The minimum value of d so that there is a dark fringe at O is ${\mathrm{d}}_{\min }$. For the value of ${\mathrm{d}}_{\min }$, the distance at which the next bright fringe is formed is x. Then 

There is a dark fringe at O if the path difference 

           $\mathrm{\delta }=\mathrm{ABO}-{\mathrm{AO}}^{\mathrm{\prime }}\mathrm{O}=\frac{\mathrm{\lambda }}{2}$

$\begin{array}{l}\Rightarrow 2\sqrt{{\mathrm{D}}^2-{\mathrm{d}}^2}-2\mathrm{D}=\frac{2{\mathrm{d}}^2}{2\mathrm{D}}=\frac{{\mathrm{d}}^2}{\mathrm{D}}=\frac{\mathrm{\lambda }}{2}\\ {\mathrm{d}}_{min}=\sqrt{\frac{\mathrm{\lambda D}}{2}}\end{array}$

The bright fringe is formed at P if the path difference 

          $\begin{array}{l}\mathrm{\delta }\text{'}=\mathrm{AO}\text{'}\mathrm{P}-\mathrm{ABP}=\mathrm{\lambda }\\ =\mathrm{D}+\sqrt{{\mathrm{D}}^2+{\mathrm{x}}^2}-\sqrt{{\mathrm{D}}^2+{\mathrm{d}}^2}-\sqrt{{\mathrm{D}}^2+(\mathrm{x}-\mathrm{d}{)}^2}=\mathrm{\lambda }\\ =\frac{{\mathrm{x}}^2}{2\mathrm{D}}-\frac{{\mathrm{d}}^2}{2\mathrm{D}}-\frac{\left({\mathrm{s}}^2+{\mathrm{d}}^2-2\mathrm{xd}\right)}{2\mathrm{D}}=\mathrm{\lambda }\end{array}$

Given $\mathrm{d}={\mathrm{d}}_{min}$

On solving, $\mathrm{x}={\mathrm{d}}_{min}=\sqrt{\frac{\mathrm{\lambda D}}{2}}$