Question

A thin lens of focal length +12cm and refractive index 1.5 is immersed in water (refractive index =1.33).What is its new focal length

Answer 1

Dear student

Formula for the focal length of the lens is :

1/f = (n_g​-n_m)(1/R₁-1/R₂)

f= focal length of the lens=12 cm
n_g= refractive index of the glass=1.5
n_​m= refrctive index of the medium=1.33
R₁&R₂ are the radius of curvature near and far from the sorce respectively.

in air , refractive index of the air is 1.
Hence, the focal length in air is given by :-
1/f_a = (1.5-1)(1/R₁-1/R₂) ------(1)

in water, refractive index of the water is 1.33.
Hence, the focal length in water is given by :​

1/f_w = (1.5-1.33)(1/R₁-1/R₂ ) -----(2)
divide equation no. (1) and (2), we get

f_w​/f_a = (1.5-1)/(1.5-1.33)
= .5/.17
= 2.94
Given :- f_a = 12cm
f_{w =}
=35.3cm
Hence, focal length of the lens in the water in 35.3cm.