JEE Main 2025 Jan 24 Shift 1, Physics Q9: Young’s Double Slit Experiment

Bright fringes coincide when the fringe positions $\frac{n\lambda D}{d}$ match for the two wavelengths:

$n_1{\lambda }_1=n_2{\lambda }_2$, i.e. $n_1(480)=n_2(600)$

Simplify: $\frac{n_1}{n_2}=\frac{600}{480}=\frac{5}{4}$. The smallest positive integer solution is $n_1=5$, $n_2=4$.

So the first coincidence is at the 5${}^{th}$ bright fringe of the 480 nm light (which overlaps the 4${}^{th}$ bright fringe of the 600 nm light).

$\therefore$ Option (4).