Question

A convex lens is made of a glass of refractive index $1.5$. The radius of curvature of each of its surface is $20cm$. Then the ratio of the power of the lens when placed in air to its power when immersed inside a liquid of refractive index $1.25$ is

• A. $2:5$
• B. $5:2$
• C. $3:2$
• D. $2:3$

Answer 1

The correct option is B $5:2$

Given - ${}_a{n}_g=1.5{,}_a{n}_l=1.25$

By using , the focal length of a glass (g) lens in a medium (m) is ,

$\frac{1}{f}={(}_m{n}_g-1\left)\right[1/R_1-1/R_2]$

When lens is in air ,

$\frac{1}{f_a}={(}_a{n}_g-1\left)\right[1/R_1-1/R_2]$ .............eq1

When lens is in liquid ,

$\frac{1}{f_l}={(}_l{n}_g-1\left)\right[1/R_1-1/R_2]$ ................eq2

now ${}_l{n}_g{=}_a{n}_g{/}_a{n}_l=1.5/1.25$

dividing eq1 by eq2 ,

$\frac{f_l}{f_a}=\frac{{(}_a{n}_g-1)}{{(}_l{n}_g-1)}$

or $\frac{f_l}{f_a}=\frac{(1.5-1)}{(1.5/1.25-1)}=5/2$

or $\frac{P_a}{P_l}=\frac{5}{2}$ (because $f=1/P$)

Hence $P_a:P_l=5:2$