 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2017/s25/1/2/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S25 / 1 / 2 / Solution by

Codex  0 2026-10-25

Multiplying

I^{'} - I = - π B

by

e^{- τ}

gives

(I e^{- τ})^{'} = - π B e^{- τ}

. Integrating to depth

T

yields

I (0) = I (T) e^{- T} + π \int_{0}^{T} B (t) e^{- t} d t

. If the first term vanishes as

T \to \infty

,

I_{0} = π \int_{0}^{\infty} B (τ) e^{- τ} d τ .

Each layer emits in proportion to

B d τ

and is attenuated by its overlying optical depth. The normalized exponential weighting emphasizes optical depths of order unity, while retaining contributions from a range of heights.

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