In biprism experiment, the $4^{\text {th }}$ dark band is formed opposite to one of the slits. The…
In biprism experiment, the $4^{\text {th }}$ dark band is formed opposite to one of the slits. The wavelength of light used is
( $\mathrm{D}=$ distance between source and screen, $d$ = distance between the slits)
$\frac{\mathrm{d}^2}{9 \mathrm{D}}$
$\frac{d^2}{11 D}$
$\frac{d^2}{14 D}$
$\frac{\mathrm{d}^2}{7 \mathrm{D}}$
Solution
Concept: In biprism experiment, location of the nth dark band is the location of the nth dark fringe given by,
$y_n=(2 n-1) \frac{\lambda D}{2 d}$
Given, slit width $=\mathrm{d}$, and $\mathrm{y}_4=\frac{\mathrm{d}}{2}$
$\begin{aligned}
& \mathrm{y}_4=\left(\frac{7 \lambda \mathrm{D}}{2 \mathrm{~d}}\right)=\left(\frac{\mathrm{d}}{2}\right) \\
& \therefore \lambda=\left(\frac{\mathrm{d}^2}{7 \mathrm{D}}\right)
\end{aligned}$
Option (D) is correct.