A thin convex lens \(L\) (refractive index \(= 1.5\)) is placed on a plane mirror \(M\). When a pin is placed at \(A\), such that \(OA = 18\, cm\), its real inverted image is formed at \(A\) itself, as shown in figure. When a liquid of refractive index \({\mu _1}\) is put between the lens and the mirror, the pin has to be moved to \(A’\). such that \(OA’ = 27\,cm\), to get its inverted real image at \(A'\) itself. The value of \({\mu _1}\) will be

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Answer & Solution

Correct Answer

(B) \(\frac{4}{3}\)

Detailed explanation

For image to form at object itself, says must retrace their path back to object. Hence must incident on mirror normally.

case \(1\) : Object will be at focus of lens

\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{1}{-18}\)

\(\Rightarrow {\mathrm{R}}=18\, \mathrm{cm}\)

case \(2\) : Retraction at \(1^{\text {st }}\) surface:

\(\frac{1}{-27}-\frac{1.5}{V_{1}}=\frac{1-1.5}{R}\) ..... \((i)\)

\(2^{\text {nd }}\) retraction:

\(\frac{1.5}{V_{1}}-\frac{\mu}{\infty}=\frac{1.5-u}{-R}\) ..... \((ii)\)

From \((i)\) and \((ii)\)

\(\mu=\frac{4}{3}\)

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