Interference fringes were produced in Young's | vedclass

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(C) The shift in the fringe pattern due to the introduction of a thin film of thickness $t$ and refractive index $\mu$ is given by $\Delta x = (\mu -

Vedclass · Accepted Answer

$1.4$

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(C) The shift in the fringe pattern due to the introduction of a thin film of thickness $t$ and refractive index $\mu$ is given by $\Delta x = (\mu - 1) t \left( \frac{D}{d} ight)$.Given that the shift is equal to $20$ fringe widths, we have $\Delta x = 20 \beta$, where $\beta = \frac{\lambda D}{d}$ is the fringe width.Equating the two expressions: $(\mu - 1) t \left( \frac{D}{d} ight) = 20 \left( \frac{\lambda D}{d} ight)$.Simplifying, we get $(\mu - 1) t = 20 \lambda$.Given $\lambda = 5000 \, Å = 5000 	imes 10^{-10} \, m = 5 	imes 10^{-7} \, m$ and $t = 2.5 	imes 10^{-3} \, cm = 2.5 	imes 10^{-5} \, m$.Substituting the values: $(\mu - 1) (2.5 	imes 10^{-5}) = 20 	imes (5 	imes 10^{-7})$.$(\mu - 1) = \frac{100 	imes 10^{-7}}{2.5 	imes 10^{-5}} = \frac{10^{-5}}{2.5 	imes 10^{-5}} = \frac{1}{2.5} = 0.4$.Therefore, $\mu = 1 + 0.4 = 1.4$.

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