# NCERT Solution | Wave Optics

Q:1. Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (i) reflected and (ii) reflected light? (μ of water is 1.33).

Sol. Wavelength, λ = 589 × 10-9 m , c = 3 × 108 m/s , μ =1.33

For reflected light,
Wavelength, λ = 589 × 10-9 m

ν = c/λ

ν =(3 × 108)/(589 × 10-9)

= 5.09 × 1014 Hz

Speed, v = c = 3 × 108 m/s

(ii) For refracted light,

λ’ = λ/μ

=(589 × 10-9)/1.33

= 4.42 × 10-7 m

As, frequency remain unaffected on entering another medium, therefore

ν’ = ν = 5.09 × 1014 Hz

Speed,v’ = c/μ

=(3 × 108)/1.33

= 2.25 × 108 m/s

Q:2. What is the shape of the wavefront in each of the following cases?
(i) Light diverging from point source.
(ii) Light emerging out of a convex lens when point source is placed at its focus.
(iii) The portion of the wavefront of light from a distance star intercepted by the earth.

Sol: (i) Geometrical Shape of the wavefront would be diverging Spherical Wavefront .

(ii) As rays emerging out of a convex lens are parallel , therefore Wavefront must be plane .

(iii) as, the star (i.e. source of light) is very far off, i.e. at infinity, the wavefront intercepted by the earth must be a plane wavefront.

Q:3. (i)The refractive index of glass is 1.5. what is the speed of light in glass? (Speed of light in vacuum is 3 ×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,

Sol:

(i) Here, refractive index, μ = 1.5

v = ?, c = 3 × 108 m/s

As,μ = c/v

⇒ v = c/μ

=(3×108)/1.5

= 2 × 108 m/s

(ii) No, the refractive index and the speed of light in a medium depend on wavelength, i.e. colour of light.

We know that, μv > μr

Therefore, vviolet < vred

Hence, violet component of white light travels slower than the red component.

Q:4. In 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 forth bright fringe is measured to be 1.2 cm, determine the wavelength of light used in the experiment.

Sol.
Here, slit width, d= 0.28 mm = 0.28 × 10-3 m

Distance between slit and screen, D = 1.4 m

y=1.2 cm= 1.2× 10-2 m,n = 4, λ = ?

For constructive interference, $\large y = \frac{n \lambda D}{d}$

$\large \lambda = \frac{y d}{n D}$

$\large \lambda = \frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}$

=6 × 10-7 m

Q: 5. In Young’s double slit experiment using monochromatic light of wavelength λ the intensity of light at a point on the screen, where path difference is λ is K unit. Find out the intensity of light at a point, where path difference is λ/3.

Sol: $\large I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} cos\phi$

Let , I1 = I2 = I

When Path difference = λ , Phase diff , φ= 2π

$\large I_R = I + I + 2\sqrt{I \times I} cos2\pi$

= 4 I = K (given)

When Path difference = λ/3 , Phase diff , φ = 2π/3

$\large I_R = I + I + 2\sqrt{I \times I} cos2\pi/3$

$\large I_R = 2I + 2I (-\frac{1}{2})$

= I = K/4

Q: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.

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

(ii) what is the least distance from the central maximum, where the bright fringes due to both the wavelengths coincide ?

Sol:
Here, λ1 = 650 nm = 650 × 10-9 m, λ2 = 520 nm =250×10-9 m

Suppose, d = distance between two slits

D = distance of screen from the slits

(i) For third bright fringe, n = 3

y= nλ1 D/d

= 3×650 D/d nm

= 1950 D/d nm

(ii) Let nth fringe due to λ2 = 520 nm coincide with (n-1)th bright fringe due to λ1 = 650 nm

2 = (n-1) λ1

or n×520 = (n-1)650

or 4n = 5n-5

n = 5

The least distance required,

y= n λ2 D/d

=5×520 D/d

=2600 D/d nm

Q:7. In a young’s double slit experiment, the angular width of the 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.

