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Q.

In Young's double slit experiment, the slits are of equal width and maximum intensity of light on the screen is found to be Im. Now one of the slits is covered with a thin sheet of thickness 15λ4, made of material of refractive index 1.2. Then intensity of light at the mid point of the source is

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a

zero

b

Im/4

c

Im

d

Im/2

answer is D.

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Detailed Solution

If I0 be the intensity of light energy from each slit, then Im=4I0.When one of the slits is covered, path difference between rays at the mid point of the screen =μ−1t=1.2−1×15λ4=3λ4∴ ϕ=phase difference =2πλ×3λ4=3π2∴ At the mid point, I=Im.cos2ϕ2=Imcos23π4=ImIV22=Im2
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In Young's double slit experiment, the slits are of equal width and maximum intensity of light on the screen is found to be Im. Now one of the slits is covered with a thin sheet of thickness 15λ4, made of material of refractive index 1.2. Then intensity of light at the mid point of the source is