NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics: Topics, Question with Solutions PDF

Q.10.1 Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.

Ans.10.1 Wavelength of incident monochromatic light,  $\lambda$ = 589 nm = 589  $\times {10}^{-9}$ m

Speed of light in air, c = 3  $\times {10}^8$ m/s

Refractive index of water,  $\mu$ = 1.33

In case of reflection, the ray goes back to the same medium. Hence wavelength, frequency and speed of reflected beam will be same as incident beam.

Frequency of light beam is given by the relation,  $\nu$ =  $\frac{c}{\lambda }$ =  $\frac{3\mathrm{}\times {10}^8}{589\mathrm{}\times {10}^{-9}}$ = 5.09  $\times {10}^{14}$ Hz.

Hence speed = 3  $\times {10}^8$ m/s, Wavelength = 589  $\times {10}^{-9}$ m, Frequency = 5.09  $\times {10}^{14}$ Hz of incident ray and reflected ray will remain unchanged.

(i) Frequency = does not depend on the property of the medium in which it is travelling, hence frequency will remain same, i.e. 5.09  $\times {10}^{14}$ Hz

(ii) Speed of light in water depends upon the refractive index of water, hence speed of light, v =  $\frac{c}{\mu }$ =  $\frac{3\mathrm{}\times {10}^8}{1.33}$ = 0.226  $\times {10}^9$ m/s

Wavelength in water,  $\lambda$ =  $\frac{v}{\nu }$ =  $\frac{0.226\mathrm{}\times {10}^9}{5.09\mathrm{}\times {10}^{14}}$ = 4.432  $\times {10}^{-7}$ m = 443.2 nm

Hence the speed, frequency and wavelength of refracted light are 0.226  $\times {10}^9$ m/s, 5.09  $\times {10}^{14}$ Hz and 443.2 nm

 

Q.10.2 What is the shape of the wave front in each of the following cases:

(a) Light diverging from a point source.

(b) Light emerging out of a convex lens when a point source is placed at its focus.

(c) The portion of the wave front of light from a distant star intercepted by the Earth.

Ans.10.2 The shape of the wave front in case of a light diverging from a point source is spherical.

The shape of the wave front in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid.

The portion of the wave front of light from a distant star intercepted by the Earth is a plane.

 

Q.10.3 (a) 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–1)

(b) 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. 10.3 The refractive index of glass,  $\mu$ = 1.5

Speed of light in vacuum, c = 3.0  $\times {10}^8$ m/s

Speed of light in glass is given by the relation,  $v$ =  $\frac{c}{\mu }$ =  $\frac{3.0\mathrm{}\times {10}^8}{1.5}$ = 2  $\times {10}^8$ m/s

The speed of light in glass is not independent of the colours of light. The refractive index of a violet component of white light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

 

Q.10.4 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.10.4 Distance between the slits, d = 0.28 mm = 0.28  $\times {10}^{-3}$ m

Distance between the slits and the screen, D = 1.4 m

Distance between the central bright fringe and the fourth ( n = 4) fringe, u = 1.2 cm

= 1.2  $\times {10}^{-2}$ m

In case of a constructive interference, we have the relation for the distance between two fringes as : u = n  $\lambda \frac{D}{d},$ where n = order of fringes = 4 and  $\lambda$ = wavelength of the light used

Hence,  $\lambda$ =  $\frac{ud}{nD}$ =  $\frac{1.2\mathrm{}\times {10}^{-2}\times 0.28\mathrm{}\times {10}^{-3}}{4\times 1.4}$ = 6  $\times {10}^{-7}$ m = 600  $\times {10}^{-9}$ m = 600 nm

Hence, wavelength of the light is 600 nm.

 

Q.10.5 In Young’s double-slit experiment using monochromatic light of wavelength  $\lambda$ , the intensity of light at a point on the screen where path difference is  $\lambda$ , is K units. What is the intensity of light at a point where path difference is  $\lambda$ /3?

Ans.10.5 Let  $I_1$ and  $I_2$ be the intensity of the two light waves. Their resultant intensity can be obtained as :

$I’$ =  $I_1+\mathrm{}{I}_2+2\sqrt{I_1{I}_2}$  $\cos \varnothing$ , where  $\varnothing =$ Phase difference between two waves

For monochromatic light waves,  $I_1=I_2$ . Hence

$I’$ =  $I_1+I_1$ + 2  $\sqrt{I_1^2}$  $\cos \varnothing$ = 2  $I_1+2I_1\cos \varnothing$

We know, phase difference  $\varnothing$ =  $\frac{2\pi }{\lambda }$  $\times$ path difference

Since path difference =  $\lambda$ , phase difference  $\varnothing$ = 2  $\pi$ , then

$I’=$ 2  $I_1+2I_1\cos 2\pi$ = 4  $l_1$

Given  $I’=K$ , so  $l_1$ =  $\frac{K}{4}$ ……….(1)

When path difference is  $\frac{\lambda }{3}$ , phase difference  $\varnothing =\frac{2\pi }{3}$ , then

$I’$ = 2  $I_1+2I_1\cos \varnothing$ = 2  $I_1+2I_1\times (-0.5)$ =  $l_1$

From equation (1), we can write l’ =  $l_1$ =  $\frac{K}{4}$

Hence, the intensity of light at a point where path difference is  $\frac{\lambda }{3}$ is  $\frac{K}{4}$ units.

