be a combination of many different frequencies. A musical note has a fundamental, or lowest frequency, and superimposed on it are higher frequencies, called overtones or partials. The number and relative strengths of the partials present determines the timbre of the note. The ear always recognizes the fundamental as determining the pitch of the note.]

Question 85.
Define intensity of sound. State its unit.
Answer:
Definition : The intensity of sound at a point is the time rate of flow of sound energy passing normally through a unit area at that point.
SI unit: the joule per second square metre (j/s.m²) or watt per square metre (W/m²).

Question 86.
What are the factors affecting the loudness of sound ? Is intensity the same as loudness ?
Answer:
(1) The factors affecting the loudness of sound are

1. the amplitude of the vibrations of the body
2. the distance of the listener from the vibrating body
3. the surface area of the vibrating body
4. the density of the medium
5. the presence of the resonating bodies
6. the sensitivity of the ear of the listener.

(2) Intensity and loudness are related, but not the same. Intensity is a measurable quantity whereas loudness is a sensation which is not measurable. Loudness depends on the intensity of sound as well as the sensitivity of the ear of the listener.

Question 87.
Explain the term decibel
Answer:
The intensity level of a sound wave, by definition, is β = log₁₀ \(\left(\frac{I}{I_{0}}\right)\)bels = 10 log₁₀ \(\left(\frac{I}{I_{0}}\right)\) decibels as one decibel is 0.1 bel. Here, I₀ (reference intensity) is taken as 10^-12 W/m².

Intensity level is expressed in decibel (dB). There is no direct relation between loudness and intensity. The decibel is not a unit of loudness.

[Note : The decibel, equal to 0.1 bel, is used for comparing two power levels, currents or voltages. The unit bel is named in honour of Alexander Graham Bell (1847-1922) British-American scientist, inventor of the telephone (1876).]

Question 88.
What is the difference between a musical sound and a noise ?
Answer:
A musical sound is pleasing to the listener while a noise is not. The pleasure derived from a musical note is because it strikes the ear as a perfectly undisturbed, uniform sound which remains unaltered as long as it exists. On the other hand, noise is accompanied by a rapid, irregular but distinct, alternations of various kinds of sounds.

A musical sound thus has a regularity or smoothness because the vibrations that cause the sound are periodic. But the converse, that if the vibrations are regular the sound is musical, is not always true. For example, a ticking clock does not produce a musical note, or the definite note produced by a card held against the teeth of a rotating toothed wheel is far from being pleasant to hear. Bearing such reservations in mind, the essential difference between music and noise is that the former is produced by periodic and continuous vibrations, while noise results from discontinuous sudden and sharp sounds with no marked periodicity.

Question 89.

1. Which quantity out of frequency and amplitude determines the pitch of the sound?
2. Which out of pitch and frequency is a measurable quantity ?

Answer:

1. The frequency of sound determines its pitch. A high pitched or shrill sound is produced by a body vibrating with a high frequency and a low pitched or flat sound is produced by a body vibrating with a low frequency.
2. Frequency is a measurable quantity whereas pitch is not a measurable quantity.

Question 90.
Write a note on the major diatonic scale.
OR
Explain what is a musical scale.
Answer:
A musical scale is constructed on the basis of certain groups of notes with simple intervals. A major chord or triad is a group of three notes with frequencies in the ratio 4:5:6 that produce a very pleasing effect when sounded together. The diatonic musical scale is composed of three sets of triads making eight notes.

Some note called the tonic, is chosen as the basis of the scale, and a triad is constructed using this note as the one of lowest frequency. Calling the tonic as the 1st, the major chords are 1st, 3rd and 5th, 4th, 6th and 8th, and 5th, 7th and 9th; 8th and 9th are respectively the octaves of the 1st and 2nd.

In addition to the eight notes of an octave, that form the major scale, five additional notes are also used. These are derived either by raising or lowering the pitch by the interval 25 / 24. If the pitch is raised the note is sharp, and when lowered, it is flat.

