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

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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.

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