When a mica sheet of thickness 7 mu m and re | vedclass

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(C) The shift in the central fringe due to the introduction of a mica sheet is given by the formula: $\Delta x = \frac{(\mu - 1) t D}{d}$.Given that t

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(C) The shift in the central fringe due to the introduction of a mica sheet is given by the formula: $\Delta x = \frac{(\mu - 1) t D}{d}$.Given that the central fringe shifts to the position of the $n^{th}$ bright fringe, the shift is equal to $n \beta$, where $\beta = \frac{\lambda D}{d}$ is the fringe width.Equating the two: $(\mu - 1) t \frac{D}{d} = n \frac{\lambda D}{d}$.This simplifies to: $\lambda = \frac{(\mu - 1) t}{n}$.Given values: $\mu = 1.6$, $t = 7 \ \mu m = 7 	imes 10^{-6} \ m$, and $n = 7$.Substituting these values: $\lambda = \frac{(1.6 - 1) 	imes 7 	imes 10^{-6}}{7}$.$\lambda = 0.6 	imes 10^{-6} \ m = 6 	imes 10^{-7} \ m$.Converting to $\mathring{A}$s: $\lambda = 6000 \ \mathring{A}$.

When a mica sheet of thickness $7 \ \mu m$ and refractive index $\mu = 1.6$ is placed in the path of one of the interfering beams in a biprism experiment, the central fringe shifts to the position of the seventh bright fringe. The wavelength of the light used is $............ \ \mathring{A}$.

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