Question 91.
Write a short note on Indian musical scale.
Answer:
Indian music is chiefly based on melody, i.e., consonant notes in suitable succession. Besides this physiological sensation, there is a deep psychological involvement.
The notes or svaras (स्वर) used in an Indian musical scale have the same musical intervals as those of the major diatonic scale. The five additional notes in pure intonation, तीव्र (sharp) and कोमल (flat) are also used. Thus, the choice is usually made from the following twelve svaras : सा (shadja), रे(को) and रे (rishabha), ग(को) and ग g\(\bar{a}\)ndh\(\bar{a}\)ra), म and म(ती) (madhyam), प (pancham), ध (को) and ध (dhaivata), नी(को) and नी (nishad), सा. However, as compared to the fixed frequency of the tonic in western music, an Indian vocalist or musician has the freedom to set any frequency as the tonic. Besides, unlike western music, dissonant intervals are sometimes introduced to enhance the musical effect.

However, the whole structure of Indian music is based on r\(\bar{a}\)gas (राग), which are well-established melody types with a wide variety of emotional content. They can be courageous, amorous, melancholy, cheerful, soothing, or ecstatic. R\(\bar{a}\)gas are capable of conveying these emotions to the listener and different r\(\bar{a}\)gas are assigned to different seasons and different parts of the day.

Question 92.
Give reasons :
The notes of a sitar and a guitar sound different even if they have the same loudness and the pitch.
Answer:
The quality or timbre of the sound of a sitar is different from that of a guitar. The number of overtones or partials present and their relative intensities determine the quality or timbre of the sound of a musical instrument. Therefore, even if the pitch and the loudness are the same, the notes of a sitar and a guitar sound different.

Question 93.
Which are the three broad types of musical instruments ?
OR
Write a short note on types of musical instruments.
Musical instruments have been classified in various ways. One ancient system that was based on the primary vibrating medium distinguished three main types of instruments : stringed, wind and percussion.
Examples :
(1) Stringed instruments (stretched strings) :
(a) Plucked : Tanpura, sitar, veena, guitar, harp
(b) Bowed : Violin
(c) Struck : Santoor, pianoforte )

(2) Wind instruments :
(a) Free (air not confined) : Harmonica or mouth organ (without keyboard), harmonium (with keyboard). (Both are reed instruments in which free brass reeds are vibrated by air, blown or compressed.)
(b) Edge (air blown against an edge) : Flute
(c) Reedpipes : Saxophone (single reed), shehnai and bassoon (double reeds)

(3) Percussion instruments :
(a) Stretched skin heads : Tabla, mridangam, drums
(b) Metals (struck against each other or with a beater) : Cymbals, Xylophone

Question 94.
A simple harmonic wave of frequency 20 Hz is travelling in the positive direction of x-axis with a velocity of 30 m/s. Two particles in the path of the wave, 0.45 m apart, differ in phase by
(A) \(\frac{\pi}{3}\) rad
(B) \(\frac{\pi}{2}\) rad
(C) 0.6 π rad
(D) π rad.
Answer:
(C) 0.6 π rad

Question 95.
What is the period of the wave given by y = 0.003 sin (\(\frac{\pi}{0.08}\)t + \(\frac{\pi}{8}\)x ) (in SI units) ?
(A) 0.08 s
(B) 0.16 s
(C) 0.32 s
(D) 0.8 s.
Answer:
(B) 0.16 s

Question 96.
The equation of a progressive wave is y = 7 sin(4t – 0.02x), where x and y are in centimetre and time t in second. The maximum velocity of a particle is
(A) 28 cm/s
(B) 32 cm/s
(C) 49 cm/s
(D) 112 cm/s.
Answer:
(A) 28 cm/s

Question 97.
When a longitudinal wave is incident at the boundary of a denser medium, then
(A) a compression reflects as a compression
(B) a compression reflects as a rarefaction
(C) a rarefaction reflects as a compression
(D) a longitudinal wave reflects as a transverse wave.
Answer:
(A) a compression reflects as a compression

Question 98.
A transverse wave travelling in a denser medium is reflected from a rarer medium. Then,
(A) an incident crest is reflected as a crest
(B) an incident crest is reflected as a trough
(C) there is a phase change of 2π rad
(D) there is a phase change of π/2 rad.
Answer:
(A) an incident crest is reflected as a crest

