# A lens made of material of refractive index $\mu =\large\frac{3}{2}$ when placed in air behaves as shown in figure (1) . But when placed in a medium of refractive index $\mu_s$ it behaves as shown in figure (2) . Then the value of $\mu_s$ is

$(a)\;\mu_s=1.75 \\ (b)\;\mu_s =1.6 \\ (c)\;\mu_s =2 \\ (d)\;\mu_s=1.45$

Focal length of a lens immersed in a medium of refractive index $\mu _s$ is given by
$\large\frac{1}{f_s} =\bigg( \large\frac{\mu_g}{\mu_s}-1\bigg) \bigg[ \large\frac{1}{R_1} -\frac{1}{R_2}\bigg]$
When $\mu_g$= refractive index of glass
Also If focal length of the lens is known in air, we have,
$f_s= \large\frac{\mu_g-1}{(\mu_g/ \mu_s-1)}f_a$
$f_s$- focal length in medium
$f_a$- focal length in air
Given ,
$f_s= +20$
$f_a= -10$
$\mu_g= \large\frac{3}{2}$
$\mu_s= ?$
$20 = \large\frac{(3/2-1)}{(3/2 \mu _s -1)} (-10)$
$\bigg ( \large\frac{3}{2 \mu _s }-1\bigg) =\large\frac{-10}{20} \times \frac{1}{2}$
$\bigg ( \large\frac{3}{2 \mu _s }-1\bigg) =\large\frac{-1}{4}$
$\bigg ( \large\frac{3}{2 \mu _s }\bigg) =\large\frac{-1}{4}$$+1 \mu _s =\large\frac{3/2}{3/4}$$=2$
Hence c is the correct answer.