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A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

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+ 160 cm

– 16 cm

– 160 cm

+ 16 cm

answer is B.

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Detailed Solution

$\begin{aligned} μ_{1} = 1.5 \\ μ_{m} = 1.6 \\ f_{a} = 20 cm \end{aligned}$

$As \frac{f_{m}}{f_{a}} = \frac{(μ_{1} - 1) μ_{m}}{(μ_{1} - μ_{m})}$

$\frac{f_{m}}{20} = \frac{(1.5 - 1) 1.6}{(1.5 - 1.6)}$

$f_{m} = - 160 cm$

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