Wave Optics: Conceptual Doubts Cleared (6)

hard • 2 min read
Tags Wave Optics

Question

In Young’s double-slit experiment, the slit separation is d=0.5d = 0.5 mm and the screen is D=1D = 1 m away. Light of wavelength λ=600\lambda = 600 nm is used. (a) Find the fringe width. (b) If the entire setup is immersed in water (μ=1.33\mu = 1.33), what is the new fringe width? (c) What happens to the fringe pattern if we cover one slit?

Solution — Step by Step

The fringe width formula is

β=λDd\beta = \frac{\lambda D}{d}

β=600×10−9×10.5×10−3=1.2×10−3\beta = \frac{600 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 1.2 \times 10^{-3} m =1.2= 1.2 mm.

Inside water, the wavelength shortens to λ′=λ/μ\lambda' = \lambda/\mu. So

β′=βμ=1.21.33≈0.9 mm\beta' = \frac{\beta}{\mu} = \frac{1.2}{1.33} \approx 0.9\,\text{mm}

With one slit blocked, there is no two-source interference. We get single-slit diffraction instead — a broad central bright band with much weaker side fringes. The crisp equally-spaced YDSE pattern disappears.

Final answers: (a) β=1.2\beta = 1.2 mm, (b) β′≈0.9\beta' \approx 0.9 mm, (c) pattern becomes single-slit diffraction.

Why This Works

Fringe width depends on wavelength because the path-difference condition for the next maximum requires an extra λ\lambda of path. A larger λ\lambda pushes that condition to a larger angle from the centre, hence a wider fringe.

When we put the apparatus in water, the frequency of light is unchanged (set by the source) but the speed drops by a factor μ\mu. Wavelength being speed/frequency, λ′=λ/μ\lambda' = \lambda/\mu — so fringes shrink.

Alternative Method

Use the path-difference geometry. For the nn-th bright fringe, dsin⁡θn=nλd\sin\theta_n = n\lambda. Small-angle gives θn≈nλ/d\theta_n \approx n\lambda/d, so position yn=Dθn=nλD/dy_n = D\theta_n = n\lambda D/d. Fringe width is yn+1−yn=λD/dy_{n+1} - y_n = \lambda D/d. Same formula, derived from first principles.

A frequent NEET trap: students write β=μλD/d\beta = \mu \lambda D/d in water, multiplying instead of dividing. Always remember — wavelength shrinks in a denser medium, so fringes shrink too.

Common Mistake

Confusing single-slit diffraction with double-slit interference. Diffraction gives unequal fringe widths and a much wider central maximum. Interference (with two slits) gives equally spaced bright/dark bands. Examiners deliberately mix these up.

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