QJEEM25Phccf9: JEE Main 2025 Physics - Class 12 JEEM25Phccf9 Solution: The Young's double slit interference experiment is performed…

$d \sin \theta = m \lambda$

For two wavelengths, $\lambda_1 = 600\,\text{nm}$ and $\lambda_2 = 480\,\text{nm}$ , the bright fringes coincide when the conditions

$m \lambda_2 = n \lambda_1$

are simultaneously satisfied for integers $m$ and $n$ , representing the bright fringe order for each wavelength.

Rearranging the equation gives:

$\frac{m}{n} = \frac{\lambda_1}{\lambda_2} = \frac{600}{480} = \frac{5}{4}$

Thus, the smallest positive integers that satisfy this ratio are:

$m = 5 \quad \text{and} \quad n = 4.$

This means the first coincidence of the bright fringes occurs at the 5th bright fringe for the 480 nm light.