The radii of curvature of a convex lens are 20 cm and 30 cm nd the refractive index of its material is 1.5 .How its nature nd focal length will change ,when it is immersed in a liquid of refractive index = 1.6?

Explanation

`R_(1)=+20cm`
`R_(2)=-30` cm
`""^(a)mu_(g)=1.5` and `""^(a)mu_(i)=1.6`
`1/(f_("air"))=(""^(a)mu_(g)-1)(1/(R_(1))-1/(R_(2)))`
`=(1.5-1)(1/20-1/(-30))`
`1/(f_("air"))=0.5(1/20+1/30)=0.5((3+2)/60)`
`1/(f_("air"))=0.5/12`
`implies1/(f_("air"))=12/0.5=24cm`
Let `f_(1)=` Focal length of lens when dipped in given liquid
`1/(f_(1))=(""^(I)mu_(g)-1)(1/(R_(1))-1/(R_(2)))`
`=((""^(a)mu_(g))/(""^(a)mu_(I))-1)(1/(R_(1))-1/(R_(2)))`
`1/(f_(1))=(1.5/1.6-1)(1/20-1/(-30))`
`=((1.5-1.6)/1.6)(1/20+1/30)`
`=-0.1/1.6((3+2)/60)`
`1/(f_(1))=-0.1/1.6xx1/12=1/192`
`impliesf_(1)=-192cm`
When convex lens is dipped in liquid of a refractive index 1.6, then it behaves as a concave lens.