An equibiconvex lens of radius of curvature 0.20 and refractive index 1.5 immersed half inside water of RI `4//3` and the rest outside in air. A parallel beam of light in air is incident on it. Find the final position of the image.

Explanation

Given `r=0.20m `
`n_(w)=(4)/(3)=1.333`
`n_(g)=1.50`
`u=oo`

For reflection through I surface
`n_(0)=1, n_(1)=1.50, u=oo, v=?`
By using `(n_(0))/(-u)+(n_(1))/(v)=(n_(0)-n_(1))/(r)`
i.e., `(1)/(oo)+(1.5)/(v)=(1.5-1)/(0.2)`
`(1.5)/(v)=(0.5)/(0.2)=2.5`
`therefore v=(1.5)/(2.5)=0.6m`
Real image serves as virtual object for II surface. Put `u.=0.6m`
`(n_(g))/(-0.6)+(n_(w))/(v)=(n_(g)-n_(w))/(r)`
`therefore (1.5)/(-0.6)+(1.333)/(v)=(1.5-1.333)/(0.2)`
`-2.5+(1.333)/(v)=(0.1667)/(0.2)=0.8335`
`(1.333)/(v)=3.333`
`v=(1.333)/(3.333)=0.3999`
`v=0.40m` from the second surface in the direction of incident light.