A telephoton lens consists of two lenses, the front converging lens and the rear diverging lens with optical power `Phi_(1) = + 10D` and `Phi_(2) = -10D`. Find:
(a) the focal length and the positions of principle axes of that system if the lenses are seperated by a distance `d = 4.0 cm`, (b) the distance `d` between the lenses at which the ratio of a focal length `f` of the system to a distance `l` between the converging lens and the rear principle focal points is the highest. what is this ratio equal to ?

Explanation

(a) Optical power of the system of combination of two lenses,
`Phi = Phi_(1) + Phi_(2) -d Phi_(1)Phi_(2)`
on putting the values,
`Phi = 4D`
or, `f = (1)/(Phi) = 25 cm`
Now, the positions of primary principle plane with respect to the vertex of converging lens,
`X = (d Phi_(2))/(Phi) = 10 cm`
Similarly, the disatnce of secondarly principle plane with respect to the vertex of diverging lens.
`'X' =-(d Phi_(1))/(Phi) =- 10 cm`, i.e. `10 cm` left to it.
(b) The distance between the rear principle focal point `F'` and the vertex of converging lens,
`l = d + ((l)/(Phi)) (-d Phi_(1)) = (Phi d)/(Phi) + ((-dPhi_(1))/(Phi))`
and `f//l = ((l)/(Phi))//(Phid)/(Phi) - (d Phi)/(Phi)`, as `f = (1)/(Phi)`
`=1//d (Phi_(1) + Phi_(2) - d Phi_(1) Phi_(2)) - d Phi_(1) = 1//d Phi_(2) - d^(2) Phi_(1) Phi_(2)`
Now, if `f//l` is maximum fro certain value of `d` then `l//f` will be minimum for the same value of `d`. And for minimum `l//f`,
`d(l//f) d d= Phi_(2) - 2d Phi_(1)Phi_(2) = 0`
or, `d = Phi_(2)//2Phi_(1) Phi_(2)`
or, `d = 1//2 Phi_(1) = 5 cm`
So, the required maximum ratio of `f//l = 4//3`.