MHT CET · Physics · Wave Optics
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)
- A \(\frac{\mathrm{d}^2}{9 \mathrm{D}}\)
- B \(\frac{d^2}{11 D}\)
- C \(\frac{d^2}{14 D}\)
- D \(\frac{\mathrm{d}^2}{7 \mathrm{D}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{d}^2}{7 \mathrm{D}}\)
Step-by-step Solution
Detailed explanation
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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