NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics – PDF Download & Explanations

10.14 The speed of light in vacuum (3 $*{10}^{8}m/s)$

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.

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 $?=\frac{v}{c}$

So from equation (1) we get

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

But $?>1,v>c$ 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.

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

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

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

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

We know, phase difference $?$ = $\frac{2?}{?}$ $\times$ path difference

Since path difference = $?$ , phase difference $?$ = 2 $?$ , then

$I’=$ 2 $I_1+2I_1\cos ?2?$ = 4 $l_1$

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

When path difference is $\frac{?}{3}$ , phase difference $?=\frac{2?}{3}$ , then

$I’$ = 2 $I_1+2I_1\cos ??$ = 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 $\frac{?}{3}$ is $\frac{K}{4}$ units.

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).

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

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

$?$ = wave length = 400 nm = 400 $*{10}^{-9}$ m

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

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

10.19 Wavelength of the light beam, $?$ = 500 nm = 500 $*{10}^{-9}$ 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 $*{10}^{-3}$ m

Let the width of the slit be = d

From the equation

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

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

Hence, the width of the slot is 0.2 mm

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,

$?=\frac{\frac{d}{d}}{d}?$ = $\frac{?}{d}$

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.

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.

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

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

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

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

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

Refractive index is related to angular width as:

$?=\frac{{?}_1}{{?}_2}$ or

${?}_2=\frac{{?}_1}{?}$ = 0.2 $*\frac{3}{4}$ = 0.15 $°$

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

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.

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.

10.9 Wave length of the incident light, $?$ = 5000 $*{10}^{-10}$ m

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

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

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

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

Hence, $?i+?r$ = 2 $?i$ = 90 $°$

So$?i=?r=45°$

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

10.16 Wavelength of the light used, $?=$ 6000 nm = 600 $*{10}^{-9}$ m

Angular width of the fringe, $?=0.1°$ = 0.1 $*\frac{?}{180}$  rad

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

d = $\frac{?}{?}$ = $\frac{600*{10}^{-9}}{0.1\mathrm{}*\frac{?}{180}}$ = 3.44 $*{10}^{-4}$ m

10.6 Wavelength of one light beam, ${?}_1$ = 650 nm

Wavelength of the other beam, ${?}_2$ = 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 ${?}_1(\frac{D}{d}$ )

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

n ${?}_2$ = (n-1) ${?}_1$

520n = 650n – 650

n = $\frac{650}{130}$ = 5

The least distance can be obtained from the relation,

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

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.

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

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

Refractive index of water, $?$ = 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, $?$ = $\frac{c}{?}$ = $\frac{3\mathrm{}\times {10}^8}{589\mathrm{}\times {10}^{-9}}$ = 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 = $\frac{c}{?}$ = $\frac{3\mathrm{}\times {10}^8}{1.33}$ = 0.226 $\times {10}^9$ m/s

Wavelength in water, $?$ = $\frac{v}{?}$ = $\frac{0.226\mathrm{}\times {10}^9}{5.09\mathrm{}\times {10}^{14}}$ = 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

10.4 Distance between the slits, d = 0.28 mm = 0.28 $*{10}^{-3}$ 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 $*{10}^{-2}$ m

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

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

Hence, wavelength of the light is 600 nm.

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 $*{10}^{4}m$

Aperture can be taken as a = h = 50 m

Fresnel's distance is given by the relation,

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

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

Therefore, the wavelength of the radio wave is 12.5 cm

10.3 The refractive index of glass,  $?$ = 1.5

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

Speed of light in glass is given by the relation,  $v$ = $\frac{c}{?}$ = $\frac{3.0\mathrm{}*{10}^8}{1.5}$ = 2 $*{10}^8$ 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.

10.11 The wavelength of $H_{?}$ line emitted by Hydrogen, $?=6563\AA$ = 6563 $*{10}^{-10}$ m$$

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

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

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

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

10.8 Refractive index of glass,  $?=1.5$

Let the Brewster angle be $?$

Brewster angle is related to $?$ by the equation

$\tan ??=?$

Or  = $?={\tan }^{-1}??$ =${\tan }^{-1}?1.5$ =56.31$°$

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