Q134 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
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$\dfrac{d^2}{9D}$
- B$\dfrac{d^2}{11D}$
- C$\dfrac{d^2}{14D}$
- Correct answerD$\dfrac{d^2}{7D}$
Solution
The position of the $n^{th}$ dark band from the central maximum is given by $y_n = (2n - 1) \dfrac{\lambda D}{2d}$.
For the $4^{th}$ dark band, $n = 4$.
$y_4 = (2(4) - 1) \dfrac{\lambda D}{2d} = \dfrac{7 \lambda D}{2d}$
The point opposite to one of the slits is at a distance of $\dfrac{d}{2}$ from the central maximum.
Equating the two distances:
$\dfrac{7 \lambda D}{2d} = \dfrac{d}{2}$
$7 \lambda D = d^2$
$\lambda = \dfrac{d^2}{7D}$
Answer: $\dfrac{d^2}{7D}$
For the $4^{th}$ dark band, $n = 4$.
$y_4 = (2(4) - 1) \dfrac{\lambda D}{2d} = \dfrac{7 \lambda D}{2d}$
The point opposite to one of the slits is at a distance of $\dfrac{d}{2}$ from the central maximum.
Equating the two distances:
$\dfrac{7 \lambda D}{2d} = \dfrac{d}{2}$
$7 \lambda D = d^2$
$\lambda = \dfrac{d^2}{7D}$
Answer: $\dfrac{d^2}{7D}$
Hence, the correct answer is option: 'D'.