A parallel beam of light strikes the first surface of a glass sphere of R.I 1.5 and radius of curvature 0.10 m. Find the position of the final image.

Explanation

`u=oo`
`n_(g)=1.5`
`n_(a)=1`
`r=0.10m`
(i) for I surface using the formula
`(n_(0))/(-u)+(n_(1))/(v)=(n_(0)-n_(1))/(r)`
`(1)/(oo) +(1.5)/(v)=(1.5-1)/(0.10)=5`
`therefore v=(1.5)/(5)=0.3m` from `p_(1)`
The image will serve as a virtual object for II surface.
Distance of I. from `P_(2)=(0.3-0.2)=0.10m`
`therefore (1.5)/(-0.10)+(1)/(v)=(1.5-1)/(0.10)=5`
`(1)/(v) - =5+15=20`
or `v=(1)/(20)=0.05m ` from `P_(2)`
`therefore ` The final image will be at 0.05 m from `P_(2)`.