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Question

In a YDSE apparatus, separation between the slits d = 1mm,λ=600 nm and D = 1m. Assume that each slit produce same intensity on the screen. The minimum distance between two points on the screen having 75% intensity of the maximum intensity is n×104 m. Find n___

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Solution

75% of Imax=75100×4I0
=3I0
3I0=4I0cos2Δϕ2
Δϕ=(π3)
Path difference Δx=Δϕ2πλ=λ6
Δxd=yD
Δx=ydD
λ6=ydD
y=λD6d
Minimum separation = 2y
=2λD6d=λD3d
=(600×1091)3×103
=200×106
=2×104 m

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