The figure displays three equidistant slits illuminated by a...

From the given figure, the line $AP_0$ is perpendicular to the plane of the slits. Thus, the distance $AP_0 = D$.

In the right-angled triangle $ABP_0$, the distance $BP_0$ is given by:

$BP_0 = \sqrt{AP_0^2 + AB^2} = \sqrt{D^2 + d^2}$

The path difference between the light rays reaching $P_0$ from slits B and A is:

$\Delta x = BP_0 - AP_0 = \sqrt{D^2 + d^2} - D$

Since $D \gg \lambda$, we can assume $D \gg d$ and use the binomial expansion for the square root:

$\Delta x = D\left(1 + \dfrac{d^2}{D^2}\right)^{1/2} - D$

$\Delta x \approx D\left(1 + \dfrac{d^2}{2D^2}\right) - D = \dfrac{d^2}{2D}$

We are given that the path difference is $BP_0 - AP_0 = \dfrac{\lambda}{3}$.

Equating the two expressions for the path difference, we get:

$\dfrac{d^2}{2D} = \dfrac{\lambda}{3}$

$d^2 = \dfrac{2\lambda D}{3}$

$d = \sqrt{\dfrac{2\lambda D}{3}}$