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`.

#### Solution

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 *R*_{1} and *R*_{2 }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