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. The intensity of light in front of one of the slits on a screen at a distance D = 10d is
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a
I02
b
34I0
c
I0
d
I04
answer is A.
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Detailed Solution
Suppose P is a point in front of one slit at which intensity is I0. From figure, it is clear that y=d2. Path difference between the waves reaching at P isΔx=ydD=d2d10d=d20=5λ20=λ4Hence corresponding phase difference is given by ϕ=2πλ×λ4=π2Resultant intensity at P is I=Imaxcos2ϕ2=I0cos2π4=I02
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