Practice portal › Wave Optics › Interference and Young's Double Slit Experiment
Asked in GSEB Board July 2016 · Fringe width and fringe position
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.
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