I just took this photo with a record-breaking absence of artistic value.
The binocular appears to increase the illumination, mostly at the exit pupil distance.
This
could indeed seem contradictory, but I guess that is because the distance to the lamp in the ceiling is so short. The illumination will decrease with the square of the distance. The binocular appears to shorten the distance with the inverted number of the magnification, hence concentrate and amplify the illumination to the paper.
I can only guess what would happen if the binoculars were directed towards the Sun. Decreasing the apparent distance with the inverted factor of the magnification would cause a dramatic increase of the illumination through the exit pupil. This is a special case, compared to customary exit pupil brightness cases.
A clear light source radiates light in all available directions, with strongly divergent rays close to it (I postulate it's a quite small one)
Farther away, a great share of the diverging beams don't go through where you stand, and if you measure them, their divergence has decreased and the illumination has decreased as well. With sufficiently long distances, the beams reaching you are almost perfectly parallel.
In every practical sense, the beams emanating from the Sun are parallel (the Sun is close enough not to appear point-shaped, but since it's so vastly bigger than the Earth, the beams coming from the small area on the Sun whose beams actually reach the Earth's surface, can safely be considered parallel).
The illumination power of a light source depends on its intensity and the distance to the surface being illuminated.
But the light source's surface brightness, assuming it's not point-shaped,is the same regardless of distance, just like with reflecting bodies like the Moon.
The same goes for any object reflecting light off its surface.
Now, do such objects with a fairly low reflection rate have an illumination power that possibly could increase through binoculars at terrestrial distances, and in particular
differ with magnification - like in my picture?
I would tend to say that the illuminating power is decided by the area alone. The photographic experiment confirmed there's no brightness difference.
Here's my (hopefully qualified) guess:
An overcast sky, as well as most reflective material found on Earth short of water, will scatter the light in all directions. When we look at such things, there's no clear vergence like from clear light sources. There's still another vergence related to the distance, coming from the object's form and structure, and parallax. That's what we pick up when we focus on it.
The scattered light with its absence of vergence cues will not make the object's apparent surface brightness change with distance or magnification.
Its only
illuminating power comes from its apparent area. So long as it is safely comparable with the adjacent image element's brightness, very little will happen when it enters the field of view.
All in all, I have tried to concentrate on terrestrial viewing throughout this particular discussion. Let me assure you that I learn the most myself and am still learning.
I dare not exclude the possibility that other physical and physiological phenomena decide brightness and detectability of faint astronomical objects with and without optical tools.
Recovered from the flu, going back to work tomorrow and I believe I might hear a quiet sigh of relief from the forum members
//L