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

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$4$

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(D) The shift in the central maximum due to the introduction of a glass plate of thickness $t$ and refractive index $\mu$ is given by the formula: $\Delta x = \frac{(\mu - 1) t D}{d}$.Given that the central maximum shifts to the position of the $4^{th}$ bright fringe, the shift is equal to the path difference of the $4^{th}$ bright fringe, which is $4\lambda$.Equating the path difference introduced by the glass plate to the shift: $(\mu - 1) t = n \lambda$.Substituting the given values: $(1.5 - 1) t = 4 	imes 500 \, nm$.$0.5 	imes t = 2000 \, nm$.$t = 4000 \, nm$.Since $1 \, \mu m = 1000 \, nm$, we have $t = 4 \, \mu m$.

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