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(B) Given: Refractive index of lens $\mu_l = 1.5$, refractive index of liquid $\mu_m = 1.6$, focal length in air $f_a = 20 \,cm$.Using the Lens Maker'

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$-160 \,cm$

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(B) Given: Refractive index of lens $\mu_l = 1.5$, refractive index of liquid $\mu_m = 1.6$, focal length in air $f_a = 20 \,cm$.Using the Lens Maker's Formula, the focal length in air is given by $\frac{1}{f_a} = (\mu_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For the liquid, the focal length $f_m$ is given by $\frac{1}{f_m} = \left( \frac{\mu_l}{\mu_m} - 1 ight) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Dividing the two equations: $\frac{f_m}{f_a} = \frac{(\mu_l - 1)}{\left( \frac{\mu_l}{\mu_m} - 1 ight)} = \frac{(\mu_l - 1) \mu_m}{(\mu_l - \mu_m)}$.Substituting the values: $\frac{f_m}{20} = \frac{(1.5 - 1) 	imes 1.6}{(1.5 - 1.6)} = \frac{0.5 	imes 1.6}{-0.1} = \frac{0.8}{-0.1} = -8$.Therefore, $f_m = 20 	imes (-8) = -160 \,cm$.

$A$ 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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