# Wave Optics

## Physics

### NCERT

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 $1.33.$

##### Solution :

Wavelength of incident monochromatic light,$\\$ $\lambda 589 nm = 589*10^{-9} m$$\\ Speed of light in air, c = 3 * 10^8 m / s$$\\$ Refractive index of water, $\mu =1.33$$\\ (a) The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray. Frequency of light is given by the relation,\\ v=\dfrac{c}{\lambda}\\ =\dfrac{3*10^8}{589*10^{-9}}\\ =5.09*10^{14} Hz$$\\$ Hence, the speed, frequency, and wavelength of the reflected light are $3 * 10^8 m / s ,5.09 * 10 Hz$ , and $589 nm$ respectively.

(b) frequency of light does not depend on the property of the medium in which it is travelling. Hence, The frequency f the refracted ray in water will be equal to the frequency of the incident or reflected light in air. $\therefore$ Refracted frequency, $v =5.09 * 10^{14} Hz$$\\ Speed of light in water is related to the refractive index of water as:\\ v=\dfrac{c}{\mu}\\ v=\dfrac{3*10^8}{1.33}\\ =2.26*10^8m/s$$\\$ Wavelength of light in water is given by the relation,$\\$ $\lambda =\dfrac{v}{u}\\ =\dfrac{2.26*10^8}{5.09*10^{14}}\\ =444.007*10^{-9}m\\ =444.01 nm$

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.

##### Solution :

(a) The shape of the wave front in case of a light diverging from a point source is spherical. The wave front emanating from a point source is shown in the given figure.$\\$ (b) 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. This is shown in the given figure.$\\$ (c) The portion of the wave front of light from a distant star intercepted by the Earth is a plane.

3   (a) The refractive index of glass is $1.5.$ What is the speed of light in glass? Speed of light in vacuum is $3.0 * 10^8 ms^{-1}$ )$\\$ (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?

##### Solution :

(a) Refractive index of glass, $\mu= 1.5$$\\ Speed of light, c = 3*10^8 m / s$$\\$ Speed Of light in glass is given by the relation,$\\$ $v=\dfrac{c}{\mu}\\ =\dfrac{3*10^8}{1.5}\\ =2*10^8 m/s$$\\ Hence, the speed of light in glass is 2 × 10^8 m/s.$$\\$ (b) The speed of light in glass is not independent 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. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

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.

##### Solution :

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 fringe and the fourth $(n= 4)$ fringe, $u = 1.2 cm = 1.2*10^{-2} m$$\\ In case of a constructive interference, we have the relation for the distance between the two fringes as:\\ u=n \lambda \dfrac{D}{d}$$\\$ where,$\\$ $n =$ Order of fringes $= 4$$\\ \lambda = Wavelength of light used\\ \therefore \lambda =\dfrac{ud}{nD}\\ =\dfrac{1.2*10^{-2}*0.28*10^{-3}}{4*1.4}\\ =6*10^{-7}\\ =600 nm$$\\$ Hence, the wavelength of the light is $600 nm.$

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 $\lambda /3$?

##### Solution :

Let $I_1$ and $I_2$ be the intensity of the two light waves. Their resultant intensities can be obtained as:$\\$ $I'=I_1+I_2+2\sqrt{I_1I_2}\cos \phi $$\\ Where,\\ \phi=Phase difference between the two waves for monochromatic light waves,\\ Since I_1=I_2$$\\$ So $I'=2I_1+2I_1 \cos \phi $$\\ Phase difference =\dfrac{2 \pi}{\lambda}*\text{Path difference}$$\\$ Since path difference $\lambda$ ,$\\$ Phase difference,$\phi =2 \pi $$\\ \therefore I'=2I_1+2I_1=4I_1$$\\$ Given,$\\$ $\therefore I_1=\dfrac{K'}{4}$ $\\$ When path difference $=\dfrac{\lambda}{3}$$\\ phase difference,\phi=\dfrac{2\pi}{3}$$\\$ Hence, resultant intensity,$\\$ $I_r=I_1+I_1+2\sqrt{I_1I_1}\cos \dfrac{2\pi}{3}\\ =2I_1+2I_2(-\dfrac{1}{2})=I_1$$\\ Using equation (1), we can write:\\ I_R=I_1=\dfrac{K}{4}$$\\$ Hence, the intensity of light at a point where the path difference is $\dfrac{\lambda}{3}$ is $\dfrac{K}{4}$units.$\\$ For monochromatic light waves,

