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

Explanation

(i) If `f_(u)` be the focal length of glass lens in air, then
`" "(1)/(f_(u))=(._(a)u_(g)-1)((1)/(R_(1))-(1)/(R_(2)))....(i)`
If the focal length of glass lens in liquid, them
`" "(1)/(f_(1))=(._(l)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))........(ii)`
Dividing (i) by (ii),
`(f_(t))/(f_(a))=(._(a)mu_(g)-1)/(._(l)mu_(g)-1) Rightarrow (f_(1))/(20)=(1.6-1)/((1.6)/(1.3)-1) or f_(1)=52"cm"`
`(ii)" " (f_(1))/(20)=(1.6-1)/((1.6)/(1.8)-1)or f_(1)=-108cm`
The negative focal length indicates that the convex lens behave as a concave lens in the liquid.