In an interference experiment, the $\mathrm{n}^{ ext {th }}$ bright fringe for light of wavelength $\lambda_1(n=0,1,2,3 \ldots)$ coincides with the $\mathrm{m}^{ ext {th }}$ dark fringe for light of wavelength $\lambda_2(m=1,2,3 \ldots)$. The ratio $\frac{\lambda_1}{\lambda_2}$ is

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In an interference experiment, the $\mathrm{n}^{	ext {th }}$ bright fringe for light of wavelength $\lambda_1(n=0,1,2,3 \ldots)$ coincides with the $\mathrm{m}^{	ext {th }}$ dark fringe for light of wavelength $\lambda_2(m=1,2,3 \ldots)$. The ratio $\frac{\lambda_1}{\lambda_2}$ is

Aicharya · Accepted Answer

As $\mathrm{n}^{	ext {th }}$ bright fringe coincides with $\mathrm{m}^{	ext {th }}$ dark fringe $\begin{array}{ll} 	herefore & \frac{\mathrm{n} \lambda_1 \mathrm{D}}{\mathrm{~d}}=\frac{(2 \mathrm{~m}-1) \lambda_2 \mathrm{D}}{2 \mathrm{~d}} \ 	herefore & \frac{\lambda_1}{\lambda_2}=\frac{(2 \mathrm{~m}-1)}{2 \mathrm{n}} \end{array}$ /

In an interference experiment, the $\mathrm{n}^{\text {th }}$ bright fringe for light of wavelength…

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