Your reasoning about the geometry is **not quite correct**. The radiation which is focussed onto the right face of the disk does not come from a circle of radius $2r$ but from an **annulus** of inner radius $r$ and outer radius $2r$. The disk itself obscures the inner area of radius $r$ of the incident light from reaching the mirror, so the area of radiation reflected onto the right face is only $\pi(2r)^2-\pi r^2=3\pi r^2$.

You have assumed that $\Delta T \propto A$ for each side of the disk, where $A$ is its effective area which collects radiant power. But you have not given any justification for this assumption. I think that you might have to consider each face as a perfect absorber-emitter (ie a black body) and use **Stephan's Law**, which relates power and temperature.

Perhaps you also need to bring the ambient temperature of the surroundings into your calculation, or demonstrate why it is not relevant.

There may be a complication : some radiation emitted by the right face of the disk is reflected back onto the same side of the disk. This does not happen for the left side of the disk, and adds to the total radiation power received by the right side. However, perhaps we are expected to assume that the mirror is highly reflecting only in a narrow band of wavelengths centred on the wavelength of the incident radiation. The disk radiates as a black body at all wavelengths, so only a very small fraction of this radiation would be reflected by the mirror. So this effect could be assumed to be negligible.

The radiant power $P$ emitted by a surface is related to its temperature $T$ by **Stephan's Law** : $$P=\sigma \epsilon AT^4$$ where $\sigma$ is Stephan's constant, $\epsilon$ is the emissivity of the surface ($\epsilon=1$ for a perfect black body emitter) and $A$ is the area. When power is absorbed from the surroundings at temperature $T$ we use the same equation except that **absorptivity** $\alpha$ replaces $\epsilon$. At thermal equilibrium, as occurs here, $\alpha=\epsilon$.

At equilibrium temperature the emitted power $P$ is also the total power received by the surface - in this case the sum of power from the source of parallel light of intensity $I$ and from the surroundings, which are at temperature $T_0$.

For the two sides (left 1 and right 2) we have $$P_1=\sigma \epsilon AT_1^4=IA+\sigma\epsilon AT_0^4$$ $$P_2=\sigma \epsilon AT_2^4=3IA+\sigma \epsilon AT_0^4$$ because (as shown above) the right side receives $3\times$ as much radiation as the left side.

Multiplying the 1st equation by 3 and subtracting the 2nd we get $$3T_1^4-T_2^4=2T_0^4$$ $$3(\frac{T_1}{T_0})^4-(\frac{T_2}{T_0})^4=2$$

In order to proceed we must assume that the faces of the disk are not much hotter than the surroundings - ie that $\delta T_1=T_1-T_0$ and $ \delta T_2= T_2-T_0$ are both very much smaller than $T_0$. This assumption is perhaps implied by the notation $\delta T$.

Then neglecting terms higher than the 1st power in $\frac{\delta T}{T_0}$ we get $$2=3(\frac{T_1}{T_0})^4-(\frac{T_2}{T_0})^4\approx 3(1+4\frac{\delta T_1}{T_0})-(1+4\frac{\delta T_2}{T_0})$$ $$\frac{\delta T_1}{\delta T_2}\approx \frac13$$

edited Oct 2 by sammy gerbil