Physics - Wave Optics Question with Solution | TestHub

The position of the point exactly in front of one slit from the central maximum is $y=\frac{d}{2}$.

The path difference at a distance $y$ from the central maximum on the screen is given by $\Delta x=\frac{yd}{D}$.

Substituting $y=\frac{d}{2}$, we get $\Delta x=\frac{d^2}{2D}$.

For the $n$-th minimum in the interference pattern, the path difference is $\Delta x=(2n-1)\frac{\lambda }{2}$.

For the second minimum, $n=2$, so $\Delta x=\frac{3\lambda }{2}$.

Equating the two expressions for path difference:

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

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

Answer: $\frac{{\mathrm{d}}^2}{3\mathrm{D}}$