In a double slit arrangement,fringes are prod | vedclass

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(A) The shift in the fringe pattern due to a glass 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 fringe pattern due to a glass 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 by plates of the same thickness $t$ but different refractive indices $\mu_1 = 1.4$ and $\mu_2 = 1.7$,the net shift is:$\Delta x = \Delta x_2 - \Delta x_1 = \frac{\beta}{\lambda}(\mu_2 - 1)t - \frac{\beta}{\lambda}(\mu_1 - 1)t$$\Delta x = \frac{\beta}{\lambda}(\mu_2 - \mu_1)t$Given that the central bright fringe shifts to the position of the fifth bright fringe,$\Delta x = 5\beta$.Equating the two expressions:$5\beta = \frac{\beta}{\lambda}(\mu_2 - \mu_1)t$$t = \frac{5\lambda}{\mu_2 - \mu_1}$Substituting the values $\lambda = 4800 	imes 10^{-10} \ m$,$\mu_2 = 1.7$,and $\mu_1 = 1.4$:$t = \frac{5 	imes 4800 	imes 10^{-10}}{1.7 - 1.4} = \frac{24000 	imes 10^{-10}}{0.3} = 80000 	imes 10^{-10} \ m = 8 	imes 10^{-6} \ m = 8 \ \mu m$.

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The fringe widths are found to be $\omega_1$ and $\omega_2$ respectively if a Young's double slit experiment is performed in media of refractive indices $n_1$ and $n_2$ respectively. The correct statement is:

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