In young double slit experiment $\lambda =500\mathrm{nm},d=1\mathrm{mm}\text{ and }D=1\mathrm{m}$ .The minimum distance from the central maximum where intensity is half of the maximum intensity is (assume intensity from each slit is same) 

Given Wavelength  $\lambda =500\text{\hspace{0.17em}nm}=5\times {10}^{-7}\text{\hspace{0.17em}m}$
Distance between two slits  $d=1{\text{\hspace{0.17em}mm=10}}^{-3}\text{\hspace{0.17em}m}$
Distance between slits and screen  $D=1\text{\hspace{0.17em}m}$
Maximum intensity  $=4I_0$
$I=4I_{\text{o}}{\cos }^2\frac{\varphi }{2}$

$\Rightarrow 2I_{\text{o}}=4I_{\text{o}}{\cos }^2\frac{\varphi }{2}$

$\Rightarrow \cos \frac{\varphi }{2}=\frac{1}{\sqrt{2}}$

$\therefore \varphi =\frac{\text{π}}{2}$

Path difference $\Delta \varphi =\frac{2\text{π}}{\lambda }\Delta x$

$\Rightarrow \Delta x=\frac{\lambda }{4}$

$\therefore y=\frac{\Delta x.D}{d}=\frac{\lambda D}{4d}=\frac{5\times {10}^{-7}\times 1}{4\times {10}^{-3}}=1.25\times {10}^{-4}\text{\hspace{0.17em}m}$