JEE Main 2024 — 1 February, Shift 2 — Question 58

Let intensity of light on screen due to each slit is $\mathrm{I}_{0}$ So internity at centre of screen is $4 \mathrm{I}_{0}$ Intensity at distance y from centre- $\mathrm{I}=\mathrm{I}_{0}+\mathrm{I}_{0}+2 \sqrt{\mathrm{I}_{0} \mathrm{I}_{0}} \cos \phi$ $\mathrm{I}_{\text {max }}=4 \mathrm{I}_{0}$ $\frac{\mathrm{I}_{\max }}{2}=2 \mathrm{I}_{0}=2 \mathrm{I}_{0}+2 \mathrm{I}_{0} \cos \phi$ $\cos \phi=0$ $\phi=\frac{\pi}{2}$ $\mathrm{K} \Delta \mathrm{x}=\frac{\pi}{2}$ $\frac{2 \pi}{\lambda} \mathrm{~d} \sin \theta=\frac{\pi}{2}$ $\frac{2}{\lambda} \mathrm{~d} \times \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2}$ $\mathrm{y}=\frac{\lambda \mathrm{D}}{4 \mathrm{~d}}=\frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}}$ $=125 \times 10^{-6}$ $=125$