Wave Optics: Numerical Problems Set (5)

Question

In Young’s double-slit experiment, the slits are $d = 0.5 \text{ mm}$ apart and the screen is $D = 1.5 \text{ m}$ away. Light of wavelength $\lambda = 600 \text{ nm}$ is used. Find (a) the fringe width, (b) the position of the 5th bright fringe from the center, (c) the new fringe width if the entire setup is dipped in water (refractive index $n = 4/3$).

Solution — Step by Step

Final answers: $\beta = \mathbf{1.8 \text{ mm}}$, $y_5 = \mathbf{9 \text{ mm}}$, $\beta' = \mathbf{1.35 \text{ mm}}$.

Why This Works

Fringe width depends on the wavelength inside the medium where interference happens. Light slows down in water (frequency unchanged, $v$ drops, so $\lambda$ drops by factor $n$). Closer-spaced wavelengths mean closer-spaced fringes.

Path difference at the screen depends on $d \sin\theta \approx dy/D$ — pure geometry, so $d$ and $D$ don’t change with the medium. Only $\lambda$ changes.

Alternative Method

For part (c), since $\beta' = \beta/n$, we can write the answer immediately as $1.8/(4/3) = 1.35 \text{ mm}$. No need to recompute $\lambda'$ separately.

Common Mistake