 

Q.10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Ans.10.6 Wavelength of one light beam,  ${\lambda }_1$ = 650 nm

Wavelength of the other beam,  ${\lambda }_2$ = 520 nm

Distance of the screen from the slits = D

Distance between two slits = d

Distance of the  $n^{th}$ bright fringe on the screen from the central maximum is given by the relation,

$x$ = n  ${\lambda }_1(\frac{D}{d}$ )

For the 3rd bright fringe, n = 3

Hence  $x$ = 3  $\times 650\times (\frac{D}{d}$ ) nm

Let  $n^{th}$ bright fringe due to wave length  ${\lambda }_2$ and  ${(n-1)}^{th}$ bright fringe due to wavelength  ${\lambda }_1$ coincide on the screen. We can equate the conditions for bright fringes as :

n  ${\lambda }_2$ = (n-1)  ${\lambda }_1$

520n = 650n – 650

n =  $\frac{650}{130}$ = 5

The least distance can be obtained from the relation,

$x$ = n  ${\lambda }_2(\frac{D}{d}$ ) = 5  $\times 520(\frac{D}{d}$ ) = 2600  $(\frac{D}{d}$ ) nm

 

Q.10.7 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.10.7 Distance of the screen from the slits, D = 1 m

Wavelength of the light used,  ${\lambda }_1$ = 600 nm

Angular width of the fringe in air,  ${\theta }_1$ = 0.2  $°$

Angular width of the fringe in water =  ${\theta }_2$

Refractive index of water,  $\mu =\frac{4}{3}$

Refractive index is related to angular width as:

$\mu =\frac{{\theta }_1}{{\theta }_2}$ or

${\theta }_2=\frac{{\theta }_1}{\mu }$ = 0.2  $\times \frac{3}{4}$ = 0.15  $°$

Hence, the angular width of the fringes in water will be 0.15  $°$

 

Q.10.8 What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)

Ans.10.8 Refractive index of glass,  $\mu =1.5$

Let the Brewster angle be  $\theta$

Brewster angle is related to  $\theta$ by the equation

$\tan \theta =\mu$

Or  =  $\theta ={\tan }^{-1}\mu$ = ${\tan }^{-1}1.5$ =56.31 $°$ ${\mathrm{<br>}}^{}$

Hence, the Brewster angle for air to glass transition is 56.31  $°$

 

Q.10.9 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.10.9 Wave length of the incident light, $\lambda$ = 5000 $\times {10}^{-10}$ m

Speed of light, c = 3 $\times {10}^8$ ${\mathrm{<br>}}^{}$m/s

Frequency of the incident light, $\nu$ =  $\frac{c}{\lambda }$  = $\frac{3\times {10}^8}{5000\times {10}^{-10}}$ Hz= 6  $\times {10}^{14}$ Hz

The wavelength and frequency of incident ray will be same as of reflected ray. So the wavelength of the reflected ray will be 5000 Å and the frequency will be 6  $\times {10}^{14}$ Hz

When reflected ray is normal to the incident ray, the sum of incidence angle $\angle i$ and the reflected angle $\angle r$ will be 90 $°$

According to the law of reflection, the incidence angle and the reflected are same, i.e. $\angle i=\angle r$

Hence, $\angle i+\angle r$ = 2 $\angle i$ = 90 $°$

So $\angle i=\angle r=45°$

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

 

Q.10.10 Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.

Ans.10.10 Fresnel’s distance ( $Z_F)$ is the distance for which the ray optics is a good approximation. It is given by the relation

$Z_F=\frac{a^2}{\lambda }$ where $a$ =aperture width = 4 mm = 4 $\times {10}^{-3}$ m

$\lambda$ = wave length = 400 nm = 400 $\times {10}^{-9}$ m

Hence $Z_F=\frac{{4\times {10}^{-3}}^2}{400\mathrm{}\times {10}^{-9}}$ = 40 m

Therefore, the distance for which ray optics is a good approximation is 40 m.

