In a biprism experiment,if the 5^{text{th}} b | vedclass

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(C) The position of the $n^{	ext{th}}$ bright band is given by $y_n = \frac{n \lambda_1 D}{d}$.For the $5^{	ext{th}}$ bright band $(n=5)$: $y_5 = \f

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$\frac{10}{11}$

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(C) The position of the $n^{	ext{th}}$ bright band is given by $y_n = \frac{n \lambda_1 D}{d}$.For the $5^{	ext{th}}$ bright band $(n=5)$: $y_5 = \frac{5 \lambda_1 D}{d}$.The position of the $n^{	ext{th}}$ dark band is given by $y_n' = \frac{(2n-1) \lambda_2 D}{2d}$.For the $6^{	ext{th}}$ dark band $(n=6)$: $y_6' = \frac{(2 	imes 6 - 1) \lambda_2 D}{2d} = \frac{11 \lambda_2 D}{2d}$.Given that the bands coincide,$y_5 = y_6'$,so:$\frac{5 \lambda_1 D}{d} = \frac{11 \lambda_2 D}{2d}$.Canceling $D$ and $d$ from both sides:$5 \lambda_1 = \frac{11 \lambda_2}{2}$.Rearranging to find the ratio $\frac{\lambda_2}{\lambda_1}$:$\frac{\lambda_2}{\lambda_1} = \frac{5 	imes 2}{11} = \frac{10}{11}$.

In a biprism experiment,if the $5^{\text{th}}$ bright band with wavelength $\lambda_1$ coincides with the $6^{\text{th}}$ dark band with wavelength $\lambda_2$,then the ratio $\left(\frac{\lambda_2}{\lambda_1}\right)$ is:

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