JEE Main 2024 — 4 April, Shift 2 — Question 36

Since, Intensity $\propto$ width of slit ( $\omega$ )

so, $\mathrm{I}_{1}=\mathrm{I}, \mathrm{I}_{2}=4 \mathrm{I}$ $\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}}\right)^{2}=\mathrm{I}$

$\mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}\right)^{2}=9 \mathrm{I}$ $\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=\frac{9 \mathrm{I}}{\mathrm{I}}=\frac{9}{1}$