QJEEM21Ph94cb: JEE Main 2021 Physics - Class 12 JEEM21Ph94cb Solution: In the Young's double slit experiment, the distance between the …

Question

Verified Solution • 2025–26 Session moderate

In the Young's double slit experiment, the distance between the slits varies in time as
d(t) = d₀ + a₀ sin

\omega

t; where d₀,

\omega

and a₀ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

Accepted Answer

Fringe Width,

\beta = {{\lambda D} \over d}

{\beta _{\max }} \Rightarrow {d_{\min }}

and

{\beta _{\min }} \Rightarrow {d_{\max }}

d = {d_0} + {a_0}\sin \omega t

{d_{\max }} = {d_0} + {a_0}

and

{d_{\min }} = {d_0} - {a_0}

\therefore

{\beta _{\min }} = {{\lambda D} \over {{d_0} + {a_0}}}

and

\therefore

{\beta _{\max }} = {{\lambda D} \over {{d_0} - {a_0}}}

{\beta _{\max }} - {\beta _{\min }} = {{\lambda D} \over {{d_0} - {a_0}}} - {{\lambda D} \over {{d_0} + {a_0}}} = {{2\lambda D{a_0}} \over {d_0^2 - a_0^2}}

Step-by-Step Verified Solution