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In a Young's double-slit experiment, the width of one slit is three times that of the other. The amplitude of light from a slit is proportional to its width. The ratio of minimum to maximum intensity in the interference pattern is x: 4. Find x.

1. 1
2. 2
3. 3
4. 0

Correct answer: 1

Solution

Since amplitude is proportional to slit width, a1: a2 = 1: 3. Imin = (a2 - a1)² proportional to (3-1)² = 4; Imax = (a2 + a1)² proportional to (3+1)² = 16. So Imin: Imax = 4: 16 = 1: 4. Given this equals x: 4, we get x = 1.

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