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In the standard Young's double-slit experiment, a point P on the screen is at distance D from the slits. At time t = 0, the 3rd bright fringe is observed at P. The screen then moves slowly away from the slits (keeping the centre of the slits and centre of the screen on the same line). At t = 1 s, the intensity at P is found to be (3/4) of the maximum intensity, and P lies between the 2nd and 3rd bright fringes. What is the speed of the screen?

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

Correct answer: 5D/13

Solution

Point P is fixed. At t=0, path difference = 3*lambda. As the screen moves away to D' = D + v, path difference decreases to 3*lambda*D / D'. Setting I = (3/4)*I_max gives cos²(pi*delta/lambda) = 3/4, so pi*delta/lambda = pi/6 + n*pi. Between 2nd and 3rd maxima: delta = 13*lambda/6. This gives D' = 18D/13, so v = 5D/13 per second.

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