A telephoton system consists of two glass balls with radii `R_(1) = 5.0 cm` and `R_(2) = 1.0 cm`. What are the distance between the centres of the balls and the magnification of the system if the bigger ball serves as no objective ?

Explanation

Let the optical powers of the first and second surfcaes of the ball of radius `R_(1)` be `Phi'_(1)` and `Phi''_(1)`, then
`Phi'_(1) = (n -1)//R_(1)` and `Phi''_(1) = (1 - n)// (-R_(1)) = ((n - 1))/(R_(1))`
This ball may be treated as a thick spherical lens of thickness `2R_(1)`. So the optical power of this sphere is,
`Phi = phi'_(1) - (2R_(1)Phi'_(1)Phi''_(1))/(n) = 2(n - 1)//nR_(1).........(1)`
Similarly, the optical power of second ball,
`Phi_(2) = 2(n - 1)//nR_(2)`
If the distance between the cnetres of these balls be `d`. Then the optical power of whole system,
`Phi = Phi_(1) + Phi_(2) - dPhi_(1)Phi_(2)`
`= (2(n - 1))/(nR_(1)) + (2(n - 1))/(nR_(2)) - (4d(n - 1)^(2))/(n^(2)R_(1)R_(2))`
`= (2(n - 1))/(nR_(1)R_(2)) [(R_(1) + R_(2)) -(2d(n - 1))/(n)]`
Now, since this system serves as telescope, teh optical power of the system must be equal to zero.
`(R_(1) + R_(2)) = (2d(n - 1))/(n)`, as `(2(n - 1))/(nR_(1)R_(2)) = 0`.
or, `d = (n(R_(1) + R_(2)))/(2(n - 1)) = 9cm`.
Since the diameter `D` of the objective is `2R_(1)` and that o fthe eye-piece is `d = 2R_(2)` So, the magnification,
`T = D//d = (2R_(1))/(2R_(2)) = R_(1)R_(2) = 5`.