So internity at centre of screen is \(4 \mathrm{I}_0\)

Intensity at distance y from centre-

\(\mathrm{I}=\mathrm{I}_0+\mathrm{I}_0+2 \sqrt{\mathrm{I}_0 \mathrm{I}_0} \cos \phi\)

\(\mathrm{I}_{\max }=4 \mathrm{I}_0\)

\(\frac{\mathrm{I}_{\max }}{2}=2 \mathrm{I}_0=2 \mathrm{I}_0+2 \mathrm{I}_0 \cos \phi\)

\(\cos \phi=0\)

\(\phi=\frac{\pi}{2}\)

\(\mathrm{~K} \Delta \mathrm{x}=\frac{\pi}{2}\)

\(\frac{2 \pi}{\lambda} \mathrm{d} \sin \theta=\frac{\pi}{2}\)

\(\frac{2}{\lambda} \mathrm{d} \times \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2}\)

\(\mathrm{y}=\frac{\lambda \mathrm{D}}{4 \mathrm{~d}}=\frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}}\)

\(=125 \times 10^{-6}\)

\(=125\)