Question

A convex lens of glass (μ= 1.5) has focal length 20 cm. The lens is immersed in water (μ = 1.33). What is the change in the power of lens ?

Answer 1

first of all, we have to find new focal length of lens when it is immersed in water.

use formula, $f_l=\frac{\mu_g-1}{\frac{\mu_g}{\mu_l}-1}\times f$

where $f_l$ denotes focal length of lens in water , f denotes focal length of lens in air , $\mu_g$ denotes refractive index of glass and $\mu_l$ denotes refractive index of liquid.

here, $\mu_g=1.5,\mu_l=1.33$ and f = 20cm

so, $f_l$ = (1.5 - 1)/(1.5/1.33 - 1) × 20

= (0.5)/(0.1278) × 20

= 10/0.1278 = 78.24 cm

now, power of lens , P' = 1/$f_l$ = 100/78.24 = 1.278 D

while power of lens when it is in air medium, P = 1/f = 100/20 = 5D

hence, change in power of lens = P' - P = 1.278 - 5 = -3.721D , hence after immerging of lens in water power of lens decreases.