A convex lens of focal length 0.24 m and of refractive index 1.5 is completely immersed in water of refractive 1.33. Find the changes in the focal length of the lens.

Explanation

Given : - `n_g=1.5 , n_w =1.33` f=0.24 m
When the lens is in air ,
`1/(f_g)=(n_g-1)[1/R_1-1/R_2]`
`1/0.24=[1.5-1][1/R_1-1/R_2]` ….(1)
When the lens is in water ,
`1/(f_(gw))=[n_g/n_w-1][1/R_1- 1/R_2] =[1.5/1.33-1] [1/R_1-1/R_2]`
`1/(f_(gw))=[1.1278-1][1/R_1 - 1/R_2]` ...(2)
Dividing eqn. (1) by eqn. (2)

`(f_(gw))/f_g=0.5/0.1278`
`(f_(gw))/0.24=0.5/0.1278`
`f_(gw)` = 0.94 m
Change in focal length = 0.94-0.24=0.70 m