 

Q.10.11 The 6563 Å H $\mathit{\alpha }$

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.10.11 The wavelength of $H_{\alpha }$ line emitted by Hydrogen, $\lambda =6563\AA$ = 6563 $\times {10}^{-10}$ m $$

Star’s red-shift, ( $\lambda \text{'}$ $-$ $\lambda )$ = 15 $\AA$ = 15 $\times {10}^{-10}$ m

Speed of light, c = 3 $\times {10}^8$ m/s

( $\lambda \text{'}$ $-$ $\lambda )$ = $\frac{v}{c}\lambda$

v = $\frac{c}{\lambda }\times$ ( $\lambda \text{'}$ $-$ $\lambda )$ = $\frac{3\mathrm{}\times {10}^8}{6563\mathrm{}\times {10}^{-10}}\times$ 15  $\times {10}^{-10}$ = 6.86 $\times {10}^5$ m/s

 

Q.10.12 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.10.12 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:

$\mathrm{c\; sin}i=v\sin r$ ………(1)

Where, i = angle of incidence, r = angle of reflection, c = velocity of light in air and v = velocity of light in water

We have the relation for relative refractive index of water with respect to air is $\mu =\frac{v}{c}$

So from equation (1) we get

  $\frac{v}{c}$ =  $\frac{\sin i}{\sin r}$= $\mu$

But $\mu >1,v>c$ This is not possible since the prediction is opposite to the experimental results of c > v. The wave picture of light is consistent with the experimental results.

 

Q.10.13 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.10.13

Let an object at O be placed in front of a plane mirror MO’ at a distance r. A circle is drawn from the centre (O) such that it just touches the plane mirror at point O’.

According to Huygens’s principle, XY is the wave front of incident light.

If the mirror is absent, then a similar wave front X’Y’ (as XY) would form behind O’ at a distance r, as shown in the 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.10.14 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/or 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.10.14 The speed of light in vacuum (3 $\times {10}^{8}m/s)$

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.

Out of these 5 factors, the speed of light in a medium depends on the wavelength of light in that medium.

 

Q.10.15 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.10.15 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 notion 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.10.16 In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?

Ans.10.16 Wavelength of the light used, $\lambda =$ 6000 nm = 600 $\times {10}^{-9}$ m

Angular width of the fringe, $\theta =0.1°$ = 0.1 $\times \frac{\pi }{180}$  rad

Angular width of a fringe is related to slit spacing (d) as $\theta =\frac{\lambda }{d}$

d = $\frac{\lambda }{\theta }$ = $\frac{600\times {10}^{-9}}{0.1\mathrm{}\times \frac{\pi }{180}}$ = 3.44 $\times {10}^{-4}$ m

 

Q.10.17 Answer the following questions:

(a) 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?

(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? $\frac{40}{2}$

(c) 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?

(d) 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.

(e) 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.10.17 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 increases up to four times.

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.

This is because light waves 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.

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.

Wavelength of light waves is too small in comparison to the size of the obstacle. Thus the diffraction angle will be small, resulting in students cannot see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves, bending of waves takes place at large angle – Resulting in students can hear each other.

The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.

 

Q.10.18 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.10.18 Distance between the towers, d = 40 km

Height of the line joining the hills, h = 50 m

Thus, the radial spread of the radio wave should not exceed 40 km

Since the hill is located halfway between the towers,

Fresnel’s distance $Z_P$  = $\frac{40}{2}$ = 20 km = 2 $\times {10}^{4}m$

Aperture can be taken as a = h = 50 m

Fresnel’s distance is given by the relation,

$Z_P$ = $\frac{a^2}{\lambda }$ or

$\lambda =\frac{a^2}{Z_P}$ =  $\frac{{50}^2}{2\mathrm{}\times {10}^4}$ = 0.125 m = 12.5 cm

Therefore, the wavelength of the radio wave is 12.5 cm

 

Q.10.19 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.10.19 Wavelength of the light beam, $\lambda$ = 500 nm = 500 $\times {10}^{-9}$ m

Distance of the screen from the slit, D = 1 m $n$

Distance of the first minimum from the centre of the screen, x = 2.5 mm = 2.5 $\times {10}^{-3}$ m

Let the width of the slit be = d

From the equation

$n\lambda =x\frac{d}{D}$ we get d = $\frac{n\lambda D}{x}$

$d=\frac{1\times 500\mathrm{}\times {10}^{-9}\times 1}{2.5\mathrm{}\times {10}^{-3}}$  = 2 $\times {10}^{-4}$ m = 0.2 mm

Hence, the width of the slot is 0.2 mm

 

Q.10.20 Answer the following questions:

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

(b) 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.10.20 Weak radar signals sent by a low flying aircraft can interfere with the TV signal received by the antenna. Hence, TV signal may get distorted, resulting in shaking of picture on the TV.

This is because superposition follows from the linear character of a differential equation that governs wave motion. If $y_1$ and $y_2$ are the solutions of the second order wave equation, then any linear combination of $y_1$ and $y_2$ will also be the solution of the wave equation.

 

Q.10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n $\mathit{\lambda }$ /a.

Justify this by suitably dividing the slit to bring out the cancellation.

Ans. 10.21 Consider that a single slit of width $d$ is divided in to $n$   smaller slits.

Therefore width of each slit,

$d^{\text{'}}=\frac{d}{n}$

Angle of diffraction is given by the relation,

$\theta =\frac{\frac{d}{d}}{d}\lambda$ = $\frac{\lambda }{d}$

Now, each of these infinitesimally small slit sends zero intensity in the direction of Ø.

Hence, the combination of these slits will give zero intensity.