KVPY SX SB Exam 2018 Question Paper

Question : 26 of 120

Marks: +1, -0

In a Young’s double slit experiment the amplitudes of the two waves incident on the two slits are A and 2A. If

I_0

is the maximum intensity, then the intensity at a spot on the screen where the phase difference between the two interfering waves is ϕ.

Solution:

I_1 = kA^2 = I ; I_2 = 4kA^2 = 4I

I_0 = (√I_1+√I_2)^2;I_0=9I

I = I + 4I + 2 √{I +4I }+ \cos θ

= 5I + 4I \cos θ =I_0/9 (5 + 4 \cos θ)

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