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

We know
`(1)/(f_(a))=(""^(a)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))`
`(1)/(20=(1.6-1)((1)/(R_(1))-(1)/(R_(2)))`
or `(1)/(R_(1))-(1)/(R_(2))=(1)/(12)`
Again `(1)/(f_(1))=(""^(1)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))=((1.6)/(1.3)-1)((1)/(12))`
`=(3)/(13) xx (1)/(12)=(1)/(sqrt(52))`
or `f_(1)=52 cm`