Two identical coherent sources produce a zero order bright fringe on a screen. If $\beta$ is the fringe width, the minimum distance between two points where the intensity is half that of maximum intensity is

$\begin{array}{l}\mathrm{I}={\mathrm{I}}_{max}{\cos }^2\mathrm{\varphi }/2\\ \Rightarrow \frac{{\mathrm{I}}_{max}}{2}={\mathrm{I}}_{max}{\cos }^2\mathrm{\varphi }/2\\ \Rightarrow {45}^{\circ }=\mathrm{\varphi }/2\\ \Rightarrow \mathrm{\varphi }={90}^{\circ }\end{array}$
$\begin{array}{l}△\mathrm{\varphi }=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}△x\\ \Rightarrow \frac{\pi }{2}=\frac{2\pi }{\mathrm{\lambda }}△x\\ \Rightarrow \frac{\mathrm{\lambda }}{4}=△\mathrm{x}\end{array}$

$\Rightarrow \frac{\mathrm{\lambda }}{4}=\mathrm{dtan\theta }$

$\Rightarrow \frac{\mathrm{\lambda }}{4}=d\frac{y}{D}$

$\Rightarrow \mathrm{y}=\frac{\mathrm{\lambda D}}{4\mathrm{d}}$

Position above central bright, $y_1=\frac{\lambda D}{4d}=\frac{\beta }{4}$

Position below central bright, $y_2=\frac{\lambda D}{4d}=\frac{\beta }{4}$

Distance between two points, $y_1+y_2=\frac{\beta }{2}$