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Question: Answered & Verified by Expert
In Young's double-slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ is K units. Then, the intensity of light at a point where path difference is λ3 is
PhysicsWave OpticsNEET
Options:
  • A K2
  • B 2K
  • C 4K
  • D K4
Solution:
2311 Upvotes Verified Answer
The correct answer is: K4
Phase difference corresponding to path difference λ is

Δϕ=2πλ×λ=2π

Intensity at the point having this phase difference 2π is

I=4I0cos2ϕ2

=4I0cos2π22

=4I0-12=4I0

Here, I0 is intensity of individual source.

It is given that

I=4I0=K

  I0=K4 ......(i)

Phase difference corresponding to path difference of λ3 is

Δϕ=2πλ×λ3=2π3

Thus, intensity of light corresponding to this phase difference of 2π3 is

I=4I0×cos2π32

=K×-122 [from Equation. (i)]

=K4

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