The width of fringe is 2 mm on the screen in a double slits experiment for the light of wavelength of 400 nm. The width of the fringe for the light of wavelength 600 nm will be:
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Question

The width of fringe is 2 mm on the screen in a double slits experiment for the light of wavelength of 400 nm. The width of the fringe for the light of wavelength 600 nm will be:

G1 Physics · Accepted Answer

In Young's double-slit experiment, fringe width β = λD/d For the same D and d, β ∝ λ Therefore, β₂/β₁ = λ₂/λ₁ = 600/400 = 3/2 Given β₁ = 2 mm: β₂ = 2×3/2 = 3 mm Hence option B follows.