In a young's double slit experiment, distance of the screen from the slits is D and separation between the slits is d. Then the minimum distance of the point from the central maximum where the intensity is same as that due to a single slit is equal to (wave length =λ).
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a
Dλd
b
Dλ2d
c
Dλ3d
d
2Dλd
answer is C.
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Detailed Solution
I=Im cos2 ϕ/2 Im=4Io ∴Io=4Iocos2ϕ2⇒cosϕ2=12⇒ϕ2=π3 Now ϕ2=πλ(d sin θ)≃πλ(d tan θ)=πλ(d.yD) ∴πλ×d.yD=π3 ⇒y=λD3d.