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.$$\\ (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? ##### Solution : Wavelength of the light beam, \lambda _1 = 650nm$$\\$ Wavelength of another light beam, $\lambda _2 = 520nm$$\\ Distance of the slits from the screen = D$$\\$ Distance between the two slits $= d$$\\ (a) Distance of the n ^{th} bright fringe on the screen from the central maximum is given by the relation,\\ x=n\lambda_1(\dfrac{D}{d})$$\\$ For third bright fringe, $n = 3$$\\ \therefore x=3*650 \dfrac{D}{d}=1950(\dfrac{D}{d})nm$$\\$ (b) Let the n th bright fringe due to wavelength $\lambda _2$ 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\\ 650=130n\\ \therefore n=5$$\\ Hence, the least distance from the central maximum cab be obtained by the relation:\\ x=n\lambda_2\dfrac{D}{d}\\ =5*520\dfrac{D}{d}=2600\dfrac{D}{d}nm$$\\$ Note: The value of d and D are not given in the question.

7   In a double-slit experiment the angular width of a fringe is found to be $0.2^o$ 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 $4/3.$

##### Solution :

Distance of the screen from the slits, $D = 1 m$$\\ Wavelength of light used, \lambda _1 = 600nm$$\\$ Angular width of the fringe in air, $\theta_1 =0.2^o$$\\ Angular width of the fringe in water =\theta_2$$\\$ Refractive index of water,$\mu =\dfrac{4}{3}$$\\ Refractive index is related to angular width as:\\ \mu =\dfrac{\theta_1}{\theta_2}\\ \theta_2=\dfrac{3}{4}\theta_1\\ =\dfrac{3}{4}*0.2=0.15$$\\$ Therefore the angular width of the fringe in water will reduce to$0.15^o$

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

##### Solution :

Refractive index of glass,$\mu = 1.5$$\\ Brewster angle =\theta$$\\$ Brewster angle is related to refractive index as:$\\$ $\tan \theta=\mu\\ \theta= \tan ^{- 1} ( 1.5)= 56.31^o$ Therefore, the Brewster angle for air to glass transition is $56.31^o$

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?

##### Solution :

Wavelength of incident light,$\lambda = 5000A = 5000 * 10^{-10} m$$\\ Speed of light, c = 3* 10^ 8 m$$\\$ Frequency of incident light is given by the relation,$\\$ $v=\dfrac{c}{\lambda}\\ =\dfrac{3*10^8}{5000*10^{-10}}\\ =6*10^{14}Hz$$\\ The wavelength and frequency of incident light is the same as that of reflected ray.\\ Hence, the wavelength of reflected light is 5000 A and its frequency is 6 * 10 ^{14} Hz.$$\\$ When reflected ray is normal to incident ray, the sum of the angle of incidence, $\angle i$ and angle of reflection, $\angle r$ is $90^o$$\\ According to the law of reflection, the angle of incidence is always equal to the angle of reflection. Hence, we can write the sum as:\\ \angle i+\angle r=90\\ \angle i+\angle i=90\\ \angle i=\dfrac{90}{2}=45^o$$\\$ Therefore, the angle of incidence for the given condition is $45^o$

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

##### Solution :

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=\dfrac{a^2}{\lambda}$$\\ Where,\\ Aperture width, a =4 mm = 4 * 10 ^{- 3 }m$$\\$ Wavelength of light $\lambda 400 nm = 400 * 10 ^{- 9} m$$\\ Z_F=\dfrac{(4*10^{-3})^2}{400*10^{-9}}=40 m$$\\$ Therefore, the distance for which the ray optics is a good approximation is$40 m$$\\$