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Q.

The maximum intensity in Young’s double-slit experiment is I0 distance between the slits is d=5λ , where λ is the wavelength of monochromatic light used in the experiment . What will be the intensity of light in front of one of the slits on a screen at a distance D=10d?

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a

I02

b

34I0

c

I0

d

I04

answer is A.

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Detailed Solution

Path difference, Δx=ydD Here, y=d2=5λ2And D=10d=50λ (as  d=5λ)So, Δx=(5λ2)(5λ50λ)=λ4Corresponding phase difference will beϕ=(2πλ)(Δx)=(2πλ)(π4)=π2Or ϕ2=π4∴I=I0cos2(ϕ2)=I0cos2(π4) =I02
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