Q 12-10-081JEE MainJEE Main 2023 (31 Jan, Shift 2)Medium
Two light waves of wavelengths $800$ and $600\ \text{nm}$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7\ \text{m}$ away from plane of slits. If the two slits are separated by $0.35\ \text{mm}$, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be ______ mm.
Numerical value type. Enter your answer.
Answer: 48
Bright fringes coincide when $n_1\lambda_1=n_2\lambda_2$: $800n_1=600n_2$, first at $n_1=3$, $n_2=4$ (i.e. $3\times800=2400\ \text{nm}$).
$$y=\frac{n_1\lambda_1D}{d}=\frac{2400\times10^{-9}\times7}{0.35\times10^{-3}}=0.048\ \text{m}=48\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics