Question

In biprism experiment, the \(4^{th}\) dark band is formed opposite to one of the slits. The wavelength of light used is (D=distance between source and screen, d= the distance between the slits)

• A. \(\frac{d^2}{9D}\)
• B. \(\frac{d^2}{11D}\)
• C. \(\frac{d^2}{14D}\)
• D. \(\frac{d^2}{7D}\)

Hint:

Dark fringes are at x = (2n - 1) lambda D / (2d); a slit is at x = d/2.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In a biprism setup the two virtual sources act as slits separated by \(d\). Dark bands are at \(x_n=\dfrac{(2n-1)\lambda D}{2d}\).

Step 2: Position of a slit:
A point opposite one of the slits is at distance \(\dfrac d2\) from the central line.

Step 3: Use n = 4:
\(\dfrac{(2\cdot4-1)\lambda D}{2d}=\dfrac d2\), so \(\dfrac{7\lambda D}{2d}=\dfrac d2\).

Step 4: Solve:
\(\lambda=\dfrac{d^2}{7D}\). Option D.

Step 5: Why the other options are wrong.
\(\dfrac{d^2}{9D}\) would be right for the 5th dark band, and \(\dfrac{d^2}{11D}\) for the 6th. Option C, \(\dfrac{d^2}{14D}\), comes from using \(2n\) instead of \(2n-1\).

Final Answer:
The wavelength is d^2 / (7D). \[ \boxed{\text{(D) }\dfrac{d^2}{7D}} \]