# NCERT Solutions for Class 12 Physics Chapter 10 - Wave Optics

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.

Sol. Given, wavelength of light λ = 589 nm = 589 × 10–9 m

Refractive index of water µw = 1.33

(a) For reflected light

(i) Wavelength of reflected light λ = 589 × 10–9 m

(ii) Frequency of reflected light

$$v=\frac{c}{\lambda}=\frac{3×10^{8}}{589×10^{\normalsize-9}}$$

where c is velocity of light

(∵ Speed of light c = 3 × 108 m/s)

v = 5.09 × 1014 Hz
(iii) In reflection the medium takes place in the same medium so speed of reflected light c = 3 × 108 m/s

(b) For refracted light

Wavelength of refracted light

$$\lambda'=\frac{\lambda}{\mu}\\=\frac{589×10^{\normalsize-9}}{1.33}=4.42×10^{\normalsize-7}\text{m}\\\text{Velocity of refracted light v =}\frac{c}{\mu}\\=\frac{3×10^{8}}{1.33}=2.25×10^{8}\text{m/s}$$

2. What is the shape of the wavefront 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 wavefront of light from a distant star intercepted by the earth.

Sol. (a) If the light diverging from a point source, then the shape of wavefront is diverging spherical as shown in figure (b) When the rays of light becomes parallel after refraction from convex lens, the plane wavefront is formed as shown in figure. (c) The light rays coming from a distant star are almost parallel to each other. So, the wavefront is plane as shown in figure. 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)

(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?

Sol. (a) Given, refractive index of glass µglass = 1.5

Speed of light c = 3 × 108 m/s

$$\text{Speed of light in glass v =}\frac{c}{\mu}=\frac{3×10^{8}}{1.5}$$

v = 2 × 108 m/s

(b) No, the speed of light is dependent on colour of light.
We know that the refractive index of violet is greater than red,

µv > µR

Thus, velocity of violet is less than the velocity of red. Therefore, violet colour travels slowers than the red colour in glass.

vv < vR

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.

Sol. Here, separation between two slits d = 0.28 mm

= 0.28 ×10–3m.

Distance between screen and slit D = 1.4 m.

The  distance between central bright and fourth bright fringe

x = 1.2 cm = 1.2 × 10–2 m

Since, number of fringes n = 4

$$\text{For constructive interference x =n}\frac{D\lambda}{d}\\1.2×10^{\normalsize-2}=\frac{4×1.4×\lambda}{0.28×10^{\normalsize-3}}\\\text{Wavelength,}\space\lambda=\frac{1.2×10^{\normalsize-2}×0.28×10^{\normalsize-3}}{4×1.4}$$

λ = 6 × 10–7m

or

λ = 600 × 10–9m

= 600 nm [∵1 nm = 10–9m]

The wavelength of light is 6 × 10–7m.

5. In Young’s double-slit experiment using mono-chromatic light of wavelength l the intensity of light at a point on the screen where path difference is λ is K units. What is the intensity of light at a point where path difference is l/3?

Sol. Given, wavelength = λ

When the path difference is λ (x), the phase difference

$$\phi=\frac{2\pi}{\lambda}.x=\frac{2\pi}{\lambda}=2\pi\\\text{Resultant intensity I}_R =\text{I}_{1}+\text{I}_{2}+2\sqrt{\text{I}_{1}\text{I}_{2}\text{cos}\phi}\\\text{I}_{R}=\text{I}+\text{I}+2\sqrt{\text{I.I}}\space\text{cos}\space2\pi=2\text{I}+ 2\text{I}= 4\text{I}=\text{K}(\text{Given})\space...(\text{i})\\(\because\text{I}_{1}=\text{I}_{2}=\text{I})\\\text{When path difference is}\space\frac{\lambda}{3},\text{then}\\\text{Phase difference =}\frac{2\pi}{\lambda}.\frac{\lambda}{3}=\frac{2\pi}{3}\\$$

The resultant intensity

$$\text{I}'_{\text{R}}=\text{I + I +2}\sqrt{\text{I.I}}\space\text{cos}\space\frac{2\pi}{3}\\2\text{I}+2\text{I}\bigg(-\frac{1}{2}\bigg)\\\bigg[\because\text{cos}\frac{2\pi}{3}=-\frac{1}{2}\bigg]\\\text{I}'_{\text{R}}=\text{I}=\frac{\text{K}}{4}\space[\text{From Eq. (i)}]\\\text{Thus, the intensity of light at a point of path difference}\space\frac{\lambda}{3}\space\text{is}\space\frac{\text{K}}{4}.$$

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?

