Proving the definition of optical throughput Optical throughout is defined by SPIE as
$$\Gamma = \frac{EP\times EW}{S^2}$$
where $EP$ is pupil area, $EW$ is window area, and $S$ is the separation between $EP$ and $EW$. I have not been able to geometrically prove this relationship and was looking for help or examples. I measured the image in the article and did the math but it did not check out.
 A: Witness that the invariance of each invariant separately follows immediately from the $\propto \,S^2$ scaling of the area of the entrance / exit window with the distance $S$. The invariance of each is really a restatement of this scaling law.
So it now remains to prove the equality of the two potentially different invariants i.e. the one calculated for the entrance pupil as opposed to the other calculated for the exit pupil. The scaling constants above are given by:
$$\frac{EW}{S^2}=\pi\,NA_i^2\propto \frac{1}{I_i}$$
$$\frac{EW^\prime}{{S^\prime}^2}=\pi\,NA_o^2\propto \frac{1}{I_o}$$
are simply the squared numerical apertures of input and output (modulo the $\pi$ scaling) and inversely proportional to the light intensities at the respective pupils in a lossless system, with proportionality constant for both inverse propotionalities. In a lossy (absorbing) system, replace light intensities with ray densities so that the argument works when the optics absorb / scatter too.
Now write down a statement of conservation of energy, equating light power (not optical power) through both pupils, using the above relationships to simplify the ratio of intensities / ray densities:
$$1=\frac{I_i\,EP}{I_o\,EP^\prime} = \frac{S^2\,{EW^\prime}^2\,EP}{{S^\prime}^2\,EW^2\,EP^\prime}$$
and there you have it.
As for your other questions: the "windows" are any planes orthogonal to the optical axis. Their areas, however, are given by the areas of the intersections between the chosen plane and the respective beam. If you take heed of this point, you can see that the ratio of the window area to the solid area subtended at the source is constant, as I talked about above. To make this definition work, you may need to extend ray bundles by linear interpolation beyond the extent of the beam (e.g. in the case of a virtual focus). So, because you need to take the window areas to be the area of the intersection between the beam and the transverse plane in question, this automatically answers your question on whether the full field of view is required. Yes, it is whatever the system sets the intersection are to be.
