Question

Consider the arrangement shown in Fig $29.15$(a). The distance $D$ is large, compared to $d$. Find minimum value of $d$ so that there is a dark fringe at $O$. For, the same value of $d$ find $x$ at which next bright fringes is formed.

Answer 1

Path difference$=AB+BO-2D$
$2\sqrt{(D^2+d^2)}-2D=\frac{\lambda }{2}$

or $2\sqrt{(D^2+d^2)}=\frac{\lambda }{2}+2D$
or $4(D^2+d^2)=\frac{{\lambda }^2}{4}+4D^2+2\lambda D$

Eliminate $\frac{{\lambda }^2}{4}$ as $\lambda <<D.$ or $d=\sqrt{\frac{\lambda D}{2}}$