A hollow sphere of glass of R.I .n has small mark M on its interior surface which is observed by an observer O from a point outside the sphere, .C is centre of the sphere. The inner cavity (air) is concentric with the external surface and thickness of the glass is every where equal to the radius of the inner surface. Find the distance by which the mark will appear nearer than it really is in terms of n and R assuming paraxial rays.

Answer

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

Explanation

`(n)/(v_(1)) + (1)/(2R) = (n-1)/(-R) implies v_(t) = -(2nR)/((2n-1))`

Now `u_(2) = R-V_(1) = R + (2nR)/((2n-1)) = ((4n-1))/((2n-1))R`
`(1)/(v_(2)) + (h)/(((4n-1)/(2n-1))R) = (1-n)/(-2R) implies v_(2) = -((4n-1)/(3n-1)) xx 2R`
Total shift = `3R - |v_(2)|`
=`3R - (4n-1)/(3n-1) xx 2R = ((n-1))/((3n-1))R`