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The internal surface of the walls of a s...

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

Text Solution

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The correct Answer is:
Image will be at `-(5 R)/(6)` from near end towards souce ; image will be at - (R)/(2)` from remote end
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.
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