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In Young’s double-slit experiment, one slit is three times as wide as the other. If the amplitude contributed by a slit is directly proportional to its width, what is the ratio of the greatest intensity to the least intensity in the resulting interference pattern?

1. 1: 4
2. 3: 1
3. 4: 1
4. 2: 1

Correct answer: 4: 1

Solution

Amplitudes are a1 = 3a, a2 = a. Imax/Imin = (a1+a2)^2/(a1-a2)^2 = (4a)^2/(2a)^2 = 16/4 = 4, i.e. 4:1.

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