Monochromatic light of wavelength $\lambda$ emerging from slit S illuminates slits $S_1$ and $S_2$ which are placed with respect to S as shown in Fig.  The distances x and D are large compared to the separation d between the slits. If x = D/2, the minimum value of d so that there is a dark fringe at the center P of the screen is

Refer to Fig. To reach point P, wave 1 has to travel a path $\left(S{S}_2+S_{2}P\right)$ while wave 2 has to travel a path $\left(S{S}_1+S_{1}P\right)$. Therefore, when the waves arrive at P, the path difference is

$\Delta =\left(S{S}_2+S_{2}P\right)-\left(S{S}_1+S_{1}P\right)$                  (1)

Now, in triangle $S{S}_2{S}_1$, we have
$$\begin{gathered} S{S}_2={\left(x^2+d^2\right)}^{1/2}=x{\left(1+\frac{d^2}{x^2}\right)}^{1/2} \\ =x\left(1+\frac{d^2}{2x^2}\right) \left[\because d<<x\right] \end{gathered}$$
Similarly, $S_{2}P={\left(D^2+d^2\right)}^{1/2}=D\left(1+\frac{d^2}{2D^2}\right)$                       $\left[\because d<<D\right]$
Also $\left(S{S}_1+S_{1}P\right)-x=D.$

Using these in Eq. (1), we have:
$$\begin{gathered} \Delta =x\left(1+\frac{d^2}{2x^2}\right)+D\left(1+\frac{d^2}{2D^2}\right)-(x+D) \\ =x + \frac{d^2}{2x}+ D + \frac{d^2}{2D}-x-D \\ \text{ or } \Delta =\frac{d^2}{2}\left(\frac{1}{x}+\frac{1}{D}\right) \end{gathered}$$
In order to have a dark fringe at $P, \Delta =\frac{\lambda }{2}$. Hence,
$$\begin{gathered} \frac{\lambda }{2}=\frac{d^2}{2}\left(\frac{1}{x}+\frac{1}{D}\right) \\ \text{ or } \dot{d}={\left[\frac{\lambda xD}{(x+D)}\right]}^{1/2} \left(2\right) \end{gathered}$$
Putting $x=\frac{D}{2}$ in Eq. (2), we find that:

$\mathrm{d} = \sqrt{\frac{\mathrm{\lambda D}}{3}}$