Physics - Wave Optics Question with Solution | TestHub

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

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

For the third minimum, $n=3$:

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

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

Equating the two expressions for $y$:

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

$5\lambda D=d^2$

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

Answer: $d^2/5D$