Telescope in a box Let’s have a box divided in 2 equal parts A and B by a wall. The wall is pierced in its center by a hole of area “s”. The normal to the hole center is the axis of a 2 mirrors telescope located in A. The primary mirror has area “S>s” and is close to the wall. The secondary mirror sends the collected photons to B through the hole. 
It seems to me that at equilibrium, the number of photon in both part are such that: ph(B)/ph(A) = S/s; hence T(B) > T(A).
My question is: where is the catch?
PS: I am looking for a direct explanation, not a “reductio ad absurdum”.
 A: The basic catch is that an optics apparatus not only changes the cross-section of the light beam, it also changes the spread of ray angles (at a given point in the beam). Liouville's theorem implies that these are balanced (that the beam maintains a constant volume in the abstract phase space that is relevant for thermodynamics).
In this case, think of how a telescope with high magnification is going to have a narrow field of view. On the objective side of the telescope, where the aperture has a small area, light rays at many different angles will be admitted through the telescope. However, on the subjective side, the beam has a larger area but the rays are all more nearly parallel to one another. Only a very narrow range of ray angles will be admitted back through to the other side. Rays entering the larger aperture at some other angle do not reach the objective and instead, in this case, are reflected back into the same side of the box where they originated. Hence, it will turn out that the energy flows are in balance when the temperatures are equal (i.e. when both sides have the same surface brightness..).
You could construct the same apparent paradox by simplifying the telescope to just a cavity that has two entrances (or even simply contains two thermally-radiating objects) of different sizes, so long as you postulate the cavity interior wall absorbs no energy (either perfectly white or perfectly reflective).
