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 = - 0.2m`
`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)`