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Q.

In the young’s double slit experiment, the distance between the slits varies in time as $d (t) = d_{0} + a_{0} \sin ω t,$ 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

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

b

$\frac{λ D}{d_{0} + a_{0}}$

c

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

d

$\frac{λ D a_{0}}{d_{0}^{2}}$

answer is A.

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Detailed Solution

Fringe width $β = \frac{λ D}{d} = \frac{λ D}{d_{0} + a_{0} \sin ω t}$

$\begin{array}{l} β_{\max} = \frac{λ D}{d_{\min}} = \frac{λ D}{d_{0} - a_{0}} \\ β_{\min} = \frac{λ D}{d_{\min}} = \frac{λ D}{d_{0} + a_{0}} \\ β_{\max} - β_{\min} = λ D (\frac{1}{d_{0} - a_{0}} - \frac{1}{d_{0} + a_{0}}) = \frac{2 λ D a_{0}}{d_{0}^{2} - d_{0}^{2}} \end{array}$

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