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Question

In a Young's double slit experiment
λ=500 nm, d=1 mm and D=1 m.
The minimum distance from the central maximum for which the intensity is half of the maximum intensity is

A
2×104 m
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B
1.25×104 m
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C
4×104 m
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D
2.5×104 m
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Solution

The correct option is B 1.25×104 m
Intensity of central maxima I=(2A0)2=4A02=4I0
Intensity at distance y from the central maxima is half of the maximum intensity if I0=4I0cos2ϕ2=4I02

If
cos2ϕ2=12ϕ2=π4ϕ=π2
Now we know
ϕ=2πλΔx
where Δx is the path difference between the two interfering waves
also Δx=ydD
ϕ=2πλydD=π2
y=λD4d500×109×14×103=1.25×104 m

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