Physics - Wave Optics Question with Solution | TestHub

The position of the point on the screen exactly in front of one of the slits is $y=\frac{d}{2}$.

The path difference at this point is given by $\Delta x=\frac{yd}{D}$.

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

For the $n^{\text{th}}$ dark fringe, the path difference is $\Delta x=(2n-1)\frac{\lambda }{2}$.

Equating the two expressions for path difference:

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

Given $\lambda =\frac{d^2}{3D}$, substituting this value:

$\frac{d^2}{2D}=(2n-1)\frac{d^2}{6D}$

$3=2n-1$

$2n=4$

$n=2$

Answer: 2