In Young's double slit experiment, two slits | vedclass

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(D) The path difference $\Delta x$ at point $P$ due to the introduction of the slabs is given by:$\Delta x = |(\mu_2 - 1)t - (\mu_1 - 1)t| = |\mu_2 -

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$10$

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(D) The path difference $\Delta x$ at point $P$ due to the introduction of the slabs is given by:$\Delta x = |(\mu_2 - 1)t - (\mu_1 - 1)t| = |\mu_2 - \mu_1|t$Given $t = 0.1 \, mm = 10^{-4} \, m$, $\mu_1 = 1.51$, and $\mu_2 = 1.55$:$\Delta x = |1.55 - 1.51| 	imes 10^{-4} \, m = 0.04 	imes 10^{-4} \, m = 4 	imes 10^{-6} \, m$The shift in the position of the central maxima $y$ is given by:$y = \frac{\Delta x D}{d} = \frac{4 	imes 10^{-6} D}{d}$The fringe width $\beta$ is given by:$\beta = \frac{\lambda D}{d} = \frac{4000 	imes 10^{-10} 	imes D}{d} = 4 	imes 10^{-7} \frac{D}{d} \, m$The number of fringes shifted $N$ is:$N = \frac{y}{\beta} = \frac{4 	imes 10^{-6} \frac{D}{d}}{4 	imes 10^{-7} \frac{D}{d}} = \frac{10^{-6}}{10^{-7}} = 10$Thus, the central bright fringe shifts by $10$ fringes.

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