To make the central fringe appear at the cent | vedclass

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(A) From the geometry of the setup,the path difference between the two rays reaching the center $O$ is $\Delta x = SS_2 - SS_1$. Given $SS_1 = d$ and

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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 between the two rays reaching the center $O$ is $\Delta x = SS_2 - SS_1$. Given $SS_1 = d$ and the distance between $S_1$ and $S_2$ is $d$,the distance $SS_2 = \sqrt{d^2 + d^2} = \sqrt{2}d$. Thus,the initial path difference is $\Delta x = \sqrt{2}d - d = d(\sqrt{2} - 1)$. To shift the central fringe to $O$,the mica sheet must introduce an additional path difference equal to the existing path difference,but in the opposite direction. The path difference introduced by a sheet of thickness $t$ and refractive index $\mu$ is $(\mu - 1)t$. Setting $(\mu - 1)t = d(\sqrt{2} - 1)$,and substituting $\mu = 1.5$: $(1.5 - 1)t = d(\sqrt{2} - 1) \Rightarrow 0.5t = d(\sqrt{2} - 1) \Rightarrow t = 2(\sqrt{2} - 1)d$. Since $SS_2 > SS_1$,the path difference is positive for the path through $S_2$. To compensate,the sheet must be placed in front of $S_1$ to increase the optical path length of the $S_1$ ray.

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

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