Interference of Light

1 Mark Questions

1.Define the term 'coherent sources' which are required to produce interference pattern in Young's double slit experiment. [Delhi 2014 c]

Ans. Two monochromatic sources, which produce light waves, having a constant phase difference are known as coherent sources.

1. How would the angular separation of interference fringes in Young's double slit experiment

change when the distance between the slits and screen is halved? [All India 2009] Ans.

The angular separation of interference fringes in Young's double slit experiment becomes double when separation between slits and screen is halved as angular separation,

<math>\theta \propto \frac{1}{D}</math> (1)

1. Why are coherent sources required to create interference of light? [Foreign 2009] Ans. To observe interference fringe pattern, there is need to have coherent sources of light which can produce light of constant phase difference
2. How does the fringe width of interference fringes change, when the whole apparatus of Young's experiment is kept in a liquid of refractive index, 1.3? [hots; Delhi 2008] Ans.

<math>{}^{\circ}</math> In Young's double slit experiment, fringe width, <math>\beta = \frac{D\lambda}{J}</math>. From the formula it is clear that when apparatus is dipped in a liquid then only wavelength <math>\boldsymbol{\lambda}</math> will change, it will become <math>\frac{1}{\mu}</math> times of its value in air.

The new fringe width becomes <math>\frac{1}{1.3}</math> times of original fringe width as

<math display="block">\frac{\beta_{air}}{\beta_{med}} = \mu \quad \Rightarrow \quad \beta_{med} = \frac{\beta_{air}}{\mu}</math> (1)

1. How does the angular separation of interference fringes change, in Young's experiment, if the distance between the slits is increased?[Delhi 2008]

Ans.

Angular separation decreases with the increase of separation between two slits as,

<math>\theta = \frac{\lambda}{d}</math>

where, d = separation between two slits. (1)

2 Marks Questions

6.Laser light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of laser light which produce interference fringes separated by 8.1 mm using same pair Of slits. [All India 2011]

Ans.

Given, <math>\beta_1 = 7.2 \times 10^{-3} \text{ m}</math>, <math>\beta_2 = 8.1 \times 10^{-3} \text{ m}</math>

and <math>\lambda_1 = 630 \times 10^{-9} \text{ m}</math>

∴ Fringe width, <math>\beta = \frac{D\lambda}{d}</math>

where, <math>\lambda</math> = wavelength, D = separation between slits and screen and <math>d = separation</math> between two slits. <math display="block">\Rightarrow \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} \quad (:D \text{ and } d \text{ are same})</math> Wavelength of another source of laser light (1) <math display="block">\Rightarrow \lambda_2 = \frac{\beta_2}{\beta_1} \times \lambda_1</math>

<math display="block">= \frac{8.1 \times 10^{-3}}{7.2 \times 10^{-3}} \times 630 \times 10^{-9} \text{ m}</math>

or <math>\lambda_2 = 708.75 \times 10^{-9} \text{ m}</math> <math>\lambda_2 = 708.75 \text{ nm}</math>

(1)

1. How will the interference pattern in Young's double slit experiment get affected, when

(i)distance between the slits S₁ and S2 reduced and

(ii)the entire set up is immersed in water? Justify your answer in each case. [Delhi 2011] Ans.

1. The fringe width of interference pattern increases with the decrease in separation between <math>S_1S_2</math> as

<math>\beta \propto \frac{1}{d}</math> (1)

1. The fringe width decrease as wavelength gets reduced when interference set up is taken from air to water. (1)
2. The figure shows a double slit experimental set up for observing interference fringes due to different interference component colours of white light.