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

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

$\frac{λ D}{d_{0} + a_{0}}$

$\frac{λ D}{d_{0}^{2}} a_{0}$

answer is B.

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Detailed Solution

Fringe Width, $β = \frac{λ D}{d}$

$β_{max} \Rightarrow d_{min} and β_{min} \Rightarrow d_{max}$

$d = d_{0} + a_{0} \sin ω t$

$d_{max} = d_{0} + a_{0} and d_{min} = d_{0} - a_{0}$

$∴ β_{min} = \frac{λ D}{d_{0} + a_{0}} and ∴ β_{max} = \frac{λ D}{d_{0} - a_{0}}$

$β_{max} - β_{min} = \frac{λ D}{d_{0} - a_{0}} - \frac{λ D}{d_{0} + a_{0}} = \frac{2 λ D a_{0}}{d_{0}^{2} - a_{0}^{2}}$

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