A corverging lens has a focal length of `20 cm` in air. It is made of a material of refractive index `1.6`. If is immersed in a liquid of refractive index `1.3`, what will be its new focal length ?

Answer

52 cm

Explanation

Here, `f_a = 20 cm, mu_g = 1.6, mu_l = 1.3, f_l = ?`
`(1)/(f_a) = ((mu_g)/(mu_a) - 1) ((1)/(R_1) - (1)/(R_2)) = 0.6 ((1)/(R_1) - (1)/(R_2))` …(i)
`(1)/(f_l) = ((mu_g)/(mu_l) - 1) ((1)/(R_1) - (1)/(R_2))`
=`((1.6)/(1.3) - 1) ((1)/(R_1) - (1)/(R_2))` ...(ii)
Divide (i) by (ii)
`(f_l)/(f_a) = (0.6)/(0.3//(1.3)`
`f_l = (0.6 xx 1.3 xx 20)/(0.3) = 52 cm`.