In an interference experiment, the m^{th} bri | vedclass

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(A) The position of the $m^{th}$ bright fringe for wavelength $\lambda_1$ is given by $y_m = m \frac{D \lambda_1}{d}$.The position of the $n^{th}$ dar

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$\frac{2m}{2n-1}$

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(A) The position of the $m^{th}$ bright fringe for wavelength $\lambda_1$ is given by $y_m = m \frac{D \lambda_1}{d}$.The position of the $n^{th}$ dark fringe for wavelength $\lambda_2$ is given by $y_n = (n - \frac{1}{2}) \frac{D \lambda_2}{d} = \frac{(2n - 1)}{2} \frac{D \lambda_2}{d}$.Since the fringes coincide, $y_m = y_n$:$m \frac{D \lambda_1}{d} = \frac{(2n - 1)}{2} \frac{D \lambda_2}{d}$.Simplifying the equation, we get $m \lambda_1 = \frac{(2n - 1)}{2} \lambda_2$.Therefore, the ratio $\frac{\lambda_2}{\lambda_1} = \frac{2m}{2n - 1}$.

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 $\frac{\lambda_2}{\lambda_1}$ is

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