Sol. Given, wavelength λ1 = 650 nm = 650 × 10–9 m
and
l2 = 320 nm = 520 × 10–9 m

(a) For third fringe bright, n = 3

The distance between third bright fringe and central maximum.

$$x=\frac{n\lambda D}{d}=3×650×10^{\normalsize-9}×\frac{\text{D}}{d}\text{m}\\=\frac{3×650×10^{\normalsize-9}×1.2}{2×10^{\normalsize-3}}$$

= 1.17 × 10–3 m

(b) Let nth bright fringe due to wavelength λ2 = (520 nm) coincide with (n + 1)th bright fringe due to wavelength λ1 (650 nm.)

$$\text{i.e.}\space n\lambda_{2}=\frac{\text{D}}{d}=(n-1)\lambda_{1}\frac{\text{D}}{d}m$$

n × 520 × 10–9 = (n – 1) 650 × 10–9

or 4n = 5n – 5

or n = 5

Thus, the least distance

$$x=n\lambda_{2}\frac{\text{D}}{d}=5×520×10^{-9}\frac{\text{D}}{d}\\x=2600\space\frac{\text{D}}{d}×10^{\normalsize-9}m\\=2600×\frac{1.2×10^{\normalsize-9}}{2×10^{\normalsize-3}}m$$

= 1.56 × 10–3m = 1.56mm

7. In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1m 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.

Sol. Given, angular width θ = 0.2°

Distance between screen and slit D = 1m

Wavelength of light l = 600 nm = 600 × 10–9 m

$$\text{Refractive index of water}\space\mu_ w=\frac{4}{3}\\\text{Using the formula of angular width}\\\theta=\frac{\lambda}{\text{D}}\qquad...\text{(i)}\\\text{and}\space\theta'=\frac{\lambda'}{\text{D}}\qquad...(\text{ii})\\\text{where,}\space\lambda'=\frac{\lambda}{\mu}$$

Dividing Eq. (ii) from Eq. (i), we get

$$\frac{\theta'}{\theta}=\frac{\lambda'}{\lambda}=\frac{\lambda'}{\mu\lambda}\\\text{or}\space\theta'=\frac{\theta}{\mu}=\frac{0.2×3}{4}=0.15\bigg(\because\mu=\frac{4}{3}\bigg)$$

Thus, the angular fringe width is 0.15°

8. What is the Brewster’s angle for air to glass transition? (Refractive index of glass = 1.5)

Sol. Given, µg = 1.5

Let ip be the Brewster’s angle.

From the Brewster’s law.

Refractive index, µ = tan ip

or tan ip = 1.5

ip = tan–1 (1.5)

ip = 56°18’

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?

Sol. Here, wavelength of light λ = 5000 Å = 5000 × 10–10 m

In case of reflection there is no change in wavelength and frequency. So, wavelength of reflected light will be 5000 Å.

$$\text{Frequency of the incident light v =}\frac{c}{\lambda}=\frac{3×10^{8}}{5×10^{\normalsize-7}}= 6 × 10^{14}\space\text{Hz}$$

When reflected ray is normal to the incident ray, AO and BO are the incident and reflected rays.

BO ⊥ AO

∴ ∠i + ∠r = 90°

For reflection, ∠i = ∠r

∴ 2 ∠i = 90°

∠i = 45° Thus, the angle of incidence is 45°.

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

Sol. Given, aperture a = 4 mm = 4 × 10–3 m

Wavelength l = 400 nm = 400 × 10–8 m

Ray optics is good approximation upto a distance equal to Fresnel’s distance (ZF).

$$b\text{Z}_{\text{F}}=\frac{a^{2}}{\lambda}=\frac{4×10^{\normalsize-3}×4×10^{\normalsize-3}}{400×10^{\normalsize-9}}$$

ZF = 40 m