The <cosmological visibility function> is the probability density for last scattering, so the stipulated two instantaneous populations give
$$
\boxed{g(\eta)=p\,\delta_D(\eta-\eta_*)
+(1-p)\delta_D(\eta-\eta_1)}.
$$
Since $g=\partial_\eta e^{-\tau}$ and $e^{-\tau}$ vanishes before the first transparent epoch,
$$
\boxed{
e^{-\tau(\eta)}=
\begin{cases}
0,&\eta<\eta_*,\\
p,&\eta_*<\eta<\eta_1,\\
1,&\eta>\eta_1.
\end{cases}}
$$
Its plot is a nondecreasing step function: it jumps from zero to $p$ at recombination and from $p$ to one at $\eta_1$. The <cosmological optical depth> interpretation is that $e^{-\tau(\eta)}$ is the probability that a photon present at time $\eta$ reaches the observer without any later scattering.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 312 / 4 / ii / Solution

{title2=$\tau(\eta)$}

The optical depth from conformal time $\eta$ to observation at $\eta_0$ is
$$
\tau(\eta)=\int_\eta^{\eta_0}\Gamma(\eta')\,d\eta'.
$$
The factor $e^{-\tau(\eta)}$ is the probability that a photon travels from $\eta$ to the observer without another scattering.


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