An optical arrangement is depicted in the figure. The distance D is...

Let the source slit be $S$, the slits in the second plane be $S_1$ and $S_2$, and the point on the screen be $O$.

From the given figure, the optical axis passes through the source slit $S$, the lower slit $S_1$, and the point $O$ on the screen. The upper slit $S_2$ is at a distance $d$ from the optical axis.

Light reaches point $O$ via two paths:
Path 1: $S \rightarrow S_1 \rightarrow O$
Path 2: $S \rightarrow S_2 \rightarrow O$

The length of Path 1 is:
$L_1 = SS_1 + S_1O = D + D = 2D$

The length of Path 2 is:
$L_2 = SS_2 + S_2O = \sqrt{D^2 + d^2} + \sqrt{D^2 + d^2} = 2\sqrt{D^2 + d^2}$

The path difference $\Delta x$ at point $O$ is:
$\Delta x = L_2 - L_1 = 2\sqrt{D^2 + d^2} - 2D$

Since $D \gg d$, we can use the binomial expansion:
$\Delta x = 2D\left(1 + \dfrac{d^2}{D^2}\right)^{1/2} - 2D$
$\Delta x \approx 2D\left(1 + \dfrac{d^2}{2D^2}\right) - 2D = \dfrac{d^2}{D}$

For a dark fringe to appear at point $O$, the path difference must be an odd multiple of $\dfrac{\lambda}{2}$:
$\Delta x = (2n - 1)\dfrac{\lambda}{2}$ where $n = 1, 2, 3, \dots$

For the minimum value of $d$, we take $n = 1$:
$\dfrac{d^2}{D} = \dfrac{\lambda}{2}$

$d^2 = \dfrac{\lambda D}{2}$

$d = \sqrt{\dfrac{\lambda D}{2}}$