Can a telescope ever increase the apparent luminance of an extended object? From what I know about common telescope designs, telescopes don't increase the apparent luminance of extended objects compared to the luminance seen with the naked eye. In this sense extended objects don't appear &quot;brighter&quot; (per unit solid angle of object/image), although the total light flux received from the object (the illuminance) can increase due to the increased magnification (the object looks bigger through the telescope). The best that can be done is keep the luminance the same, which requires eliminating transmission losses. In addition to transmission losses, the light received by the eye per unit solid angle of image is further reduced at high magnifications when the exit pupil is smaller than the viewer's pupil.
Can there be no telescope design (purely optical, e.g. without using electronic eyepieces) that makes extended objects appear &quot;brighter&quot; (in the above sense) than with the naked eye, by somehow overcoming the limitation of the observer's pupil? If so, can this be proven? If not, what would such a design look like?
Can there be no telescope design... that makes extended objects appear &quot;brighter&quot;... than with the naked eye, by somehow overcoming the limitation of the observer's pupil?

I've left out all of the parenthetical qualifications and I'll answer this, let's see if it gets at the heart of the question.
tl;dr: No, because etendue; the same reason that a wall doesn't get brighter when we walk towards it and why we can't go outside with a magnifying glass and concentrate blue sky.


I thought I'd leave the other answer in place for contrast and as a teaching moment, but the down voters had a go at it, so now you can't see it unless you have 10k reputation.
I'd claimed that a pair of 7x50 binoculars would make the object 49 times larger in solid angle but collect $(50/6)^2$ or about 69 times more light.

If my fully dark-adapted pupil is 6mm in diameter then the aperture is 8.3 times larger in diameter than my pupil, but the image is only 7 times larger. We square the ratio to get the ratio of surface brightnesses, so it will appear to be
$$\left( \frac{50/6}{7/1} \right)^2 \approx 1.42$$

However the OP pointed out in a comment that this would produce an exit pupil larger than the 6 mm entrance pupil of the eye.

Thanks for the answer. I'm not sure about the conclusion though. With 7x50 binoculars, the exit pupil is 7.14 mm in diameter, meaning not all of the incident light falls on the retina since the eye pupil is smaller. Specifically, the fraction of of light that does enter the eye is $(6/7.14)^2=0.705$, which is precisely the reciprocal of the factor 1.42 that you calculated. So in the absence of transmission losses I think the luminance remains the same as that seen with the naked eye.

That was an Aha! moment, nature is smart, or at least I'm not.
I replied:

omg I think that I have I failed to recognize something as fundamental as conservation of etendue. Now it looks like my answer is wrong. :-( ...

This failed for the same reason that a wall doesn't get brighter when we walk towards it and why we can't go outside with a magnifying glass and concentrate blue sky on a sheet of paper. In classical mechanics the analogy is conservation of phase space and Liouville's theorem