A convex lens made up of glass of refractive index `1.5` is dippedin turn
(i) in a medium of refractive index `1.65`
(ii) in a medium of refractive index `1.33`
(a) Will it behave as converging or diverging lens in the two cases ?
(b) How will its focal length changes in the two media ?

Answer

(i) diverging
(ii) converging

Explanation

Here, `mu_g = 1.5`.
The focal length of the lens in air is
`(1)/(f_a) = ((mu_g)/(mu_a) - 1)((1)/(R_1) -(1)/(R_1))`
=`((1.5)/(1) - 1) ((1)/(R_1) -(1)/(R_2))`
`(1)/(R_1) - (1)/(R_2) = (2)/(f_a)`
(i) When lens is dipped in medium `A` of `mu_A = 1.65`,
`(1)/(f_A) = ((mu_g)/(mu_A) - 1) ((1)/(R_1) - (1)/(R_2))`
=`((1.5)/(1.65) -1) xx (2)/(f_a) = (-0.15 xx 2)/(1.65 f_a)`
`f_A = (1.65 f_a)/(0.15 xx 2) = -5.5 f_a`.
:. In medium `A`, the lens will behave as adiverging lens, of `f_A = -5.5 f_A`
(ii) When lens is dipped in medium `B` of `mu_B = 1.33`.
`(1)/(f_B) = ((mu_g)/(mu_B) - 1) ((1)/(R_1) - (1)/(R_2))`
`(1)/(f_B) = ((1.55)/(1.33) -1) xx (2)/(f_a) = (0.17 xx 2)/(1.33 f_a)`
`f_B = (1.33 f_a)/(0.34) = 3.91 f_a`
`:.` In medium `B`, the lens behaves as a converging lens of `f_B = 3.91 f_a`.