At $20\,\mu\mathrm m$,
$$
E_3(0.1)=0.4163,
\qquad
E_3(1)=0.1097.
$$
The normalized flux weights of the $400$, $600$, and $800\,\mathrm K$ layers are therefore
$$
2(1/2-E_3(0.1))=0.1674,
$$
$$
2(E_3(0.1)-E_3(1))=0.6132,
\qquad
2E_3(1)=0.2194.
$$
Using the <Planck law> at the band centre and $R_p/R_*=0.0354$ gives
$$
\boxed{F_p/F_*\simeq7.45\times10^{-5}simeq74.5\,\mathrm{ppm}}.
$$
This is only slightly larger than the $71.7\,\mathrm{ppm}$ isothermal $600\,\mathrm K$ value because the cool upper-layer suppression and hot deep-layer enhancement nearly cancel in this broad weighting.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 2 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-315/2/b/ii/solution