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(C) The focal length of a lens in a medium is given by the lens maker's formula: $\frac{1}{f_m} = (_{	ext{medium}}\mu_{	ext{lens}} - 1) \left( \frac

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$_a \mu_g < _a \mu_l$

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(C) The focal length of a lens in a medium is given by the lens maker's formula: $\frac{1}{f_m} = (_{	ext{medium}}\mu_{	ext{lens}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a concave lens,the term $(\frac{1}{R_1} - \frac{1}{R_2})$ is negative.In air,the lens is divergent $(f  1$.For the lens to behave as a convergent lens in a liquid,the focal length $f_m$ must be positive.Since $(\frac{1}{R_1} - \frac{1}{R_2})$ is negative,the term $(_{	ext{liquid}}\mu_{	ext{glass}} - 1)$ must also be negative for $f_m$ to be positive.This implies $_{	ext{liquid}}\mu_{	ext{glass}} Since $_{	ext{liquid}}\mu_{	ext{glass}} = \frac{_a\mu_g}{_a\mu_l}$,we have $\frac{_a\mu_g}{_a\mu_l} < 1$,which means $_a\mu_g < _a\mu_l$.

$A$ glass concave lens is placed in a liquid in which it behaves like a convergent lens. If the refractive indices of glass and liquid with respect to air are $_a \mu_g$ and $_a \mu_l$ respectively,then

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