JEE Main 2025 — Wave Optics Question with Solution

$$d\sin \theta =m\lambda$$

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:

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

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