In the Young's double slit experiment, the distance between the slits… | JEE Main 2021 (25 Jul, Shift 1)

Q57JEE Main 2021 · 25 JUL · SHIFT 1Medium

YDSE: Fringe Width & Fringe Positions

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:

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Answer: (b)

$\beta = \frac{\lambda D}{d}$ ; $d$ ranges from $d_0 - a_0$ to $d_0 + a_0$
$\beta_{max} = \frac{\lambda D}{d_0 - a_0}$ , $\beta_{min} = \frac{\lambda D}{d_0 + a_0}$
$\beta_{max} - \beta_{min} = \lambda D \frac{(d_0 + a_0) - (d_0 - a_0)}{d_0^2 - a_0^2} = \frac{2\lambda D a_0}{d_0^2 - a_0^2}$
So the answer is (b).

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