Q129 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$ . The ratio $\dfrac{\lambda_2}{\lambda_1}$ is
- A$\dfrac{m}{n-1}$
- B$\dfrac{m}{(2n-1)}$
- Correct answerC$\dfrac{2m}{(2n-1)}$
- D$\dfrac{2m}{(2n+1)}$
Solution
The position of the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ is given by:
$y_1 = \dfrac{m \lambda_1 D}{d}$
The position of the $n^{th}$ dark fringe for light of wavelength $\lambda_2$ is given by:
$y_2 = \dfrac{(2n - 1) \lambda_2 D}{2d}$
Since the two fringes coincide, we equate their positions ($y_1 = y_2$):
$\dfrac{m \lambda_1 D}{d} = \dfrac{(2n - 1) \lambda_2 D}{2d}$
$m \lambda_1 = \dfrac{(2n - 1) \lambda_2}{2}$
$\dfrac{\lambda_2}{\lambda_1} = \dfrac{2m}{2n - 1}$
Answer: $\dfrac{2m}{(2n-1)}$
$y_1 = \dfrac{m \lambda_1 D}{d}$
The position of the $n^{th}$ dark fringe for light of wavelength $\lambda_2$ is given by:
$y_2 = \dfrac{(2n - 1) \lambda_2 D}{2d}$
Since the two fringes coincide, we equate their positions ($y_1 = y_2$):
$\dfrac{m \lambda_1 D}{d} = \dfrac{(2n - 1) \lambda_2 D}{2d}$
$m \lambda_1 = \dfrac{(2n - 1) \lambda_2}{2}$
$\dfrac{\lambda_2}{\lambda_1} = \dfrac{2m}{2n - 1}$
Answer: $\dfrac{2m}{(2n-1)}$
Hence, the correct answer is option: 'C'.