Hello,
I am new here. I believe that the exit pupil is a plane where PARY (0,1) (0,0) is exactly zero (or a very small numerical error). However, in the simple example attached, I read EXPP = -25.26 µm. This difference from zero is bigger than a numerical error. Could you explain why the exit pupil is not exactly where is should be?
Thank you,
Olivier
+2Hi Olivier,
There are some subtle assumptions that OpticStudio makes under-the-hood when calculating entrance and exit pupils. The Entrance Pupil is indeed a paraxial calculation that only looks at the base radius of curvature, the conic constant and for some surfaces, the A2 even aspheric coefficient. You can manually calculate the ENPP location by using the simple ABCD matrix ray trace from Geometric Optics.
The Exit Pupil on the other hand uses 2 real parabasal rays (rays extremely close to the optical axis) in the +y and -y direction. These rays are traced from near the center of the Stop and their closest approach to each other will be the Exit Pupil. When you have a Hy of 0.1 for your REAY operand, this is way too large of field height for the parabasal rays. OpticStudio uses a value of around 1e-3 for the parabasal rays (I can’t remember if this is normalized field height or direct ray tracing).
So ENPP uses paraxial rays and the EXPP uses real rays. One other thing to note is the Chief Ray Height (C) solve uses paraxial rays while the Pupil Position (U) uses real rays.
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