In a Young's double-slit experiment,light of | vedclass

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(A) The shift produced by a plate of thickness $t$ and refractive index $n$ is given by $\Delta x = \frac{(n-1)tD}{d}$.Since both slits are covered by

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(A) The shift produced by a plate of thickness $t$ and refractive index $n$ is given by $\Delta x = \frac{(n-1)tD}{d}$.Since both slits are covered by plates of the same thickness $t$,the net shift in the central bright fringe is given by:$\Delta x_{net} = |\Delta x_2 - \Delta x_1| = \left| \frac{(n_2-1)tD}{d} - \frac{(n_1-1)tD}{d} ight| = \frac{(n_2 - n_1)tD}{d}$.Given that this shift is equal to the position of the fifth bright fringe,we have $\Delta x_{net} = 5 \frac{\lambda D}{d}$.Equating the two expressions:$\frac{(n_2 - n_1)tD}{d} = \frac{5 \lambda D}{d}$.$t = \frac{5 \lambda}{n_2 - n_1}$.Substituting the given values: $\lambda = 4800 	imes 10^{-10} \, m$,$n_1 = 1.4$,$n_2 = 1.7$.$t = \frac{5 	imes 4800 	imes 10^{-10}}{1.7 - 1.4} = \frac{24000 	imes 10^{-10}}{0.3} = 80000 	imes 10^{-10} \, m = 8 	imes 10^{-6} \, m = 8 \, \mu m$.

Similar Questions

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:

Each plate reflects $25\%$ of the incident light intensity. When $AB$ and $A'B'$ are taken as the two slits in Young's double-slit experiment,what is the ratio of maximum to minimum intensity ${I_{\max }}/{I_{\min }}$ (in $: 1$)?

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