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: Use the path difference:
Opposite slit $S_1$, the path difference to the two slits is $\Delta=\dfrac{xd}{D}=\dfrac{d}{2}\cdot\dfrac dD=\dfrac{d^2}{2D}$.

Step 2: Dark band condition:
For the 4th dark band, $\Delta=\dfrac{7\lambda}{2}$.

Step 3: Solve:
$\dfrac{d^2}{2D}=\dfrac{7\lambda}{2}$, so $\lambda=\dfrac{d^2}{7D}$. Option D.

Final Answer:
At x = d/2 the path difference is d^2/(2D) = 7 lambda/2. \[ \boxed{\text{(D) }\dfrac{d^2}{7D}} \]