The physical widths of the two narrow slits in a Young's double-slit experiment are in the ratio 1:2. Assume the wave amplitude contributed by a slit is proportional to its width. What is I(max) ÷ I(min) for the interference pattern?
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Question

The physical widths of the two narrow slits in a Young's double-slit experiment are in the ratio 1:2. Assume the wave amplitude contributed by a slit is proportional to its width. What is I(max) ÷ I(min) for the interference pattern?

Options

G1 Physics · Accepted Answer

Amplitude ratio is 1:2. Let the amplitudes be a and 2a. Then, I(max) ∝ (3a)² = 9a², I(min) ∝ (a)² = a². Therefore, I(max) ÷ I(min) = 9. Correct option: C.