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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)
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( $\mathrm{D}=$ distance between source and screen, $d$ = distance between the slits)
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Verified Answer
The correct answer is:
$\frac{\mathrm{d}^2}{7 \mathrm{D}}$
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.
$y_n=(2 n-1) \frac{\lambda D}{2 d}$
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.
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