\question
{Shotaro Maruyama}       
{2/3/12}          
{shotaro.maruyama@cmpt.phys.tohoku.ac.jp}  
{5-5}
{The Poynting vector, $\vec{S}=\vec{E} \times \vec{H}$ is the power density per unit area of the electromagnetic field. In a typical micro-Raman measurement system, the diameter of the light beam is about 1$\rm{\mu m}$ and the laser power is 1$\rm{mW}$. Estimate the power density of this micro-Raman setup and caliculate $\vec{E}$. Show that the electric field thus obtain is not strong enough to be in the non-liniear regime.}
{
Using $|\vec{B}|=|\vec{E}|/\rm{c}$, we get a formula for $|\vec{S}|$.
\begin{eqnarray}
|\vec{S}| &=& |\vec{E}| |\vec{H}|\,{\sin\frac{\pi}{2}}\nonumber\\
&=& \frac{1}{\mu_0} |\vec{E}| |\vec{B}|\nonumber\\
&=& \frac{1}{\mu_0 \rm{c}}|\vec{E}|^2,
\end{eqnarray}
where $\mu_0$ is absolute permeability of vacuum. We also get the following equation from the numerical value given by Q5-5 as follows:
\begin{eqnarray}
|\vec{S}|=\frac{10^{-3}}{\pi (0.5\times10^{-6})^2} \rm{Wm}^{-2}.
\end{eqnarray}
Using Eqs.\,(5-5-1) and (5-5-2), we obtain the magnitude of $\vec{E}$ as follows:
\begin{eqnarray}
|\vec{E}|^2 &=& \mu_0 \rm{c} \times \frac{10^{-3}}{\pi (0.5\times10^{-6})^2}\nonumber\\
&=& (4\pi \times 10^{-7})\times \rm{c} \times \frac{10^{-3}}{\pi (0.5\times10^{-6})^2}\nonumber\\
&=& 4.78 \times 10^{11}\,\,{\rm V^{2}m^{-2} }.
\end{eqnarray}
Using Eq.\,(5-5-3), we get  
\begin{eqnarray}
|\vec{E}| &=& 6.9\times 10^{5}\,\,{\rm Vm^{-1}},
\end{eqnarray} 
where $\mu_0 = 4\pi \times 10^{-7}\,{\rm N/A^{2}}$,\,\,$\rm{c}=2.99\,{\rm m/s}$. In the previous problem Q5-4,  we show the numerical value where we can neglect the non-linear term of $\vec{E}$. According to the result, the electric field is not strong enough to be in the non-linear term when the magnitude of $\vec{E}$ is smaller than $10^{9}\,{\rm Vm^{-1}}$. The numerical value in Eq.\,(5-5-3) satisfies the condition and we can neglect the higher term of $|\vec{E}|$.
}