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: 


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*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?
 A: The exit pupil of any optical system is defined as the image of the controlling aperture stop (AS) formed by the subsequent optical components. The controlling aperture stop is the optical component which limits the maximum cone of rays from an object point which can actually be processed by the whole system (basically it controls how much light from each point actually makes it all the way through). In a system like a camera we deliberately set the AS by having an iris which opens and closes, but in the absence of an iris one of the lenses limits the maximum angle. In the case of a simple telescope this is generally the main front lens, and the exit pupil is its image formed by the eyepiece lens as your formula states. Though if you put too small an eyepiece lens then this could be the AS itself in which case it will also be the exit pupil. 
In general the exit pupil is designed to form at the position where the eye will be placed and further that it will be the same size as the pupil of the eye itself, this ensures that the maximum possible amount of light can be transmitted from the object to the viewer and that the image fills the eye. The exit pupil limits the cone angle of rays illuminating each point of the image.
The aperture stop should not be confused with the field stop which controls the field of view of the system. This aperture limits the maximum angular distance an object can be from the optical axis and still be seen. Its image in the subsequent components is called the Exit Window and effectively defines the total size of the final image. To an observer on the image side this window appears to limit the area of the of the image.