Sol.
Here, θ1 = 0.2°,D = 1 m , λ1 =600 nm

Angular width of a fringe, θ=λ/d

θ2 = ?, μ = 4/3

θ21 = λ21 = 1/μ =3/4

θ2=(3/4)×θ1

=(3/4)× 0.2° =0.15°

Q:8. What is Brewster angle for air to glass transition ? (μ of glass = 1.5)

Sol: $\large \mu = tani_p$

$\large i_p = tan^{-1}(\mu)$

$\large i_p = tan^{-1}(1.5)$

= 56.3°

Q:9. Light of wavelength 5000 A° 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: λ = 5000 A°

On reflection , there is no change in wavelength or frequency

wavelength of the reflected light = λ’ = λ = 5000 A°

frequency of the reflected light , ν = c/λ = 3×108/5000×10-10 = 6 × 1014Hz

When reflected ray is normal to the incident ray

∠i + ∠r = 90°

As , ∠r = ∠i

2∠i = 90°

∠i = 45°

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

Sol: a = 4 mm = 4 × 10-3 m

λ = 400 nm = 400 ×10-9 m

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

$\large Z_F = \frac{a^2}{\lambda}$

$\large Z_F = \frac{(4 \times 10^{-3})^2}{4 \times 10^{-7}}$

= 40 m

Q:11. The 6563 Ao H2 line emitted by hydrogen in a star is found to be red shifted by 15 Ao . Estimate the speed with which the star is receding from earth .

Sol: λ = 6563 Ao , Δλ = + 15 Ao , c = 3 × 108 m/s

Since star is receding away , hence its velocity is negative i.e. if Δλ is positive , v is negative .

$\displaystyle \Delta \lambda = – \frac{v \lambda}{c}$

$\displaystyle v = – \frac{c \Delta \lambda}{\lambda}$

$\displaystyle v = – \frac{3 \times 10^8 \times 15}{6563}$

v = – 6.86 × 105 m/s

-ve sign shows the recession of star .

Q:16.In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1 Ao. Find the spacing between the two slits.

Sol: λ = 600 × 10-9 m = 6 × 10-7 m

θ = 0.1° $= \frac{0.1 \pi}{180} rad$

Angular width , $\theta = \frac{\lambda}{d}$

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

$d = \frac{6 \times 10^{-7}}{\frac{\pi}{180}\times 0.1}$

d = 3.44 × 10-4 m

Q:18. Two towers on top of two hills are 40 m 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 .

Sol: In order that the hill may not obstruct the spreading radio beam , the radial speed of beam over the hill 20 km away must not exceed 50 km i.e. ZF = 20 km = 2 × 104 m ; a = 50 m ; λ = ?

As $Z_F = \frac{a^2}{\lambda}$

$\lambda = \frac{a^2}{Z_F}$

$\lambda = \frac{50^2}{2 \times 10^4}$

λ = 1250 × 10-4 m

= 0.125 m

Q: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.

Sol: λ = 500 × 10-9 m = 5 × 10-7 m

D = 1 m , n =1 ; x = 2.5 mm = 2.5 × 10-3 m ; a = ?

$x = \frac{n \lambda D}{a}$

$a = \frac{n \lambda D}{x}$

$a = \frac{1 \times 5 \times 10^{-7} \times 1}{2.5 \times 10^{-3}}$

a = 2 × 10-4 m = 0.2 mm

Q:21. In deriving the single slit diffraction pattern , it was stated that the intensity is zero at angles nλ/a . Justify this suitably dividing the slit to bring out the cancellation .

Sol: Let the slit be divided into n smaller slits each of width a’ = a/n

The angle ,$\theta = \frac{n \lambda}{a}$

$\theta = \frac{n \lambda}{a’ n} = \frac{\lambda}{a’}$

Therefore , each of the smaller slits would send intensity in the direction of θ . Hence for the entire single slit , intensity at nλ/a would be zero .