Interference and Diffraction: Edge Cases and Subtle Traps (3)

hard • 3 min read
Tags Interference And Diffraction

Question

In a Young’s double-slit experiment, slits separated by d=1d = 1 mm are illuminated by light of wavelength λ=600\lambda = 600 nm. The screen is D=1.5D = 1.5 m from the slits. A thin glass plate of thickness t=5 μt = 5~\mum and refractive index n=1.5n = 1.5 is placed in front of one slit. Find (a) the shift of the central fringe and (b) the new position of the central maximum on the screen.

Solution — Step by Step

When light passes through a slab of thickness tt and refractive index nn, the optical path length increases by (n−1)t(n-1)t relative to the same physical distance through air.

Δ=(n−1)t=(1.5−1)(5×10−6)=2.5×10−6 m\Delta = (n-1)t = (1.5 - 1)(5 \times 10^{-6}) = 2.5 \times 10^{-6} \text{ m}

The central fringe is where the path difference between the two slits is zero. After the plate is inserted, the slit with the plate has extra optical path Δ\Delta. To compensate, the geometric path from that slit to the central fringe must shorten — i.e., the central fringe shifts toward the slit with the plate.

The geometric path difference equals y⋅d/Dy \cdot d / D where yy is the position on the screen. Setting this equal to Δ\Delta:

y=Δ⋅Dd=2.5×10−6×1.510−3=3.75×10−3 m=3.75 mmy = \frac{\Delta \cdot D}{d} = \frac{2.5 \times 10^{-6} \times 1.5}{10^{-3}} = 3.75 \times 10^{-3} \text{ m} = 3.75 \text{ mm}

The central fringe shifts by 3.753.75 mm toward the slit covered by the glass plate. The fringe spacing β=λD/d=0.9\beta = \lambda D/d = 0.9 mm is unchanged — only the entire pattern slides.

Why This Works

A thin transparent slab doesn’t deflect light noticeably (assuming normal incidence) — but it slows light down inside the slab, increasing optical path. Interference depends on optical path difference, not geometric. So the zero-path-difference point shifts to wherever the other path picks up an equal extra length.

The fringe spacing is independent of any path-length offsets that affect both slits the same way. So β=λD/d\beta = \lambda D / d stays at 0.9 mm — the pattern translates without stretching.

Alternative Method

Equivalent fringe-shift formula: shift =(n−1)t⋅Dd= \frac{(n-1)t \cdot D}{d}. Plug in directly to skip writing Δ\Delta separately. Many JEE Advanced setups use this one-shot version.

Common Mistake

Two traps. First, students use ntnt instead of (n−1)t(n-1)t — the extra optical path is the additional delay due to refraction, which is the difference relative to traveling the same distance through air. Second, students think the shift changes the fringe spacing β\beta. It doesn’t — only a wavelength change or a dd, DD change can affect β\beta.

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