To make the central fringe at the centre O,a | 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 the source $S$ is at a d

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 the source $S$ is at a distance $d$ from $S_1$ and the geometry implies $SS_2 = \sqrt{2}d$,the path difference is $\Delta x = \sqrt{2}d - d = d(\sqrt{2} - 1)$.To make the central fringe at $O$,the path difference must be zero. Introducing a mica sheet of thickness $t$ and refractive index $\mu = 1.5$ in front of $S_1$ introduces an additional path difference of $(\mu - 1)t$.Setting the net path difference to zero: $(\mu - 1)t = \Delta x$.Substituting the values: $(1.5 - 1)t = d(\sqrt{2} - 1)$.$0.5t = d(\sqrt{2} - 1)$.$t = 2d(\sqrt{2} - 1)$.Thus,the thickness of the sheet is $2(\sqrt{2} - 1)d$ in front of $S_1$.

To make the central fringe at the centre $O$,a mica sheet of refractive index $1.5$ is introduced. Choose the correct statement$(s)$.

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