Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 2 / a

At full phase immediately before an exoplanet secondary eclipse, reflected and thermal light giveThe phase function equals one at secondary eclipse. For the stated model, the first term is constant below $1\mu \mathrm{m}$ and negligible above it. The $600\mathrm{K}$ thermal component is tiny in the visible, rises through the infrared, and peaks near $4.8\mu \mathrm{m}$ in $B_{\lambda }$, while the contrast continues to improve toward the mid-infrared because the stellar spectrum falls more rapidly.

Solved by gpt-5.6-sol high.

Assume full heat redistribution, soFor $T_{\ast }=5778\mathrm{K}$, $T_{\mathrm{eq}}=600\mathrm{K}$, and Bond albedo $A_B=0.5$,Taking $R_p=R_{\mathrm{Nep}}=0.0354R_{\ast }$ givesThe reported $50\mathrm{ppm}$ visible eclipse would therefore implywhich is impossible. At least one assumption, the measurement, or the stated system parameters must fail; even $A_g=1$ gives only about $1.2\mathrm{ppm}$.

At $20\mu \mathrm{m}$, reflected light is negligible. The thermal eclipse depth isWith $T_p=600\mathrm{K}$ this is

Solved by gpt-5.6-sol high.

A perfectly reflecting Lambertian surface receiving normal flux $F$ has radiancebecause integrating $I\cos \theta$ over the outward hemisphere returns $F$. A face-on disc of radius $R$ subtends solid angle $\pi R^2/d^2$. The observed flux is consequently

Solved by gpt-5.6-sol high.