Analysis of Power Absorption
Let incident intensity be $I_0$. The temperature difference $\Delta T$ is proportional to the power absorbed $P$ (assuming Newton’s law of cooling for small differences, $P \propto \Delta T$).
1. Left Surface (Side 1)
This side faces the incoming parallel beam directly. It absorbs light over its area $\pi r^2$.
$$ P_1 \propto I_0 (\pi r^2) $$2. Right Surface (Side 2)
This side receives light reflected from the mirror. We must determine which part of the mirror contributes to illuminating the back of the disc.
- The reflected rays converge to the focus $F$ at distance $R/2$ from the pole.
- The disc is at distance $R/4$ from the pole. By similar triangles (or linear geometry), the beam width at the disc ($x=R/4$) is exactly half the beam width at the mirror ($x=0$).
- Conversely, to hit the disc of radius $r$, the light must have reflected from a spot on the mirror of radius $2r$.
The Shadow Effect: The incident light traveling towards the center of the mirror is blocked by the disc itself! The disc casts a shadow of radius $r$ on the mirror (since the beam is parallel).
Therefore, the mirror reflects light only from the annular region between radius $r$ and radius $2r$.
$$ \text{Effective Area on Mirror} = \pi (2r)^2 – \pi (r)^2 $$ $$ \text{Effective Area} = 4\pi r^2 – \pi r^2 = 3\pi r^2 $$Because energy is conserved along the flux tubes, the power hitting the back of the disc is equal to the power incident on this annular region of the mirror.
$$ P_2 \propto I_0 (3\pi r^2) $$Conclusion
The ratio of temperature excesses corresponds to the ratio of absorbed powers:
$$ \frac{\Delta T_1}{\Delta T_2} = \frac{P_1}{P_2} = \frac{I_0 \pi r^2}{I_0 (3\pi r^2)} $$