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

Table of contents
- Solution i

Solution

Let

τ_{1}

and

τ_{2}

be the top-down optical depths at

P_{1}

and

P_{2}

, and let

μ

be the outward direction cosine. The formal solution of the radiative transfer equation gives

I_{ν} (0, μ) = B_{ν} (T_{1}) (1 - e^{- τ_{1} / μ}) + B_{ν} (T_{2}) (e^{- τ_{1} / μ} - e^{- τ_{2} / μ}) + I_{ν, 0} e^{- τ_{2} / μ} .

For a semi-infinite lower layer in local thermodynamic equilibrium,

I_{ν, 0} = B_{ν} (T_{3})

.

E_{3} (x) = \int_{0}^{1} μ e^{- x / μ} d μ .

The emergent planetary surface flux is

F_{p, ν} = 2 π [B_{1} (\frac{1}{2} - E_{3} (τ_{1})) + B_{2} (E_{3} (τ_{1}) - E_{3} (τ_{2})) + B_{3} E_{3} (τ_{2})] .

Hence the band-centre planet-star ratio is

\frac{F _{p}}{F _{*}} = (\frac{R _{p}}{R _{*}})^{2} \frac{2 [ B _{1} ( 1/2 - E _{3} ( τ _{1} )) + B _{2} ( E _{3} ( τ _{1} ) - E _{3} ( τ _{2} )) + B _{3} E _{3} ( τ _{2} )]}{B _{ν} ( T _{*} )} .

If

T_{1} = T_{2} = T_{3} = T_{p}

, the weights telescope to

1/2

, so

F_{p, ν} = π B_{ν} (T_{p})

and the expression reduces to the blackbody thermal eclipse depth, about

72 ppm

for

T_{p} = 600 K

.

Solved by gpt-5.6-sol high.

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