= Single-scattering forward-plus-isotropic flux partition
Consider unit <energy> of incident <radiation> travelling downward into a conservatively scattering slab. Let $t$ be its unscattered transmission fraction. Of the fraction $1-t$ that scatters, suppose a fraction $a$ continues exactly forward and the remaining fraction $1-a$ has an <isotropic> outgoing distribution. Approximate every scattered contribution as escaping without a second interaction.
The isotropic part splits equally between upward and downward directions. The escape fractions are therefore
$$
f_\uparrow=\frac{(1-a)(1-t)}2,\qquad
f_\downarrow=t+\left[a+\frac{1-a}{2}\right](1-t).
$$
They sum to one, as required by <conservation of energy> without <electromagnetic absorption>. For example, $t=0.6$ and $a=0.5$ give $f_\uparrow=0.1$ and $f_\downarrow=0.9$. Exactly forward scattering does not remove downward energy in this model; treating all scattering as backward loss would give a different result.
For a specified direct path, <exponential attenuation> gives $t=e^{-\tau_s}$ with scattering <optical depth> $\tau_s$. The single-escape assumption becomes unreliable when repeated scattering is important. A thick slab generally requires the angular <scattering phase function> and the full <radiative transfer equation>, rather than this two-branch bookkeeping.
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