In the Young's double slit experiment, the distance between the slits varies in time as $d(t)=d_{0}+a_{0} \sin \omega t$; where $d_{0}, \omega$ and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

(C) $\frac{2 \lambda D a_{0}}{(d_{0}^{2}-a_{0}^{2})}$

(A) $\frac{\lambda D}{d_{0}+a_{0}}$

(B) $\frac{\lambda D}{d_{0}^{2}} a_{0}$

(D) $\frac{2 \lambda D(d_{0})}{(d_{0}^{2}-a_{0}^{2})}$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Wave Optics

In the Young's double slit experiment, the distance between the slits varies in time as

d (t) = d_{0} + a_{0} sin ω t

; where

d_{0}, ω

and

a_{0}

are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

Answer: (C)

The fringe width is

β = \frac{λ D}{d}

, so it is largest when the separation is smallest and smallest when the separation is largest. Over a cycle the sine runs between

- 1

and

+ 1

, so

d

runs between

d_{0} - a_{0}

and

d_{0} + a_{0}

and

Δ β = \frac{λ D}{d _{0} - a _{0}} - \frac{λ D}{d _{0} + a _{0}} = λ D \frac{( d _{0} + a _{0} ) - ( d _{0} - a _{0} )}{d _{0}^{2} - a _{0}^{2}}

The difference between the largest and the smallest fringe width is therefore

\frac{2 λ D a _{0}}{d _{0}^{2} - a _{0}^{2}}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option