Wave Optics: Real-World Scenarios (4)

easy • 3 min read
Tags Wave Optics

Question

In a Young’s double-slit experiment used in a physics demo, the slit separation d=0.5d = 0.5 mm and the screen is D=1.0D = 1.0 m away. Light of wavelength λ=600\lambda = 600 nm is used. Find (a) the fringe width, (b) the position of the third bright fringe from the centre, and (c) the new fringe width if the apparatus is immersed in water (refractive index n=1.33n = 1.33).

Solution — Step by Step

β=λDd\beta = \frac{\lambda D}{d}

Plug in: β=600×10−9×1.00.5×10−3=1.2×10−3\beta = \frac{600 \times 10^{-9} \times 1.0}{0.5 \times 10^{-3}} = 1.2 \times 10^{-3} m =1.2= 1.2 mm.

The nn-th bright fringe is at yn=nβy_n = n \beta. For n=3n = 3:

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

λwater=λ/n=600/1.33≈451\lambda_{\text{water}} = \lambda / n = 600 / 1.33 \approx 451 nm. Slit separation and screen distance stay the same.

 β′=λwaterDd=βn=1.21.33≈0.902 mm\beta' = \frac{\lambda_{\text{water}} D}{d} = \frac{\beta}{n} = \frac{1.2}{1.33} \approx 0.902 \text{ mm}

Final answers: β=1.2\beta = 1.2 mm, y3=3.6y_3 = 3.6 mm, β′≈0.9\beta' \approx 0.9 mm.

Why This Works

Fringe width is directly proportional to wavelength. When light enters a denser medium, frequency stays the same but speed drops, so wavelength shrinks by a factor of nn. Fringes get closer together by the same factor.

Slit-to-screen geometry doesn’t change (it is just lengths of glass and water in between), so DD and dd stay constant. Only λ\lambda changes — that’s the only thing you need to track.

Alternative Method

Use the path difference condition. For the nn-th bright fringe, dsin⁡θ=nλd \sin\theta = n\lambda. With small angles, sin⁡θ≈y/D\sin\theta \approx y/D, giving yn=nλD/dy_n = n \lambda D / d. Same answer, different starting point.

For “fringes in a medium” questions, the shortcut is βmedium=βair/n\beta_{\text{medium}} = \beta_{\text{air}} / n. Memorise it — saves the rederivation.

Common Mistake

Multiplying instead of dividing by nn. Students sometimes write β′=nβ\beta' = n \beta, reasoning that the medium “stretches” something. Actually wavelength shrinks, so β\beta shrinks too.

Mixing up which nn in yn=nβy_n = n\beta. For bright fringes, n=0,1,2,...n = 0, 1, 2, ... counting from the central maximum. For dark fringes, the position is yn=(n+12)βy_n = (n + \frac{1}{2})\beta. Students sometimes use the dark-fringe formula by mistake.

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