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Pol Grasland-Mongrain edited The_absorption_of_the_laser__.tex
over 8 years ago
Commit id: 033d3e74b0d87c1eec539a01ed21a32a04d6386c
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The absorption of the laser beam by the medium gives then rise to an absorbed optical energy $\gamma I$. Assuming that all the optical energy is converted to heat, a local increase of temperature appears. Temperature distribution $T$ in absence of convection and of phase transition can be computed using the heat equation:
\begin{equation}
\rho C \frac{\partial T}{\partial t} = \kappa \nabla ^2 T +
\frac{\gamma I}{\rho C} \gamma I
\label{eq:eqChaleur}
\end{equation}
where $\rho$ is the density, $C$ the heat capacity and $\kappa$ the thermal diffusivity. The thermal diffusion path, equal to $\sqrt{4\kappa \tau}$, with $\tau$ = 10 ns the laser emission duration and $\kappa$ = 1.43 10$^{-7}$ m$^2$.s$^{-1}$ for water \cite{Blumm_2003}, is approximately equal here to 80 nm. As $\gamma^{-1} \gg \sqrt{4\kappa t}$, propagation of heat is negligible during laser emission, so that equation \ref{eq:eqChaleur} can be simplified as: