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In Young's double-slit experiment, both the slits produce equal intensities on a screen. A 100% transparent thin film of refractive index μ=1.5 is kept in front of one of the slits, due to which the intensity at the point O on the screen becomes 75% of its initial value. If the wavelength of monochromatic light is 720 nm, then what is the minimum thickness (in nm) of the film?

PhysicsWave OpticsJEE Main
Solution:
2251 Upvotes Verified Answer
The correct answer is: 240

The path difference at the point O is

Δx=μ-1t

I=4l0cos2ϕ2

0.754I0=4I0cos2ϕ2

cosϕ2=±32

ϕmin=60°

ϕmin=2πλΔx=2πλμ-1t=π3

t=λ6μ-1=λ061.5-1=240 nm

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