Wave Optics: Speed-Solving Techniques (2)

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Tags Wave Optics

Question

In a Young’s double-slit experiment, the slits are 0.5 mm apart, the screen is 1 m away, and light of wavelength 500 nm is used. Find: (a) fringe width, (b) distance of the fourth bright fringe from the central maximum, (c) shift in fringe pattern if a glass slab of thickness 10 µm and refractive index 1.5 is placed in front of one slit. Solve in under 90 seconds using direct formulas.

Solution — Step by Step

Fringe width formula:

β=λDd\beta = \frac{\lambda D}{d}

Plug in: λ=500×10−9\lambda = 500 \times 10^{-9} m, D=1D = 1 m, d=0.5×10−3d = 0.5 \times 10^{-3} m.

β=500×10−9⋅10.5×10−3=10−3 m=1 mm\beta = \frac{500 \times 10^{-9} \cdot 1}{0.5 \times 10^{-3}} = 10^{-3} \text{ m} = 1 \text{ mm}

The nn-th bright fringe is at distance nβn\beta from the central maximum. So:

y4=4⋅1=4 mmy_4 = 4 \cdot 1 = 4 \text{ mm}

When a slab of thickness tt and refractive index μ\mu is placed in front of one slit, the fringe pattern shifts toward that slit by:

Δy=(μ−1)t⋅Dd\Delta y = \frac{(\mu - 1) t \cdot D}{d}

Plug in: μ−1=0.5\mu - 1 = 0.5, t=10×10−6t = 10 \times 10^{-6} m, D=1D = 1 m, d=0.5×10−3d = 0.5 \times 10^{-3} m.

Δy=0.5⋅10×10−6⋅10.5×10−3=10−2 m=10 mm\Delta y = \frac{0.5 \cdot 10 \times 10^{-6} \cdot 1}{0.5 \times 10^{-3}} = 10^{-2} \text{ m} = 10 \text{ mm}

(a) β=1\beta = 1 mm, (b) y4=4y_4 = 4 mm, (c) shift = 10 mm.

Why This Works

Fringe width comes from the small-angle path difference dsin⁡θ≈d⋅y/D=nλd\sin\theta \approx d \cdot y/D = n\lambda. The slab introduces an extra optical path length of (μ−1)t(\mu - 1)t, which shifts the central maximum by Δy=(μ−1)t⋅D/d\Delta y = (\mu - 1)t \cdot D/d. The pattern moves bodily toward the slit covered by the slab.

These three formulas — β\beta, yny_n, and Δy\Delta y — handle most YDSE questions in JEE Main directly.

If the question asks “how many fringes shift?”, divide Δy\Delta y by β\beta. Here: 10/1=1010/1 = 10 fringes. JEE loves this style.

Alternative Method

Optical path difference approach: the slab adds (μ−1)t(\mu - 1)t of path in the slit it covers. To restore the central maximum (zero path difference), the screen position must shift by an amount that introduces −(μ−1)t-(\mu - 1)t from the geometry — which works out to Δy=(μ−1)t⋅D/d\Delta y = (\mu - 1)t \cdot D/d. Same answer.

Common Mistake

Students use μ⋅t\mu \cdot t instead of (μ−1)t(\mu - 1)t for the extra path. Remember: the slab replaces the air path, so the additional path is (μ−1)t(\mu - 1)t, not μt\mu t. This single sign-shift mistake costs the entire question.

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