Question

A convex lens of refractive index 1.5 has a focal length of 18 cm in air .Calculate the change in its focal length when it is immersed in water of refractive index `4/3`.

Answer 1

Reflective index of convex lens in air, μ = 1.5

Focal length of convex lens, f_a = 18 cm

Reflective index of water, μ′ = `4/3`

When the lens is in air:

`I/f_a  = (mu - 1) (1/R_1 - 1/R_2)`

Where R₁ and R₂ are the radius of curvature of the convex lens

`1/18 = (1.5 - 1) (1/R_1  - 1/R_2)`

`(1/R_1  - 1/R_2)  = 1/9`

When the lens is in air:

`1/f_w  = (mu' - 1) (1/R_1  -1/ R_2)`

Where, f_w is the focal length of the lens, when immersed in water

`1/f_w  = ((1.5)/(4/3)  - 1 ) (1/R_1  - 1/R_2)`

`1/f_w  = 1/8 xx 1/9`

`f_w =72 cm`

Change in focal length = f_w − f_a = 72 cm − 18 cm = 54 cm