Pol Grasland-Mongrain edited Let_s_describe_the_phenomenon__.tex  over 8 years ago

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I(z)=(1-R) I_0 \exp(- \gamma z)  \label{eq:expontentialDecay}  \end{equation}  where $R$ is the reflection coefficient of the material (supposed negligible on a black mat material as the one used here) and $\gamma$ the absorption coefficient of the medium. The absorption coefficient $\gamma$ can be estimated in two ways: by measuring the fraction of light going through different thickness of the medium or by calculating the skin depth $\delta = \frac{2}{\gamma}$ (the factor 2 is due to the fact that $\delta$ is related to magnitude of the electrical field while $\gamma$ is related to the magnitude of the optical energy, which is the square of the electrical field magnitude). Skin depth $\delta$ is equal to $(\pi \sigma \mu_r \mu_0 \nu)^{\frac{1}{2}}$, where $\sigma$ is the electrical conductivity of the medium, $\mu_r \mu_0$ its permeability and $\nu$ the frequency of the radiation. Substituting $\sigma \approx$ 0.1 S.m$^{-1}$, $\mu_r \mu_0$ = 4 $\pi \times 10^{-7} H.m^{-1}$ and $\nu$ = 3 10$^8$ / 532 10$^{-9}$ = 5.6 10$^{14}$ Hz, the skin depth for our medium is about 70 $\mu$m: it means that about 63\% of the radiation is absorbed in the first 70 micrometers of the sample. We have validated experimentally this value by measuring the fraction of light which go through different thicknesses of the medium (respectively 0, 30, 50 and 100 $\mu$m) with a laser beam power measurement device (QE50LP-S-MB-D0 energy detector, Gentec, Qu\'ebec, QC, Canada). We found respective power of 100\%, 88\%, 71\% and 57\%. An exponential fit indicated that $\gamma^{-1} \approx$ 50 40  $\mu m$ in our sample, as illustrated in the little graph in Figure 1. 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 the heat equation:  \begin{equation}