How to find out the temperature of phonons using the Raman anti-Stokes and Stoke intensities ratio?

Question

Basically the title, the concept is simple, but I`m having a hard time trying to9 find an explicit solution.

Answer 1

The Stokes signal intensity $I_S \propto n+1$, while the anti-Stokes intensity $I_A \propto n$, where $n$ is the phonon population following the Bose-Einstein distribution, and the constants of proportionality are the same. $$ n = \frac{1}{e^{\epsilon/kT}-1} $$ Here, $\epsilon$ is the energy of the phonon (= $h c$ times measured wavenumber), and $k$ is the Boltzmann constant. I will leave it as an exercise for the reader to determine $T$ from $I_S/I_A$.