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

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Question

(A) The position of the $n^{	ext{th}}$ bright fringe is given by $y_{n} = \frac{n \lambda_{1} D}{d}$. For the $5^{	ext{th}}$ bright band,$y_{5} = \f

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$\frac{10}{11}$

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(A) The position of the $n^{	ext{th}}$ bright fringe is given by $y_{n} = \frac{n \lambda_{1} D}{d}$. For the $5^{	ext{th}}$ bright band,$y_{5} = \frac{5 \lambda_{1} D}{d}$.The position of the $m^{	ext{th}}$ dark fringe is given by $y'_{m} = \frac{(2m - 1) \lambda_{2} D}{2d}$. For the $6^{	ext{th}}$ dark band,$y'_{6} = \frac{(2 	imes 6 - 1) \lambda_{2} D}{2d} = \frac{11 \lambda_{2} D}{2d}$.Since the bands coincide,$y_{5} = y'_{6}$.$\frac{5 \lambda_{1} D}{d} = \frac{11 \lambda_{2} D}{2d}$.Simplifying,$5 \lambda_{1} = \frac{11 \lambda_{2}}{2}$.Therefore,$\frac{\lambda_{1}}{\lambda_{2}} = \frac{11}{10}$,which implies $\frac{\lambda_{2}}{\lambda_{1}} = \frac{10}{11}$.

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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