The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:
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The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:

G1 Physics · Accepted Answer

According to the source solution: Intensity is taken proportional to slit width. Let: I₁ = I I₂ = 4I Then: I(max) = (√I₁ + √I₂)² = (√I + 2√I)² = 9I I(min) = (√I₂ - √I₁)² = (2√I - √I)² = I Therefore: I(max) : I(min) = 9 : 1