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When a lens of power P (in air) made of material of refractive index 3 is dipped in a liquid of refractive index 4, the new power of lens is

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$- \frac{4 P}{3}$

$\frac{P}{8}$

$\frac{3 P}{4}$

$- \frac{P}{8}$

answer is D.

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Detailed Solution

$P = (3 - 1) \{\frac{1}{R_{1}} - \frac{1}{R_{2}}\} = 2 \{\frac{1}{R_{1}} - \frac{1}{R_{2}}\} P^{'} = (\frac{3}{4} - 1) \{\frac{1}{R_{1}} - \frac{1}{R_{2}}\} = - \frac{1}{4} \{\frac{1}{R_{1}} - \frac{1}{R_{2}}\} = - \frac{P}{8}$

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