Interesting question! I always enjoying trying to see where the second law is hiding.
Rather than considering the details of the CPCs, I'm going to steel man the concept and consider a perfect optical element which conserves optical etendue.
Energy balance from the perspective of plate 1
At best, etendue ($\epsilon$) of heat fluxes will be conserved between the plates,
$$ \epsilon_1 = \epsilon_2 \ A_1 n_1^2 \sin(\theta_1) = A_2 n_2^2 \sin(\theta_2) $$
Note, for the actual configuration the optical elements are not perfect, $\epsilon_1 > \epsilon_2$. So actual system will not reach this limit.
Top plate has subscript 1 and bottom subscript 2.
The optical energy balance between the plates is the etendue multiplied by the power emitted by the blackbody (<PERSON>–Boltzmann law).
For plate 1,
$$ \epsilon_1 \sigma T_1^4 $$
For plate 2, $$ \epsilon_2 \sigma T_2^4 $$
At equilibrium, the rates of energy exchanged between the plates must be equal to each other, because otherwise temperatures will carry on increasing forever — you have a perpetual motion machine.
As you already said the emission angles are $\theta_1 = \theta_2 = \pi/2$, which leads to the condition that $\sin(\theta_1) = \sin(\theta_2) = 1$.
From the perspective of plate 1 the heat is exchanged in vacuum so $n_1 = n_2 = 1$, and by definition this exchange occurs over the surface area of plate 1, therefore, the energy balance is,
$$ 0 = -A_1 \sigma T_1^4 + A_1 \sigma T_2^4 $$
The first term is the energy emitted by plate 1 and the seconds term is the energy absorbed by plate 1.
This leads directly to,
$$T_1 = T_2$$
So if etendue is conserved (the ideal case) the plates will have equal temperatures.
In practice any optical element introduces loses so,
$$ A_1 n_1^2 \sin(\theta_1) > A_2 n_2^2 \sin(\theta_2) $$
leading to
$$T_1 > T_2$$
Energy balance from the perspective of plate 2
Now let's apply energy balance from the perspective of plate 2.
Plate 2 absorbs the heat flux from plate 1,
$$ A_2 n_2^2 \sin^2(\theta_c) \sigma T_1^4 $$
The critical angles appears here for the reasons shown below: rays hitting the optical element over all external angles get transformed to rays with internal angles which cover the acceptance/escape cone. This is just the application of Snell's Law.
The heat flux emitted by plate 2 which can escape and be absorbed by plate 1 must be restricted to angles inside the escape cone,
$$ A_2 n_2^2 \sin^2(\theta_c) \sigma T_2^4 $$
Light that is emitted outside the escape cone is totally internally reflected and reabsorbed by plate 2.
I think this is where a lot of people get struck. They don't realise that their idea has introduced an irreversibility in the direction of travel of light: there are no optical diodes.
Simplifying the above using,
$$ \sin^2(\theta_c) = \frac{1}{n_2^2} $$
The energy balance from the perspective of plate 2 is, therefore,
$$ 0 = A_2 \sigma T_1^4 - A_2 \sigma T_2^4 $$
Again, this leads to the same results as before,
$$ T_1 = T_2 $$
for the ideal case and
$$ T_1 > T_2 $$
when etendue is not conserved and losses are introduced. 749