Question 99.
Two simple harmonic waves of the same amplitude and frequency, but 90° out of phase, pass through the same region in a medium. The resultant wave has
(A) an amplitude greater than either of the component waves
(B) an amplitude smaller than either of the component waves
(C) zero amplitude
(D) an amplitude slowly varying with time.
Answer:
(A) an amplitude greater than either of the component waves

Question 100.
At a given instant two vibrating particles in the same loop of a stationary wave have
(A) the same phase
(B) opposite phases
(C) slightly different phases
(D) opposite velocities.
Answer:
(A) the same phase

Question 101.
Two vibrating particles in the adjacent loops of a stationary wave have ….. at a given instant.
(A) the same phase
(B) opposite phases
(C) slightly different phases
(D) the same velocity
Answer:
(B) opposite phases

Question 102.
A stretched string, 2 m long, vibrates in its third overtone. The distance between consecutive nodes is
(A) 40 cm
(B) 50 cm
(C) 66.7 cm
(D) 100 cm.
Answer:
(B) 50 cm

Question 103.
A stretched string of length l vibrates in the third overtone. The wavelength of stationary wave formed is
(A) \(\frac{l}{2}\)
(B) \(\frac{l}{4}\)
(C) l
(D) 21.
Answer:
(A) \(\frac{l}{2}\)

Question 104.
A stretched string tied between two rigid supports vibrates with a frequency double the fundamental frequency. The point midway between the supports is
(A) a node
(B) an antinode
(C) either a node or an antinode
(D) neither a node nor an antinode.
Answer:
(A) a node

Question 105.
A travelling wave of frequency 100 Hz along a string is reflected from a fixed end. The stationary wave formed has the nearest node at a distance of 10 cm from the fixed end. The speed of the travelling wave was
(A) 40 m/s
(B) 20 m/s
(C) 10 m/s
(D) 5m/s.
Answer:
(B) 20 m/s

Question 106.
Stationary waves are produced on a 10 m long stretched string fixed at both ends. If the string vibrates in 5 segments and the wave velocity is 20 m/s, the frequency of the waves is
(A) 10 Hz
(B) 5 Hz
(C) 4 Hz
(D) 2 Hz.
Answer:
(B) 5 Hz

Question 107.
The fundamental frequency of transverse vibrations of a stretched string of radius r is proportional to
(A) r^-2
(B) r^-1
(C) \(r^{-\frac{1}{2}}\)
(D) r².
Answer:
(B) r^-1

Question 108.
A stretched string of length 50 cm vibrates in five segments when stationary waves are formed on it. If the wave speed is 14 m/s, its frequency of vibration is
(A) 28 Hz
(B) 35 Hz
(C) 70 Hz
(D) 140 Hz.
Answer:
(C) 70 Hz

Question 109.
Two strings A and B are identical except that the diameter of A is twice the diameter of B. The ratio of the frequency of sound from A to that from B is
(A) 2 : 1
(B) \(\sqrt{2}\) : 1
(C) 1 : \(\sqrt{2}\)
(D) 1 : 2.
Answer:
(D) 1 : 2.

Question 110.
Two strings, A and B, have the same tension and length. The string A has a mass m while the string B has a mass Am. If the speed of the waves in string A is v, that on string B is
(A) \(\frac{1}{2}\)v
(B) v
(C) 2v
(D) v.
Answer:
(A) \(\frac{1}{2}\)v

Question 111.
An organ pipe is closed at one end. The pth overtone is the ….. th harmonic.
(A) 2p + 1
(B) 2p – 1
(C) p + 1
(D) P – 1
Answer:
(A) 2p + 1

Question 112.
Of two narrow organ pipes A and B, A is open at one end and B at both ends. Both the pipes have the same fundamental frequency. If A is 1.2 m long, how long is B?
(A) 0.8 m
(B) 1.8 m
(C) 2.4 m
(D) 3.0 m
Answer:
(C) 2.4 m

Question 113.
The value of end correction for an open organ pipe of radius r is
(A) 0.3 r
(B) 0.6 r
(C) 0.9 r
(D) 1.2 r.
Answer:
(D) 1.2 r.

