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(D) When a thin sheet of thickness $t$ and refractive index $\mu$ is placed in the path of one of the beams,the additional path difference introduced

Vedclass · Accepted Answer

$1.5$

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(D) When a thin sheet of thickness $t$ and refractive index $\mu$ is placed in the path of one of the beams,the additional path difference introduced is $\Delta x = (\mu - 1)t$.The shift in the central bright fringe is given by $S = \frac{(\mu - 1)tD}{d}$.The fringe width is given by $\beta = \frac{\lambda D}{d}$.According to the problem,the shift is equal to the fringe width,so $S = \beta$.Therefore,$\frac{(\mu - 1)tD}{d} = \frac{\lambda D}{d}$.This simplifies to $(\mu - 1)t = \lambda$.Substituting the given values: $(\mu - 1) 	imes 12 	imes 10^{-7} = 6 	imes 10^{-7}$.$\mu - 1 = \frac{6 	imes 10^{-7}}{12 	imes 10^{-7}} = 0.5$.$\mu = 1 + 0.5 = 1.5$.

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