Along a ray, the <radiative transfer equation> has the <formal solution of the radiative transfer equation>
$$
I_{\nu,i}=I_{\nu,b}e^{-\tau_{\nu,i}/\mu}
+B_\nu(T_i)(1-e^{-\tau_{\nu,i}/\mu})
$$
for each isothermal region in <local thermodynamic equilibrium>, neglecting scattering. For a self-luminous atmosphere with no incident intensity from below at the relevant photosphere, and in the optically thick limit, $I_{\nu,i}=B_\nu(T_i)$.

The observed <radiative flux> is the projected-disc integral
$$
F_{\nu,p}=\frac{2\pi R_p^2}{d^2}
\int_0^{\pi/2}I_\nu(\theta)\cos\theta\sin\theta\,d\theta.
$$
The inner region occupies projected fraction $f=\sin^2\theta_T$, hence
$$
\boxed{
F_{\nu,p}=\frac{\pi R_p^2}{d^2}
\left[
B_\nu(T_1)\sin^2\theta_T
+B_\nu(T_2)\cos^2\theta_T
\right]}.
$$
For finite optical depths, each <Planck law> function in this formula is replaced by its corresponding emergent intensity above.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 1 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-315/1/b/i/solution