The intensity at the maximum in a Young's double slit experiment is I 0 . Distance between two slits is d = 5 λ , where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d ?

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Aicharya · Accepted Answer

In YDSE I m a x = I 0 Path difference at a point in front of one of shifts is ∆ x = d y D = d d 2 D = d 2 2 D ( H e r e y = d 2 ) ∆ x = d 2 2 10 d = d 20 = 5 λ 20 = λ 4 Path difference is ϕ = 2 π λ Δ x = 2 π λ λ 4 ϕ = π 2 So intensity at that point is I = I m a x cos 2 θ 2 I = I 0 cos 2 π 4 = I 0 2

The intensity at the maximum in a Young's double slit experiment is I 0 . Distance between two slits is d =…

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