A lens which has focal length of 4 cm and refractive index of 1.4 is immersed in a liquid of refractive index 1.6, then the focal length will be

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– 12.8 cm

12.8 cm

– 32 cm

32 cm

answer is A.

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Detailed Solution

$\frac{f_{l}}{f_{a}} = (\frac{{aμ}_{g} - 1}{{lμ}_{g} - 1}) \Rightarrow \frac{f_{l}}{4} = \frac{(1 .4 - 1)}{\frac{1 .4}{1 .6} - 1} \Rightarrow f_{l} = - 12 .8 cm$

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