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

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(B) The path difference introduced by the slabs is given by $\Delta x = (\mu_2 - 1)t - (\mu_1 - 1)t = (\mu_2 - \mu_1)t$.Substituting the given values:

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$10$

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(B) The path difference introduced by the slabs is given by $\Delta x = (\mu_2 - 1)t - (\mu_1 - 1)t = (\mu_2 - \mu_1)t$.Substituting the given values: $\Delta x = (1.55 - 1.51) 	imes 0.1 	ext{ mm} = 0.04 	imes 0.1 	ext{ mm} = 0.004 	ext{ mm} = 4 	imes 10^{-6} 	ext{ m}$.The shift in the central fringe in terms of fringe width $\beta$ is given by $n = \frac{\Delta x}{\lambda}$.Given $\lambda = 4000 	ext{ Å} = 4000 	imes 10^{-10} 	ext{ m} = 4 	imes 10^{-7} 	ext{ m}$.Therefore, $n = \frac{4 	imes 10^{-6}}{4 	imes 10^{-7}} = 10$.Thus, the central bright fringe shifts by $10$ fringes.

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