In Young's double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are $1.0 \mathrm{~mm}$ apart and screen is placed at $1.0 \mathrm{~m}$ away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is _________ $\times 10^{-6}$ $\mathrm{m}$.

Let intensity of light on screen due to each slit is $\mathrm{I}_0$ <br/><br/>So internity at centre of screen is $4 \mathrm{I}_0$ <br/><br/>Intensity at distance y from centre- <br/><br/>$$ \begin{aligned} & I=I_0+I_0+2 \sqrt{I_0 I_0} \cos \phi \\\\ & I_{\max }=4 I_0 \\\\ & \frac{I_{\max }}{2}=2 I_0=2 I_0+2 I_0 \cos \phi \end{aligned} $$ <br/><br/>$\begin{aligned} & \cos \phi=0 \\\\ & \phi=\frac{\pi}{2} \\\\ & K \Delta \mathrm{x}=\frac{\pi}{2} \\\\ & \frac{2 \pi}{\lambda} \mathrm{d} \sin \theta=\frac{\pi}{2} \\\\ & \frac{2}{\lambda} \mathrm{d} \times \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2}\end{aligned}$ <br/><br/>$\begin{aligned} & \mathrm{K} \Delta \mathrm{x}=\frac{\pi}{2} \\\\ & \frac{2 \pi}{\lambda} \mathrm{d} \sin \theta=\frac{\pi}{2} \\\\ & \frac{2}{\lambda} \mathrm{d} \times \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2} \\\\ & \mathrm{y}=\frac{\lambda \mathrm{D}}{4 \mathrm{~d}}=\frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}} \\\\ & =125 \times 10^{-6} \\\\ & =125\end{aligned}$