# Keplerian Telescope Exit Pupil Location - Whats the Basis for It's Formula?

For a simple two lens Keplerian telescope, this is the formula for the location of the exit pupil:

$$z'=\frac{f_2}{f_1}(f_1+f_2)$$

Where $z'$ is the distance to the exit pupil location (i.e. eye relief), $f_1$ is the focal length of the objective, and $f_2$ is the focal length of the eye peice lens.

I've found the equation, in one incarnation or another, in the following three sources:

• Handbook of Optics, Third Edition, Volume 1
• Handbook of Optical Design, Third Edition
• an opti202L lab manual of unknown origins

I've done some work on the equation and it appears to be applying the thin lens equation to the eye piece lens:

$$\frac{1}{f_2}=\frac{1}{z'}-\frac{1}{z}$$

Where $z=-(f_1+f_2)$. This is the part I dont understand. The only way I've seen the thin lens equations used for systems of lenses, $z$ was treated as the distance to the previous lens' image formation point. This, however, suggests that the rear eye peice is looking at the objective lens itself and not the image formed by the objective.

How does this work and whats the rationale for using it to find the exit pupil location?