To make the central fringe appear at the cent | vedclass

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(A) From the geometry of the setup,the path difference at point $O$ without the mica sheet is $\Delta x = SS_2 - SS_1$. Given $SS_1 = d$ and the verti

Vedclass · Accepted Answer

The thickness of the sheet is $2(\sqrt{2}-1)d$ in front of $S_1$.

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(A) From the geometry of the setup,the path difference at point $O$ without the mica sheet is $\Delta x = SS_2 - SS_1$. Given $SS_1 = d$ and the vertical distance between $S_1$ and $S_2$ is $d$,we have $SS_2 = \sqrt{d^2 + d^2} = \sqrt{2}d$.Thus,the initial path difference is $\Delta x = \sqrt{2}d - d = d(\sqrt{2}-1)$.To shift the central fringe to $O$,we must introduce a mica sheet of thickness $t$ and refractive index $\mu$ in front of $S_1$ such that the optical path difference introduced by the sheet compensates for the existing path difference.The path difference introduced by the sheet is $(\mu - 1)t$.Setting this equal to the initial path difference: $(\mu - 1)t = d(\sqrt{2}-1)$.Given $\mu = 1.5$,we have $(1.5 - 1)t = d(\sqrt{2}-1)$.$0.5t = d(\sqrt{2}-1) \Rightarrow t = 2(\sqrt{2}-1)d$.Since the path $SS_2$ is longer,the sheet must be placed in front of $S_1$ to increase the optical path of the $S_1$ ray.

To make the central fringe appear at the centre $O,$ a mica sheet of refractive index $\mu = 1.5$ is introduced. Choose the correct statement:

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