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In Young's experiment the distance between the two slits is halved and the distance between the screen and the slits is doubled. The width of the fringe ________.

Asked in GSEB Board July 2016 · Fringe width and fringe position

Answer: (4) becomes 4 times

Step-by-step solution

Given: d → d/2 and D → 2D, with the wavelength unchanged.

Idea: the fringe width is β = (λ D)/d — directly proportional to the screen distance, inversely proportional to the slit separation.

Halving the slit separation on its own doubles β.

Doubling the screen distance on its own doubles β again.

Together, β' = (λ(2D))/(d/2) = 4 × (λ D)/d = 4β.

The fringe width becomes 4 times what it was.

Why the other options are wrong

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