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Question: Answered & Verified by Expert
The intensity at the maximum in a Young's double slit experiment is I0. Distance between two slits is d=5λ, 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?
PhysicsWave OpticsNEETNEET 2016 (Phase 1)
Options:
  • A I0
  • B I04
  • C 34I0
  • D I02
Solution:
2535 Upvotes Verified Answer
The correct answer is: I02
In YDSE Imax=I0

Path difference at a point in front of one of shifts is

x=dyD=dd2D=d22D     ( Herey= d 2 )

x=d2210d=d20=5λ20=λ4

Path difference is

ϕ=2πλΔx=2πλλ4

ϕ=π2

So intensity at that point is

I=Imaxcos2θ2

I=I0cos2π4=I02

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