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- $$ \begin{aligned} & \mathrm{I}=\mathrm{I}_0+\mathrm{I}_0+2 \sqrt{\mathrm{I}_0 \mathrm{I}_0} \cos \phi \\ & \mathrm{I}_{\max }=4 \mathrm{I}_0 \\ & \frac{\mathrm{I}_{\max }}{2}=2 \mathrm{I}_0=2 \mathrm{I}_0+2 \mathrm{I}_0 \cos \phi \\ & \cos \phi=0 \\ & \phi=\frac{\pi}{2} \\ & \mathrm{~K} \Delta \mathrm{x}=\frac{\pi}{2} \\ & \frac{2 \pi}{\lambda} \mathrm{~d} \sin \theta=\frac{\pi}{2} \\ & \frac{2}{\lambda} \mathrm{~d} \times \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2} \\ & \mathrm{y}=\frac{\lambda \mathrm{D}}{4 \mathrm{~d}}=\frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}} \\ & =125 \times 10^{-6} \\ & =125 \end{aligned} $$