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Q.

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

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a

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

b

$\frac{2 {λDa}_{0}}{(d_{0}^{2} - a_{0}^{2})}$

c

$\frac{λD}{d_{0}^{2}} a_{0}$

d

$\frac{λD}{d_{0} + a_{0}}$

answer is B.

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Detailed Solution

Fringe width $β = \frac{Dλ}{d}$
$\begin{array}{l} β = \frac{Dλ}{(d_{0} + a_{0} \sin ωt)} \\ [∴ d = d_{0} + a_{0} \sin ωt] \\ β_{max} - β_{\min} \Rightarrow \frac{Dλ}{d_{0} - a_{0}} - \frac{Dλ}{d_{0} + a_{0}} \\ \Rightarrow Dλ [\frac{d_{0} + a_{0} - d_{0} + a_{0}}{d_{0} + a_{0}}] = \frac{2 a_{0} Dλ}{d_{0}^{2} - a_{0}^{2}} \end{array}$

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