The central fringe of the interference patter | vedclass

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(A) The shift in the central fringe due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by the formula: $\De

Vedclass · Accepted Answer

$4.8$

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(A) The shift in the central fringe due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by the formula: $\Delta x = \frac{t(\mu - 1)D}{d}$.Given that this shift is equal to the position of the $4^{th}$ bright fringe,we have $\Delta x = 4\beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Equating the two: $\frac{t(\mu - 1)D}{d} = 4 \frac{\lambda D}{d}$.Simplifying,we get $t(\mu - 1) = 4\lambda$.Substituting the given values: $t(1.5 - 1) = 4 	imes 6000 	imes 10^{-10} \, m$.$t(0.5) = 24000 	imes 10^{-10} \, m$.$t = 48000 	imes 10^{-10} \, m = 4.8 	imes 10^{-6} \, m$.Since $1 \, \mu m = 10^{-6} \, m$,the thickness $t = 4.8 \, \mu m$.

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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