Leif,
Thanks for the response. Please let me go a bit further into my thinking, which I hope isn't considered being argumentative.
When I wrote the equation I was thinking in vector/matrix terms, since lateral CA is not a single quantity, but rather a non-linear function of field angle. I believe this function is symmetric around the principle axis, but it has a different shape for each lens design.
An aerial image produced by the objective appears at its image plane inside the binocular, and contains lateral CA (including any contributions by the prism). This image is input to the eyepiece, which magnifies the aerial image for projection onto the retina. Along with that, the CA vector is also magnified by m, as noted by Dorian in an earlier post, i.e., m*CA(obj)
Could not the design of the eyepiece, which involves many lens elements, attempt to make (by design optimization):
—CA(eyepiece) = m*CA(obj)
again, thinking of these as vectors? Note that the sign is opposite, and if the two non-linear functions were matched the CA of the eyepiece would exactly cancel that of the objective. If the two functions were not matched exactly, but of opposite sign, then only the residual would be seen at the eye.
Continuing merrily along this happy line of thought, and assuming the eyepiece did correct some/all of CA(obj), would not the information lost at the image plane of the objective also be recovered in the final image at the retina?
Ed
