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At a point on the screen in YDSE, 3rd maxima is observed at $t = 0$ . Now screen is slowly moved with constant speed away from the slits in such a way that the centre of slits and centre of screen lie on same line always and at $t = 1$ s, the intensity at that point is observed to be ${(\frac{3}{4})}^{t h}$ of maximum intensity in between 2^nd and 3^rd maxima. The maximum speed of screen in m/s will be____
(Distance between slits and the screen at t=0 is 1m)

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answer is 0.38.

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Detailed Solution

$\frac{2 π}{λ} (\frac{y d}{D + v}) = 4 π + \frac{π}{3}$ ; $\frac{2 π}{λ} \frac{D (3 λ)}{D + v} = 4 π + \frac{π}{3}$

$\Rightarrow v = \frac{5}{13}$ m/s

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