Q 12-10-046JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
The Young's double slit interference experiment is performed using light consisting of $480\ \text{nm}$ and $600\ \text{nm}$ wavelengths to form interference patterns. The least number of the bright fringes of $480\ \text{nm}$ light that are required for the first coincidence with the bright fringes formed by $600\ \text{nm}$ light is
Answer: (A) $5$
Bright fringes coincide where $n_1\lambda_1 = n_2\lambda_2$:
$$480\,n_1 = 600\,n_2 \Rightarrow \frac{n_1}{n_2} = \frac{5}{4}$$
The first coincidence is the $5^{\text{th}}$ bright fringe of $480\ \text{nm}$ with the $4^{\text{th}}$ of $600\ \text{nm}$. So 5 fringes.
Solution by Sreeraj P, M.Sc Physics