A hollow sphere of glass of refractive index `mu` has a small mark on its interior surface which is observed from a point outside the sphere on the side opposite the centre. The inner cavity is concentric with external surface and the thickness of the glass is everywhere equal to the radius of the inner surface. Prove that the mark will appear nearer than it really is, by a distance `((mu-1)R)/(3mu-1) ,` where R is the radius of the inner surface.

Answer

`(n-1) R//(3n-1)`

Explanation

For lst refraction
`(n)/(v_(1))-(1)/(-2R)=(n-1)/(-R) ………….(i)`
For llnd refraction
`(1)/(v_(2))-(n)/(v_(1)-R)=(1-n)/(-2R)……….(ii)`
From `(i)` and `(ii)`
`v_(2)=(2R(4n-1))/(1-3n)`
so shift `=|v_(2)-(-3R)|`
`=3R+2((4n-1)R)/(1-3n)=((n-1)R)/((3n-1))`