At a fixed point on the screen, the 3rd maximum is observed when the screen is at distance D from the slits in a standard YDSE. The screen is then moved away from the slits with constant speed, keeping the centres of the slit plane and screen collinear. After 1 s, the intensity at that same point is $\frac{3}{4}$ of the maximum intensity at a position between the 2nd and 3rd maxima. Find the speed of the screen.
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Question

At a fixed point on the screen, the 3rd maximum is observed when the screen is at distance D from the slits in a standard YDSE. The screen is then moved away from the slits with constant speed, keeping the centres of the slit plane and screen collinear. After 1 s, the intensity at that same point is $\frac{3}{4}$ of the maximum intensity at a position between the 2nd and 3rd maxima. Find the speed of the screen.

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G1 Physics · Accepted Answer

At the fixed point, the initial order is 3. When the screen moves away, the order becomes m=3D/D′. The source-key branch for intensity 3Iₘₐₓ/4 between the 2nd and 3rd maxima is m=13/6. Hence 3D/D′=13/6, so D′=18D/13. In 1 s the screen moves D′−D=5D/13. Therefore the speed is 5D/13. Option A.