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In YDSE set up, the distance between slits varies as $d = d_{0} + A \sin ωt .$ What is the difference between maximum and minimum fringe width? (D= distance between slits and screen $λ$ = wavelength of light used)

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$\frac{2 λ d_{0} D}{d_{0}^{2} - A^{2}}$

$\frac{2 λ A D}{d_{0}^{2} - A^{2}}$

$\frac{2 λ A D}{d_{0}^{2} + A^{2}}$

$\frac{2 λ A D}{d_{0}^{2}}$

answer is B.

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Detailed Solution

$β_{\max} = \frac{λ D}{d_{0} - A}$

$β_{\min} = \frac{λ D}{d_{0} + A}$

$β_{\max} - β_{\min} = \frac{λ D}{d_{0} - A} - \frac{λ D}{d_{0} + A}$

$= \frac{λ D (d_{0} + A - d_{0} + A)}{{d_{0}}^{2} - A^{2}}$

$= \frac{2 λ D A}{{d_{0}}^{2} - A^{2}}$

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