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(A) The shift in the central bright fringe due to a plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{D}{d}(\mu - 1)t =

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(A) The shift in the central bright fringe due to a plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{D}{d}(\mu - 1)t = \frac{\beta}{\lambda}(\mu - 1)t$.When both slits are covered,the net shift is $\Delta x_{net} = \frac{\beta}{\lambda}(\mu_2 - \mu_1)t$.Given that the central fringe shifts to the position of the $5^{th}$ bright fringe,the net shift is $5\beta$.Therefore,$5\beta = \frac{\beta}{\lambda}(\mu_2 - \mu_1)t$.$5 = \frac{1}{\lambda}(\mu_2 - \mu_1)t$.$t = \frac{5\lambda}{\mu_2 - \mu_1} = \frac{5 	imes 4800 	imes 10^{-10} \, m}{1.8 - 1.5} = \frac{24000 	imes 10^{-10}}{0.3} \, m$.$t = 80000 	imes 10^{-10} \, m = 8 	imes 10^{-6} \, m = 8 \, \mu m$.

Similar Questions

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