At a point on the screen in YDSE experiment $3^{r d}$ 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=1sec the intensity at that point is observed ${(3 / 4)}^{t h}$ of maximum intensity in between $2^{n d}$ and $3^{r d}$ maxima. The speed of screen can be (D=separation between the screen and slits d= separation between the slits, $d < < D, λ = 500 \overset{\circ}{A}$ )

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$\frac{17 D}{5}$

$\frac{13 D}{5}$

$\frac{5 D}{13}$

$\frac{D}{17}$

answer is A, D.

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Detailed Solution

$\frac{2 λ}{λ} (\frac{d y}{D + v}) = 4 π + \frac{π}{3} \frac{2 π}{λ} (\frac{d y}{D + v}) = 6 π - \frac{π}{3} \frac{d y}{D} = 3 λ \frac{d y}{D} = 3 λ \frac{2 π .3 D}{d x} (\frac{d y}{D + v}) = \frac{13 π}{3} \frac{2 π .3 D}{d x} (\frac{d y}{D + v'}) = \frac{17 π}{3} \frac{6 D}{D + v} = \frac{13}{3} \frac{6 D}{D + v'} = \frac{17}{3} 18 D = 13 D = 13 v 18 D = 17 D = 17 v' \frac{5 D}{13} = v v' = \frac{D}{17}$