Consider a two-slit interference arrangement (see figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre $O$.

(D) The situation described is depicted in the figure. According to the problem,the distance of the screen from the slits is $D$,and the distance between the slits is $d$. We are given that $D = d/2$,which implies $d = 2D$.
The path difference between the waves superposing at point $P$ is given by $\Delta x = r_2 - r_1 = S_2P - S_1P$.
From the geometry of the figure:
$S_2P = \sqrt{D^2 + (d/2 + D)^2} = \sqrt{D^2 + (D + D)^2} = \sqrt{D^2 + 4D^2} = \sqrt{5}D$.
$S_1P = \sqrt{D^2 + (d/2 - D)^2} = \sqrt{D^2 + (D - D)^2} = D$.
Therefore,the path difference is $\Delta x = \sqrt{5}D - D = D(\sqrt{5} - 1)$.
For the first order minima (destructive interference),the condition is $\Delta x = \frac{\lambda}{2}$.
Equating the two expressions:
$D(\sqrt{5} - 1) = \frac{\lambda}{2}$.
$D = \frac{\lambda}{2(\sqrt{5} - 1)}$.
Using $\sqrt{5} \approx 2.236$:
$D = \frac{\lambda}{2(2.236 - 1)} = \frac{\lambda}{2(1.236)} = \frac{\lambda}{2.472} \approx 0.404\lambda$.