Q157 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
In Young's double slit experiment, in an interference pattern, third minimum is observed exactly in front of one of the slits. If '$d$' is the distance between the slits and '$D$' is the distance between the slits and the screen, the wavelength of light used is
- A$d^2/3D$
- B$d^2/4D$
- Correct answerC$d^2/5D$
- D$d^2/6D$
Solution
The position of the $n$-th minimum from the central maximum in Young's double slit experiment is given by:
$y_n = (2n - 1) \dfrac{\lambda D}{2d}$
For the third minimum, $n = 3$:
$y_3 = (2(3) - 1) \dfrac{\lambda D}{2d} = \dfrac{5 \lambda D}{2d}$
Since the third minimum is observed exactly in front of one of the slits, its distance from the central maximum is $y = \dfrac{d}{2}$.
Equating the two expressions for $y$:
$\dfrac{5 \lambda D}{2d} = \dfrac{d}{2}$
$5 \lambda D = d^2$
$\lambda = \dfrac{d^2}{5D}$
Answer: $d^2/5D$
$y_n = (2n - 1) \dfrac{\lambda D}{2d}$
For the third minimum, $n = 3$:
$y_3 = (2(3) - 1) \dfrac{\lambda D}{2d} = \dfrac{5 \lambda D}{2d}$
Since the third minimum is observed exactly in front of one of the slits, its distance from the central maximum is $y = \dfrac{d}{2}$.
Equating the two expressions for $y$:
$\dfrac{5 \lambda D}{2d} = \dfrac{d}{2}$
$5 \lambda D = d^2$
$\lambda = \dfrac{d^2}{5D}$
Answer: $d^2/5D$
Hence, the correct answer is option: 'C'.