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In a standard Young's double-slit experiment, point P on the screen is located at a distance D from the slits and shows the 3rd bright fringe at t = 0. The screen then moves away from the slits at a constant speed v along the axis joining the centre of the slits and the centre of the screen. At t = 1 s the intensity at P is (3/4) of the maximum intensity, and P lies between the 2nd and 3rd bright fringes. What is the speed v of the screen?

1. 5D/13 m/s
2. 13D/5 m/s
3. 17D/5 m/s
4. 7D/17 m/s

Correct answer: 5D/13 m/s

Solution

At t=0, y_P = 3*lambda*D/d. At t=1, the screen is at distance D+v, so the path difference at P is 3*lambda*D/(D+v). Setting I/I_max = 3/4 gives path difference = 13*lambda/6 between the 2nd and 3rd maxima, which yields v = 5D/13.

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