It has been claimed from time to time that the image in each half of the linear BV can be brighter than the image in each half of a conventional BV. Obviously the light in either type of BV is split between the two sides. It is also the case that the conventional BV has two half-intensity full-circles where as the linear BV has two full-intensity half-circles for the exit pupil. I can think of one way in which the linear BV can produce an image that gets around twice the photons. Please correct any mis-statements of what I say below to try to explain how this could occur.
Consider the black circles below as the telescope/eyepiece exit-pupil and the red circles as the entrance pupil of the eye. I'm looking at a case where the eye is not dilating wide enough to take full advantage of the telescope/eyepiece exit pupil. (forgive any non-standard nomenclature).
Relative to the single eyepiece, half the light would be going into each eye for the case of the conventional BV.
But for the case of linear BV, if the same-size entrance pupil of each eye could be positioned to be in the full intensity illuminated half of the half-pupil, then each eye is getting the full intensity light - thus brighter. Another way of thinking about it would be that in this case more light is wasted (not getting into the eye) for the case of the conventional BV as compared to the linear BV.
It could be that the tables could be turned with different pupil sizes relative to the telescope/eyepiece pupil.
I have both a linear BV as well as a Williams Optics conventional BV but I don't have the necessary eyepieces to get the exact same magnification in each set-up to do my own experiment. An experiment could be done with a manual camera but I'm not yet set-up for that.
So please correct any flaws in my reasoning above. I'm curious.
(Note that one sees the full image in each eye of the linear when the eye is positioned at the correct distance (like looking through a non-circular keyhole).