A Young's double-slit experimental set up is kept in a medium of refractive index $\left(\frac{4}{3}\right)$. Which maximum in this case will coincide with the ${6}^{\text{th }}$ maximum obtained if the medium is replaced by air?

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Correct answer: c) ${8}^{\text{th }}$. For the ${6}^{th}$Sv6Kpe[] maximum in air to coincide with the ${n}^{th}$Sv6Kpe[] maximum in the medium, their fringe positions must be equal:Sv6Kpe[] Sv6Kpe[] Sv6Kpe[] ${y}_{6,air}=\frac{6{\lambda }_{air}D}{d}$ ${y}_{n,medium}=\frac{n{\lambda }_{medium}D}{d}=\frac{n(\frac{{\lambda }_{air}}{4/3})D}{d}$ Solving for $n$ Sv6Kpe[]Sv6Kpe[] : Setting the two equations equal:Sv6Kpe[] $\frac{6{\lambda }_{air}D}{d}=\frac{n(\frac{{\lambda }_{air}}{4/3})D}{d}$ $6=\frac{n}{(4/3)}$ $6=n\times \frac{3}{4}$ $n=6\times \frac{4}{3}=8$

A Young's double-slit experimental set up is kept in a medium of refractive index \(\left(\frac{4}{3}\right)\). Which ma…

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