Question 114.
Of two long narrow organ pipes A and B, A is open at one end and B at both ends. If both the pipes have the same fundamental frequency, the first overtone of A is ….. the first overtone of B.
(A) half of
(B) \(\frac{2}{3}\) of
(C) equal to
(D) twice
Answer:
(B) \(\frac{2}{3}\) of

Question 115.
In an open organ pipe, the first overtone produced is of such frequency that the length of the pipe is equal to
(A) \(\frac{\lambda}{4}\)
(B) \(\frac{\lambda}{3}\)
(C) \(\frac{\lambda}{2}\)
(D) λ
Answer:
(D) λ

Question 116.
A sonometer wire vibrates with three nodes and two antinodes. The corresponding mode of vibration is
(A) the first overtone
(B) the second overtone
(C) the third overtone
(D) the fourth overtone.
Answer:
(A) the first overtone

Question 117.
Velocity of a transverse wave along a stretched string is proportional to [T = tension in the string]
(A) \(\sqrt{T}\)
(B) T
(C) \(\frac{1}{\sqrt{T}}\)
(D) \(\frac{1}{T}\)
Answer:
(A) \(\sqrt{T}\)

Question 118.
The frequency of the second overtone of the vibration of a stretched string is
A. \(\frac{1}{l} \sqrt{\frac{T}{m}}\)
B. \(\frac{3}{2 l} \sqrt{\frac{T}{m}}\)
C. \(\frac{1}{2 l} \sqrt{\frac{T}{m}}\)
D. \(\frac{2}{3 l} \sqrt{\frac{T}{m}}\)
Answer:
B. \(\frac{3}{2 l} \sqrt{\frac{T}{m}}\)

Question 119.
When the air column in a pipe closed at one end vibrates such that three nodes are formed in it, the frequency of its vibrations is ….. times the fundamental frequency.
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(D) 5

Question 120.
One beat means that the intensity of sound should be
(A) once maximum
(B) once minimum
(C) once maximum and once minimum
(D) twice maximum and twice minimum.
Answer:
(C) once maximum and once minimum

Question 121.
Let n₁ and n₂ be two slightly different frequencies of sound waves. The time interval between a waxing and the immediate next waning is
(A) \(\frac{1}{n_{1}-n_{2}}\)
(B) \(\frac{2}{n_{1}-n_{2}}\)
(C) \(\frac{n_{1}-n_{2}}{2}\)
(D) \(\frac{1}{2\left(n_{1}-n_{2}\right)}\)
Answer:
(D) \(\frac{1}{2\left(n_{1}-n_{2}\right)}\)

Question 122.
In the formation of beats, the resultant amplitude varies with a frequency equal to
(A) the beat frequency
(B) the average frequency
(C) half the beat frequency
(D) double the beat frequency.
Answer:
(C) half the beat frequency

Question 123.
A tuning fork A of frequency 512 Hz produces 3 beats per second with another tuning fork B of frequency 515 Hz. If the prongs of B are filed a little, the number of beats produced per second will
(A) increase
(B) decrease
(C) remain the same
(D) increase or decrease.
Answer:
(A) increase

Question 124.
A tuning fork gives 1 beat in 2 seconds with a timing fork of frequency 341.3 Hz. If the beat frequency decreases when the first fork is filed a little, its original frequency was
(A) 336.3 Hz
(B) 340.8 Hz
(C) 341.8 Hz
(D) 346.3 Hz.
Answer:
(B) 340.8 Hz

Question 125.
In a set of 25 tuning forks, arranged in order of increasing frequency, each fork gives 3 beats per second with the succeeding one. If the frequency of the 10th fork is 127 Hz, the frequency of the 16th fork is
(A) 139 Hz
(B) 145 Hz
(C) 148 Hz
(D) 151 Hz.
Answer:
(B) 145 Hz

Question 126.
If two sound waves with the same amplitude but slightly different frequencies n₁ and n₂ superpose to produce beats, the resultant wave motion has frequency
(A) |n₁ – n₂|
(B) n₁ + n₂
(C) \(\frac{\left|n_{1}-n_{2}\right|}{2}\)
(D) \(\frac{n_{1}+n_{2}}{2}\)
Answer:
(D) \(\frac{n_{1}+n_{2}}{2}\)

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.