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The central fringe of interference pattern produced by light of wavelength $6000$ A Is found to shift to the position of 4th bright fringe, after a glass plate of $\mu = 1.5$ is introduced. The thickness of the glass plate is:A) $4.8\mu m$B) $8.23\mu m$C) $14.98\mu m$D) $3.78\mu m$

Last updated date: 13th Jun 2024
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Hint: Shift at ${n^{th}}$ bright fringe is $(\mu - 1)t = n\lambda$ (where, $\mu$ is the refractive index of the glass plate, $t$ is the thickness of the glass plate, $\lambda$ is the wavelength of the light).

According to the formula, shift at ${n^{th}}$ bright fringe is $(\mu - 1)t = n\lambda$ (where, $\mu$ is the refractive index of the glass plate, $t$ is the thickness of the glass plate, $\lambda$ is the wavelength of the light.)
$(1.5 - 1)t = 4(6000 \times {10^{ - 10}})$
$\implies t = \dfrac{{4(6000 \times {{10}^{ - 10}})}}{{0.5}} = 4.8 \times {10^{ - 10}}m = 4.8\mu m$
So, the thickness of the glass plate is $4.8\mu m$.
$\therefore$ The correct option is A.