Askiitians Expert Soumyajit IIT-Kharagpur
Last Activity: 15 Years ago
Dear Arun Dash,
Ans: B=λD/d................(1)
Now at a distance B/4 from the central maxima the angle made by the light ray is θ, then
tanθ=B/4D≈θ≈sinθ (as θ is very small).......................(2)
Now path difference is d sinθ=d B/4D and hence phase difference is (2∏/λ)d (λD/d)/4D=∏/2
Now intensity at this point is=Icos²(∏/2)=I/2
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Soumyajit Das IIT Kharagpur