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 ?

Explanation

`"Given, "`
`"Refractive index of glass w.r.t. air, """_(a)mu_(g)=1.6`
`"Refractiven index of liquid w.r.t air, """_(a)mu_(1)=1.3`
`"Refractive index of glass w.r.t. to liquid is "`
`""_(1)mu_(g)=(""_(a)mu_(g))/(""_(a)mu_(1))=(1.6)/(1.3)=1.23`
`"Focal length of lens in air, "f_(a)=20m=0.2m`
`"Focal length of lens in air, "f_(a)=20cm=0.2m`
`therefore" "(1)/(f_(a))=(""_(a)mu_(g)-1)[(1)/(R_(1))-(1)/(R_(2))]`
where `R_(1) and R_(2)` are radii of curvature of curvature of two faces
`rArr" "(1)/(0.20)=(1.6-1)[(1)/(R_(1))-(1)/(R_(2))]`
`rArr" "(1)/(R_(1))-(1)/(R_(2))=(100)/(12)`
When the lens is immersed in liquid, focal length is
`(1)/(f_(1))=(""_(1)mu_(g)-1)[(1)/(R_(1))-(1)/(R_(2))]`
`=(1.23-1)[(100)/(12)]`
`(1)/(f_(1))=(0.23xx100)/(12)=(23)/(12)`
`rArr" "f_(1)=(12)/(23)=0.52m`
It is a convex lens inside the liquid.