JEE Main 2025 Apr 3 Shift 2, Physics Q4: Young’s Double Slit Experiment

Intensity at a slit is proportional to its width, so if one slit has half the width of the other, $I_1:I_2=1:2$.

Maximum and minimum intensities in two-source interference are $(\sqrt{I_1}+\sqrt{I_2}{)}^2$ and $(\sqrt{I_1}-\sqrt{I_2}{)}^2$.

$\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{I_1}+\sqrt{I_2}{)}^2}{(\sqrt{I_1}-\sqrt{I_2}{)}^2}=\frac{(1+\sqrt{2}{)}^2}{(\sqrt{2}-1{)}^2}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$.

$\therefore$ The ratio is $(3+2\sqrt{2}):(3-2\sqrt{2})$.