# NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

• Q. Let us list some of the factors, which could possibly influence the speed of wave propagation :
• (i) Nature of the source.
• (ii) Direction of propagation.
• (iii) Motion of the source and observer.
• (iv) Wavelength.
• (v) Intensity of the wave.
• On which of these factors, if any, does :
• (a) The speed of light in vacuum.
• (b) The speed of light in a medium (say, glass or water) depend ?
• Ans. (a) The speed of light in a vacuum i.e., 3 × 108 m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum.
• (b) Out of the listed factors, the speed of light in a medium depends on the wavelength of light in that medium.

Q. In dividing the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitable dividing the slit to bring out the cancellation.

Ans Consider that a single slit of width d is divided into n smaller slits.

$$\therefore\space\text{Width of each slit,}d'=\frac{d}{n}\\\text{Angle of diffraction is given by the relation,}\\\theta=\frac{\frac{d}{d'}\lambda}{d}=\frac{\lambda}{d'}$$

Now, each of these infinitesimally small slit sends zero intensity in direction q}.Hence, the combination of these slits will give zero intensity.

Q. What is the shape of the wavefront in each of the following cases :

• (i) Light diverging from a point source.
• (ii) Light emerging out of a convex lens when a point source is placed at its focus.
• (iii) The portion of the wavefront of light from a distance star intercepted by the Earth.
• Ans. (i) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure. (ii) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure. wavefront

(iii) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

Q. What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm ?

Ans. As  c=vλ

$$\frac{\Delta V}{V}=\frac{\Delta \lambda}{\lambda}(\text{for small changes in v and }\lambda).\\\text{For}\space \Delta\lambda=589.6-589.0=+0.6\text{nm}\\\frac{\Delta v}{v}=-\frac{\Delta \lambda}{\lambda}=-\frac{v_{radial}}{c}\\\text{or}\space v_{radial}\cong+c\bigg(\frac{0.6}{589.0}\bigg)\\=+3.06×10^{5}\text{ms}^{-1}\\=306\space\text{km/s}\\\text{Therefore, the galaxy is moving away from us.}$$

Q. The 6563 Å Hα line emitted by hydrogen in a star is found to be red shifted by 15 Å. Estimate the speed with which the star is receding from the earth.
• Ans Given : Wavelength or Hα line emitted by
• Hydrogen (λ) = 6563 Å
• = 6563 × 10–10 m.
• Star’s red-shift (λ′ – λ) = 15 Å = 15 × 10–10 m
• Speed of light (c) = 3 × 108 m/s
• Let the velocity of the star receding away from the Earth be v.
• The red shift is related with velocity by

$$\Rarr\lambda’-\lambda=\frac{v}{c}\lambda\\v=\frac{c}{\lambda}×(\lambda^{‘}-\lambda)=\frac{3×10^{8}×15×10^{-10}}{6563×10^{-10}}\\=6.87×10^{5}\text{m/s}$$

Hence, the speed with which the star is receding away from the earth is 6.87 × 105 m/s.

Q. Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light ? For what
angle of incidence is the reflected ray normal to the incident ray ?

Ans. Given, wavelength of incident light (λ)

= 5000 Å = 5000 × 10–10 m

Speed of light (c) = 3 × 108 m/s

Frequency of incident light is given by

$$v=\frac{c}{\lambda}\\=\frac{3×10^{8}}{5000×10^{-10}}\\=6×10^{14}\text{Hz}$$

The wavelength and frequency of incident light is the same as that of reflected ray.Thus, the wavelength of reflected light is 5000 Å and its frequency is} 6 × 1014Hz When reflected ray is normal to incident ray, the sum of the angle of incidence angle i and angle of reflection ∠r is 90°

$$\Rarr\space\angle i+\angle r=90\degree\\\Rarr\angle i+\angle i=90\degree\\\angle i=\frac{90\degree}{2}=45\degree$$

Therefore, the angle of incidence for the given condition is 45°

Q. In double slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distance screen is 0.1°. What is the spacing between the two slits ?

Ans. Wavelength of light used,600nm=600×10-9m

Angular width of fringe,

$$\theta=0.1\degree=0.1×\frac{\pi}{180}=\frac{3.14}{1800}\text{rad}\\\text{Angular width of a fringe is related to slit spacing (d) as :}\\\theta=\frac{\lambda}{d}\\d=\frac{\lambda}{\theta}=\frac{600×10^{-9}}{\frac{3.14}{1800}}\\=3.44×10^{-4}\text{m}\\\text{Therefore, the spacing between the slits is}\space 3.44×10^{-4}m.$$

Q. In a double slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe. If the entire experimental apparatus is immersed in water ? Take refractive index of water to be 4/3.

• Ans. Given, distance of the screen from the slits (D) = 1 m
• Wavelength of light used (λ1) = 600 nm
• Angular width of the fringe in air (θ1) = 0.2°
• Angular width of the fringe in water (θ2) = ?

$$\text{Refractive index of water}(\mu)=\frac{4}{3}\\\text{Refractive index is related to angular width by}\\\mu=\frac{\theta_1}{\theta_2}\\\theta_2=\frac{3}{4}\space\theta_1=\frac{3}{4}×0.2=0.15\degree\\\text{Hence, the angular width of the fringe in water will reduce to}\space 0.15\degree.$$

Q. In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

• Ans. Given, Distance between the slits (d) = 0.28 mm = 0.28 × 10–3 m
• Distance between the slit and the screen (D) = 1.4 m
• Distance between the central fringe and the fourth (n = 4) fringe,
• (u) = 1.2 cm = 1.2 × 10–2 m
• In case of a constructive interference, the distance between the two fringes is

$$u=n\lambda\frac{D}{d}\\\text{where,}\space n=\text{Order of fringes}=4\\\lambda=\text{Wavelength of light}\\\therefore\space\lambda=\frac{ud}{nD}=\frac{1.2×10^{-2}×0.28×10^{-3}}{4×1.4}\\=6×10^{-7}\\=600\text{nm}\\\text{Thus, the wavelength of the light is 600 nm.}$$

Q. For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm ?

$$\textbf{Ans.}\space \text{Z}_F=\frac{a^{2}}{\lambda}=\frac{(3×10^{-3})^{2}}{5×10^{-7}}=18\text{m}$$

This example shows that even with a small aperture, diffraction spreading can be neglected for rays many metres in lengths. Thus, ray optics is valid in many common situations.

