In a biprism experiment, the distance between | vedclass

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(C) The distance of the $n^{th}$ bright band from the central fringe is given by $x_n = n \cdot \frac{D \lambda}{d}$, where $D$ is the distance betwee

Vedclass · Accepted Answer

$5400$

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(C) The distance of the $n^{th}$ bright band from the central fringe is given by $x_n = n \cdot \frac{D \lambda}{d}$, where $D$ is the distance between the source and the screen, and $d$ is the distance between the two coherent sources.For the first case:$\lambda_1 = 6000 	ext{ Å}$The distance between the $4^{th}$ and $13^{th}$ bright band is $y = x_{13} - x_4 = (13 - 4) \frac{D \lambda_1}{d} = 9 \frac{D \lambda_1}{d}$.For the second case:$\lambda_2 = \lambda$The distance between the $6^{th}$ and $16^{th}$ bright band is $y = x_{16} - x_6 = (16 - 6) \frac{D \lambda_2}{d} = 10 \frac{D \lambda_2}{d}$.Since the distance $y$ is the same in both cases:$9 \frac{D \lambda_1}{d} = 10 \frac{D \lambda_2}{d}$$9 \lambda_1 = 10 \lambda_2$$9 	imes 6000 = 10 	imes \lambda$$\lambda = \frac{54000}{10} = 5400 	ext{ Å}$.

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