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(D) When a thin transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits in a Young's double slit experime

Vedclass · Accepted Answer

$\frac{n\lambda}{(\mu - 1)}$

Vedclass · Answer

(D) When a thin transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits in a Young's double slit experiment,the path difference introduced is $\Delta x = (\mu - 1)t$.The shift in the central maximum position $y$ is given by $y = \frac{D}{d} \Delta x = \frac{D}{d}(\mu - 1)t$,where $D$ is the distance between the slits and the screen,and $d$ is the slit separation (given as $a$ in the figure).The fringe width is $\beta = \frac{\lambda D}{d}$.According to the problem,the shift is equal to $n$ fringe widths,so $y = n\beta$.Substituting the expressions,we get:$\frac{D}{a}(\mu - 1)t = n \frac{\lambda D}{a}$Canceling $\frac{D}{a}$ from both sides:$(\mu - 1)t = n\lambda$Therefore,$t = \frac{n\lambda}{(\mu - 1)}$.Comparing this with the given options,option $D$ is $\frac{n\lambda}{(\mu - 1)}$. Thus,option $D$ is correct.

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