In Young's double slit experiment, in an interference pattern,...

The position of the $n$-th minimum from the central maximum in Young's double slit experiment is given by:

$y_n = (2n - 1) \dfrac{\lambda D}{2d}$

For the third minimum, $n = 3$:

$y_3 = (2(3) - 1) \dfrac{\lambda D}{2d} = \dfrac{5 \lambda D}{2d}$

Since the third minimum is observed exactly in front of one of the slits, its distance from the central maximum is $y = \dfrac{d}{2}$.

Equating the two expressions for $y$:

$\dfrac{5 \lambda D}{2d} = \dfrac{d}{2}$

$5 \lambda D = d^2$

$\lambda = \dfrac{d^2}{5D}$

Answer: $d^2/5D$