Convex lens is made of glass of refractive index `1.5` If the radius of curvature of each of the two surfaces is `20 cm` find the ratio of the powers of the lens, when placed in air to its power, when immersed in a liquid of refractive index `1.25`.

Answer

`5//2`

Explanation

Here, `mu_g = 1.5, R_1 = 20 cm = 0.2 m`
`R_2 = -20 cm = -0.2 m`
`P_1 = (1)/(f_1) = ((mu_g)/(mu_a) - 1) ((1)/(R_1) - (1)/(R_2))`
=`(1.5 - 1)((1)/(0.2) + (1)/(0.2)) = 5 D`
When the same lens is placed in liquid,
`P_2 = (1)/(f_2) = ((mu_g)/(mu_l) - 1) ((1)/(R_1) - (1)/(R_2))`
=`((1.5)/(1.25) - 1)((1)/(0.2) + (1)/(0.2)) = (0.25)/(1.25) xx (2)/(0.2) = 2 D`
`(P_1)/(p_2) = (5)/(2)`.