A corverging lens has a focal length of `20 cm` in air. It is made of a material of refractive index `1.6`. If is immersed in a liquid of refractive index `1.3`, what will be its new focal length ?

Answer

`52 cm`

Explanation

Here, `f_(a) = 20 cm, mu_(g) = 1.6, mu_(l) = 1.3, f_(l) = ?`
`(1)/(f_(a)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2))) = 0.6((1)/(R_(1)) - (1)/(R_(2)))` …(i)
`(1)/(f_(l)) = ((mu_(g))/(mu_(l)) - 1)((1)/(R_(1)) - (1)/(R_(2)))`
`= ((1.6)/(1.3) - 1)((1)/(R_(1)) - (1)/(R_(2)))` ...(ii)
Divide (i) by (ii)
`(f_(l))/(f_(a)) = (0.6)/(0.3//1.3)`
`f_(l) = (0.6 xx 1.3 xx 20)/(0.3) = 52 cm`