Young's double slit experiment is performed in a medium of refractive index of 1.33. The maximum intensity is I_0 . The intensity at a point on the screen, where path difference between the light coming out from slits

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Correct answer: B. (I_0)/(2)

Given, path difference, Δ x=(λ)/(4) Phase difference, φ=((2 π)/(λ)) × (Path difference) =(2 π)/(λ) × (λ)/(4)=(π)/(2)...(i) It is given that, maximum intensity is I_0 . So, intensity at a point on the screen, where phase difference φ is given by aligned I & =I_0 ^2((φ)/(2)) & =I_0 × ^2((π / 2)/(2)) [ using Eq.(i) ] & =I_0 ×( 1 2 )^2 & =(I_0)/(2) aligned

Young's double slit experiment is performed in a medium of...

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