In a Young's double slit experiment with light of wavelength $\lambda$ the separation of slits is d and distance of screen is D. Such that $D >> d >> \lambda$ . If the Fringe width is $\Beta$ the distance from point where intensity of the point where intensity falls to half of maximum intensity on either side is :

Question

$\begin{array}{1 1} \frac{\beta}{2} \\ \frac{\beta}{4} \\ \frac{\beta}{3} \\ \frac{\beta}{6} \end{array} $