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When a converging lens of refractive index $n_{1}$ is placed in a liquid of refractive index $n_{2}$ , the lens behaves as diverging , then

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$n_{2} > n_{1}$

$n_{2} < n_{1}$

$n_{1} = n_{2}$

$n_{2} = 0$

answer is B.

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Detailed Solution

$\frac{1}{f} = (μ - 1) \frac{2}{R}$

$\frac{1}{f_{m}} = (\frac{μ - μ_{l}}{μ_{l}}) \frac{2}{R}$

$d i v i d i n g w e g e t \frac{f_{m}}{f} = \frac{μ_{l} (μ - 1)}{μ - μ_{l}}$

$i f μ_{l} > μ f_{m} w i l l b e o p p o s i t e i n n a t u r e o f f$

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