In the Young's double slit | Wave Optics |JEE-MAIN 2021

JEE MAIN 2021

25-07-2021 S1

Question

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 :

Select the correct option:

$\frac{2 \lambda \mathrm{D}\left(\mathrm{d}_0\right)}{\left(\mathrm{d}_0^2-\mathrm{a}_0^2\right)}$

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

$\frac{\lambda \mathrm{D}}{\mathrm{d}_0^2} \mathrm{a}_0$

$\frac{\lambda D}{d_0+a_0}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution