A thin equiconvex lens is made of glass of refractive index 1.5 and its focal length is 0.2 m, if it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the refractive index of the liquid is

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$\frac{13}{8}$

$\frac{17}{8}$

$\frac{9}{8}$

$\frac{15}{8}$

answer is B.

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Detailed Solution

$\frac{f_{l}}{f_{a}} = \frac{{aμ}_{g} - 1}{{lμ}_{g} - 1} \Rightarrow \frac{- 0 .5}{0 .2} = \frac{1 .5 - 1}{{lμ}_{g} - 1} \Rightarrow_{l} μ_{g} - 1 = - 0 .2 \Rightarrow_{l} μ_{g} = 0 .8 = \frac{4}{5} \Rightarrow \frac{^{a} μ_{g}}{_{a} μ_{l}} = \frac{4}{5} \Rightarrow \frac{1 .5}{_{a} μ_{l}} = \frac{4}{5} \Rightarrow {aμ}_{l} = \frac{15}{8}$

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