The central fringe of the interference patter | vedclass

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(A) The shift in the interference pattern due to the introduction of a glass plate is given by the formula: $\Delta x = \frac{t(\mu - 1)D}{d}$.Given t

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$4.8$

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(A) The shift in the interference pattern due to the introduction of a glass plate is given by the formula: $\Delta x = \frac{t(\mu - 1)D}{d}$.Given that the central fringe shifts to the position of the $4^{th}$ bright fringe,the shift is equal to $4 \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Equating the shift: $\frac{t(\mu - 1)D}{d} = 4 \frac{\lambda D}{d}$.This simplifies to: $t(\mu - 1) = 4 \lambda$.Given $\lambda = 6000 \, \mathring{A} = 6 	imes 10^{-7} \, m$ and $\mu = 1.5$,we have:$t(1.5 - 1) = 4 	imes 6 	imes 10^{-7} \, m$.$t(0.5) = 24 	imes 10^{-7} \, m$.$t = \frac{24 	imes 10^{-7}}{0.5} = 48 	imes 10^{-7} \, m$.$t = 4.8 	imes 10^{-6} \, m = 4.8 \, \mu m$.

The central fringe of the interference pattern produced by light of wavelength $6000 \, \mathring{A}$ is found to shift to the position of the $4^{th}$ bright fringe after a glass plate of refractive index $1.5$ is introduced. The thickness of the glass plate would be ........ $\mu m$.

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