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

Optical depth increases downward because $d\tau =-k\rho dz$. Along an upward ray the beam loses irradiance by absorption, as in Beer-Lambert attenuation, and gains it from local thermal emission. Kirchhoff's law equates emissivity and absorptivity in local thermodynamic equilibrium, giving source term $\pi B$. Thus $-dI/d\tau =-I+\pi B$, equivalently the given equation. The factor $\pi$ converts isotropic radiance into hemispheric irradiance in the stipulated diffuse approximation. The following two limits hold for the same downward optical-depth coordinate.

With negligible atmospheric emission, $dI/d\tau =I$. If the warm surface is at ${\tau }_s$, $\boxed{I(\tau )=I_s{e}^{-({\tau }_s-\tau )}}$. Irradiance decreases as the beam travels upward to smaller $\tau$, and $I(0)=I_s{e}^{-{\tau }_s}$. The positive derivative with respect to downward optical depth is consistent with attenuation, not amplification along propagation.

For an infinitely deep isothermal atmosphere the bounded solution is $\boxed{I(\tau )=\pi B}$. A more general formal solution contains $C{e}^{\tau }$, excluded by the bounded deep boundary condition. Emission exactly replenishes absorption, so the emergent spectrum is blackbody in this idealized wavelength-by-wavelength thermal model.

Multiplying $I^{\prime }-I=-\pi B$ by $e^{-\tau }$ gives $(I{e}^{-\tau }{)}^{\prime }=-\pi B{e}^{-\tau }$. Integrating to depth $T$ yields $I(0)=I(T)e^{-T}+\pi {\int }_0^{T}B(t)e^{-t}dt$. If the first term vanishes as $T\to \infty$,Each layer emits in proportion to $Bd\tau$ 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.

Hydrostatic balance and the ideal gas law give $dp/dz=-pg/(R_{a}T)$, hence approximately $p=p_0{e}^{-z/H}$ with $H=R_{a}T/g$ for weak temperature variation. For well-mixed absorber abundance, ${\rho }_{\mathrm{abs}}\propto p/T$, while pressure broadening gives $k\propto p$. Thus $k{\rho }_{\mathrm{abs}}\propto e^{-2z/H}$ and its integral above height $z$ has the same scale. ThereforeThe factor two depends on both pressure dependence of absorption and constant mixing ratio; it is not the ordinary density scale height.

Let ${\tau }_t$ be the tropopause optical depth and linearize $B$ with positive derivative $B_T$. In the troposphere $T=T_t+\Gamma H_{\tau }\log (\tau /{\tau }_t)$, so the requested schematic isIt is constant in the isothermal stratosphere and rises linearly with $\log \tau$ below it. At unchanged temperatures and a multiplicative abundance change $a$, optical depth at each height changes to $a\tau$, hence $\boxed{B_a(\tau )=B_{\mathrm{old}}(\tau /a)}$. Doubling moves the tropopause kink to $2{\tau }_t$ and gives colder emission temperatures at a fixed optical depth; halving does the opposite.