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This is quite higher than metals where the radiation is absorbed within a few nanometres. The absorption of the laser beam by the medium gives then rise to an absorbed optical energy $q = \gamma I$. 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 heat equation:  \begin{equation}  \frac{\partial T}{\partial t} = \kappa \nabla ^2 T + \frac{q}{\rho C}  %\label{eq:eqChaleur} \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 $\tau$ = 20 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 0.1 $\mu$m. 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 $\frac{\partial T}{\partial t} = \frac{q}{\rho C}$. 
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