JEE PHYSICS MCQ: WAVE OPTICS

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Application +4 | -1

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

View answer and explanation

Correct option: B

\frac{{2\lambda D{a_0}}}{{\left( {d_0^2 - a_0^2} \right)}}

Option B is correct because the difference between the largest and smallest fringe widths is derived by considering the maximum and minimum values of the time-varying slit separation in Young's double-slit experiment.

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