Wave Optics: Edge Cases and Subtle Traps (1)

easy • 2 min read
Tags Wave Optics

Question

In a Young’s double-slit experiment, the slit separation is d=0.5 mmd = 0.5\text{ mm} and the screen is at D=1 mD = 1\text{ m} from the slits. Light of wavelength λ=600 nm\lambda = 600\text{ nm} is used. (a) Find the fringe width. (b) Find the position of the third bright fringe. (c) If the entire setup is dipped in water (refractive index n=4/3n = 4/3), find the new fringe width.

Solution — Step by Step

For double-slit interference, fringe width:

β=λDd=600×10−9×10.5×10−3=1.2×10−3 m=1.2 mm\beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 1.2 \times 10^{-3}\text{ m} = 1.2\text{ mm}

Bright fringes are at yn=nβy_n = n\beta from the central maximum. For n=3n = 3:

y3=3×1.2=3.6 mmy_3 = 3 \times 1.2 = 3.6\text{ mm}

Inside water, wavelength shrinks: λwater=λ/n=600/(4/3)=450 nm\lambda_{\text{water}} = \lambda/n = 600/(4/3) = 450\text{ nm}. Frequency does not change. New fringe width:

βwater=λwaterDd=450×10−9×10.5×10−3=0.9 mm\beta_{\text{water}} = \frac{\lambda_{\text{water}} D}{d} = \frac{450 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 0.9\text{ mm}

β=1.2 mm\beta = 1.2\text{ mm}, y3=3.6 mmy_3 = 3.6\text{ mm}, βwater=0.9 mm\beta_{\text{water}} = 0.9\text{ mm}.

Why This Works

The fringe pattern arises from path difference Δ=dsin⁡θ\Delta = d\sin\theta. Constructive interference requires Δ=nλ\Delta = n\lambda. For small angles, sin⁡θ≈y/D\sin\theta \approx y/D, giving yn=nλD/dy_n = n\lambda D / d. Adjacent fringes are separated by β=λD/d\beta = \lambda D / d.

Immersing in water reduces the wavelength (because speed reduces, frequency stays put), so all fringe spacings shrink by the same factor 1/n1/n.

Alternative Method

Path-difference reasoning directly: in water, the optical path remains the same, but geometric path is less. Equivalently, the wavelength in the medium is λ/n\lambda/n. Substitute into β=λD/d\beta = \lambda D / d and you get the same βwater=β/n\beta_{\text{water}} = \beta/n relation.

Common Mistake

Confusing wavelength and frequency behaviour in a medium. Wavelength changes with nn; frequency does not. Students sometimes use f′=f/nf' = f/n — wrong. The source sets frequency, the medium sets speed, and wavelength adjusts to match: λ=v/f\lambda = v/f.

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