Fringe shift due to slab in Young's double slit experiment is given by Δx=dD(μ−1)t
Since, fringe width β0=dDλ
Then, Δx=λβ0(μ−1)t
Putting the values, we have
=5×10−710×10−6(1.2−1)β0
=5×10−710−5×0.2β0=4β0
Hence, the value of x=4.
JEE Main 2023 — Physics Optics
2023integerhard
As shown in the figure, in Young's double slit experiment, a thin plate of thickness t=10μm and refractive index μ=1.2 is inserted infront of slit S1. The experiment is conducted in air (μ=1) and uses a monochromatic light of wavelength λ=500nm. Due to the insertion of the plate, central maxima is shifted by a distance of xβ0. β0 is the fringe-width before the insertion of the plate. The value of the x is ______.
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Held on 1 Feb 2023 · Verified 6 Jul 2026.
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