The internal surface of the walls of a sphere is specular. The radius of the sphere is `R=36cm.` A point source S is placed at a distance `R//2` from the cneter of the sphere and sends light to the remote part of the sphere. Where will the image of the source be after two successive reflections from the remote and then nearest wall of the sphere? How will the position of the image change if the source sends light to the nearest wall first?Consider paraxial rays.

Answer

Image will be at `-(5 R)/(6)` from near end towards souce ; image will be at - (R)/(2)` from remote end

Explanation

Case- I

Applying mirror formula for remote pair
`(u=-(3R)/(2),f=-(R)/(2))`
`rArr v=(uf)/(u-f)=(-3)/(4)R`
Now consider reflection from the nearer part for nearer part
`u=(3R)/(4)-2R=-(5R)/(4) rArr v=(-5R)/(6)`
Case II
Consider reflection form nearer part first
`u=-(R)/(2)`
`v= (uf)/(u-f)=(-3)/(4)R`
Now consider reflection from the nearer part for nearer part
`u=(3R)/(4)-2R =-(5R)/(4) rArr v=(-5R)/(6)`
Case II
Consider reflection from nearer part first
`u=-(R)/(2)`
`v=(uf)/(u-f)=infty`
Reflection from remote part gives `v= - (R)/(2)`
Image will be at `-(5R)/(6)` from near end towards source , image will be at `-(R)/(2)` from remote end.