In a Young's double slit experiment λ=500nm,d=1mmandD=1m.
The minimum distance from the central maximum for which the intensity is half of the maximum intensity is
A
2×10−4m
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B
1.25×10−4m
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C
4×10−4m
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D
2.5×10−4m
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Solution
The correct option is B1.25×10−4m 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=12⇒ϕ2=π4⇒ϕ=π2
Now we know ϕ=2πλΔx
where Δx is the path difference between the two interfering waves
also Δx=ydD ⇒ϕ=2πλydD=π2 ⇒y=λD4d⇒500×10−9×14×10−3=1.25×10−4m