Visible light of wavelength 6000×10−8cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 600 from the central maximum. If the first minimum is produced at θ1, then θ1, is close to,
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a
250
b
450
c
200
d
300
answer is A.
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Detailed Solution
For 2nd minima dsinθ=2λ Since ,θ=600 , sinθ=32given ⇒λd=34______i So for 1st minima is dsinθ1=λ ⇒sinθ1=λd=34 from equation (1) ⇒θ=25.65º⇒θ≈25º