Question

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

• A. $\frac{d^2}{14D}$
• B. $\frac{d^2}{4D}$
• C. $\frac{d^2}{7D}$
• D. $\frac{d^2}{3.5D}$

Hint:

Whenever a fringe forms "opposite one of the slits," always set its position equal to $\frac{d}{2}$. For the $n^{\text{th}}$ dark band, this gives $\frac{d}{2} = \frac{(2n-1)\lambda D}{2d}$, which simplifies to the shortcut formula $\lambda = \frac{d^2}{(2n-1)D}$. Plucking in $n=4$ immediately yields $\frac{d^2}{7D}$.

Answer 1

The Correct Option is C

Step 1: Locate a slit relative to the centre.
The two coherent slits are separated by $d$, so the central axis of symmetry runs midway between them. Each slit therefore lies at a distance $d/2$ from the centre of the pattern.
Step 2: Write the position of a dark band.
The $n^{\text{th}}$ dark fringe lies at $$x_n = (2n - 1)\frac{\lambda D}{2d},$$ measured from the central bright fringe.
Step 3: Apply the given geometric condition.
The $4^{\text{th}}$ dark band falls opposite a slit, so its position equals the slit offset: $$x_4 = \frac{d}{2}.$$
Step 4: Substitute $n = 4$.
$$x_4 = (2\times 4 - 1)\frac{\lambda D}{2d} = \frac{7\lambda D}{2d}.$$
Step 5: Equate the two expressions for $x_4$.
$$\frac{d}{2} = \frac{7\lambda D}{2d}.$$ Cancelling the $2$'s gives $d = \dfrac{7\lambda D}{d}.$
Step 6: Solve for the wavelength.
Cross-multiplying, $d^2 = 7\lambda D$, so $$\lambda = \frac{d^2}{7D}.$$
\[ \boxed{\lambda = \frac{d^2}{7D}} \]