In Young's double slit experiment the fringe pattern is observed on a screen placed at a distance D. The slits are illuminated by light of wavelength λ . The distance from the central point where the intensity falls to one fourth of the Maximum is
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a
λD3d
b
λD2d
c
λDd
d
λD4d
answer is A.
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Detailed Solution
Distance of the screen = D Wavelength =λ Separation between the two slits = dWhen intensity is 14th of the maximum IImax=14 4 I0cos2(ϕ2)4I0=14 (∵ I=4 I0cos2(ϕ2), Imax=4 I0) cos2(ϕ2) =14 cos(ϕ2) =12 ϕ2=π3 ϕ=2π3 Path difference x=λϕ2π =λ2π32π=λ3 y=x Dd =λ3 Dd ∴ y= λD3d