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Refractive index and absorption

$\displaystyle E(z,t)=Re(E_0 e^{j(-kz+\omega t)})= Re(E_0 e^{j(\frac{-2 \pi (n+j...
...i\kappa z}{\lambda}}Re(E_0 e^{\frac{j(-2 \pi (n+j\kappa)}{\lambda}z +\omega t})$ (104)

And because the intensity is proportional to the square of the electric field the absorption coefficient becomes

$\displaystyle e^{-\alpha x}=e^{\frac{2\pi\kappa z}{\lambda}}$ (105)

$\displaystyle \alpha=-\frac{4\pi\kappa}{\lambda_0}$ (106)



rod 2017-12-08