Interference and Diffraction: Exam-Pattern Drill (2)

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Tags Interference And Diffraction

Question

In a Young’s double-slit experiment, the slit separation is 0.50.5 mm and the screen is 1.51.5 m from the slits. Light of wavelength 600600 nm is used. Find (a) the fringe width, and (b) the distance of the third bright fringe from the central maximum. (c) If the same setup is immersed in water (n=4/3)(n = 4/3), what is the new fringe width? JEE Main 2024 pattern.

Solution — Step by Step

β=λDd\beta = \frac{\lambda D}{d}

With λ=600×10−9\lambda = 600 \times 10^{-9} m, D=1.5D = 1.5 m, d=0.5×10−3d = 0.5 \times 10^{-3} m:

β=600×10−9×1.50.5×10−3=9×10−75×10−4=1.8×10−3 m=1.8 mm\beta = \frac{600 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}} = \frac{9 \times 10^{-7}}{5 \times 10^{-4}} = 1.8 \times 10^{-3} \text{ m} = 1.8 \text{ mm}

The nthn^{\text{th}} bright fringe is at yn=nβy_n = n\beta from the central maximum.

y3=3×1.8=5.4 mmy_3 = 3 \times 1.8 = 5.4 \text{ mm}

When the apparatus is immersed in water, the wavelength changes: λ′=λ/n=600/(4/3)=450\lambda' = \lambda/n = 600/(4/3) = 450 nm. The slit separation and screen distance are unchanged.

β′=λ′Dd=βn=1.84/3=1.35 mm\beta' = \frac{\lambda' D}{d} = \frac{\beta}{n} = \frac{1.8}{4/3} = 1.35 \text{ mm}

Final answers: β=1.8\beta = 1.8 mm, y3=5.4y_3 = 5.4 mm, βwater′=1.35\beta'_{\text{water}} = 1.35 mm.

Why This Works

Fringe width depends linearly on wavelength: shorter wavelength produces tighter fringes. When light enters a medium of refractive index nn, its wavelength shrinks by a factor of nn (frequency is unchanged), so the fringe pattern compresses by the same factor.

This is also why immersion microscopy and oil-objective lenses work: a higher-index medium effectively shortens the wavelength inside, allowing finer details to be resolved.

Alternative Method

Memorise the shortcut: βmedium=βvacuum/n\beta_{\text{medium}} = \beta_{\text{vacuum}}/n. Saves time in MCQs that test the immersion variant.

Students sometimes change DD or dd when the apparatus is immersed in water. Don’t — slit separation and screen distance are physical lengths, unaffected by the medium. Only λ\lambda changes inside the medium.

For diffraction (single slit), the analogous formula is angular width θ=λ/a\theta = \lambda/a for the central maximum. Both interference and diffraction widths shrink by 1/n1/n in a medium.

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