In a Young's double-slit experiment, the screen distance is 1 m and the slit separation is 1 mm. A transparent slab of thickness 10 μm and refractive index 1.5 is introduced in front of one slit. If the central maximum shifts by N mm, find N.
Physics · Wave Optics · young's dubble slit experiments
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Question

In a Young's double-slit experiment, the screen distance is 1 m and the slit separation is 1 mm. A transparent slab of thickness 10 μm and refractive index 1.5 is introduced in front of one slit. If the central maximum shifts by N mm, find N.

Physics · Wave Optics · young's dubble slit experiments

G1 Physics · Accepted Answer

The slab introduces extra optical path (μ−1)t. The central-fringe shift is y₀ = D(μ−1)t ÷ d. Thus, y₀ = 1×0.5×10×10⁻⁶ ÷ 10⁻³ = 5×10⁻³ m = 5 mm. Hence N = 5.