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Q.

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
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