Q. Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.

Ans. Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation.

$$\text{Z}_F=\frac{a^{2}}{\lambda}\\\text{where, aperture width,} a = 4 \text{mm} = 4 × 10^{\normalsize–3} m\\\text{Wavelength of light},\lambda=400\text{nm}=400×10^{-9}m\\\text{Z}_F=\frac{(4×10^{-3})^{2}}{400×10^{-9}}=40\text{m}\\\text{Therefore, the distance for which the ray optics is a good approximation is 40 m.}$$

Q. Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects ?

• Ans. Given,
• Distance between the towers (d) = 40 km
• Height of the line joining the hills (d′) = 50 m.
• Since the hill is located halfway between the towers. Fresnel’s distance can be obtained as :
• (ZF) = 20 km = 2 × 104 m
• Aperture can be taken by
• a = d' = 50 m
• Fresnel’s distance is given by

$$\text{Z}_F=\frac{a^{2}}{\lambda}\\\text{where} (\lambda)=\text{Wavelength of radio waves}\\\therefore\space\lambda=\frac{a^{2}}{Z_{F}}=\frac{(50)^{2}}{2×10^{4}}\\=1250×10^{-4}=0.1250m\\=12.5\text{cm}\\\text{Therefore, the wavelength of the radio waves is 12.5 cm.}$$

Q. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

• Ans. Wavelength of light beam, λ = 500 nm = 500 × 10–9 m
• Distance of screen from the slit, D = 1 m
• For first minima, n = 1
• Distance between the slits = d
• Distance of the first minimum from the centre of
• the screen can be obtained as :
• x = 2.5 mm = 2.5 × 10–3 m
• It is related to the order of minima as :

$$\text{n}\lambda=x\frac{d}{D}\\\Rarr d=\frac{n\lambda D}{x}\\=\frac{1×500×10^{-9}×1}{2.5×10^{-3}}\\=2×10^{-4}\text{m}=0.2\text{mm}\\\text{Therefore, the width of the slits is 0.2 mm.}$$

Q. Explain how corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment ?

Ans. No, Wave theory

Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged. Hence, we can write the expression.

c sin i = v sin r …(i)

where, i = Angle of incidence

r = Angle of reflection

c = Velocity of light in air

v = velocity of light in water

We have the relation for relative refractive index of water with respect to air as—

$$\mu=\frac{v}{c}\\\text{Hence, equation (i) reduces to}\\\frac{v}{c}=\frac{\text{sin}i}{\text{sin}r}=\mu\space\text{…(ii)}\\\text{But}\space\mu\text{\textgreater}1$$

Hence, it can be inferred from equation
(ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v.The wave picture of light is consistent with the experimental results.

Q. You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Ans. Let an object at O be placed in front of a plane mirror of MO′ at a distance r (as shown in the given figure).

A circle is drawn from the centre O such that it just touches the planes mirror at point O′.
According to Huygens’ principle, XY is the wavefront of incident light. If the mirror is absent, then a similar wavefront X′Y′ (as XY) would form behind O′ at distance r (as shown in the given figure). X′Y′ can be considered as a virtual reflected ray for the plane mirror.

Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).

Q. (i) The refractive index of glass is 1.5. What is the speed of light in glass ? Speed of light in vacuum is 3.0 × 108 m/s.

(ii) Is the speed of light in glass independent of the colour of light ? If not, which of the two colours red and violet travels slower in a glass prism ?

Ans. (i) Refractive index of glass, µ = 1.5.

Speed of light, c = 3 × 108 m/s

$$\text{v}=\frac{c}{\mu}=\frac{3×10^{8}}{1.5}=2×10^{8}\text{m/s}\\\text{Thus, the speed of light in glass is}\space 2×10^{8}\text{m/s}$$

(ii) The speed of light in glass is dependent of the colour of light. The refractive index of a violet component of white light is greater than the refractive index of a red component.Therefore, the speed of violet light is less than the speed of red light in glass.Thus, violet light travels slower than red light in a glass prism.

Q. (i) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.

(ii) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle ?

Ans. (i) Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the antenna. This may distort TV signals. Thus, when a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen.

(ii) The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because
superposition follows from the linear character of a differential equation that governs wave motion. If y1 and y2are the solutions of the second order wave equation,
then any linear combination of y1 and y2 will also be the solution of the wave equation.

Q. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations : (i) source at rest; observer moving, and (ii) source moving; observer at rest.
The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium ?

Ans. No, sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same.

In case of light waves, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.

Q. Answer the following questions :

• (i) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band ?
• (ii) In what way is diffraction from each slit related to the interference pattern in a double slit experiment ?
• (iii) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why ?
• (iv) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
• (v) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification ?
• Ans. (i) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.
• (ii) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.
• (iii) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This is because light wave are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.
• (iv) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.
• On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.
• (v) The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.