Questions
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 made of material of refractive index 1.2. Then intensity of light at the mid point of the source is
detailed solution
Correct option is D
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=Im2Talk to our academic expert!
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