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Optical model

On the left of the interface the electric field is given by

$\displaystyle E_{1}=E^{+}_{1} e^{-j k_1 z}+E^{-}_{1} e^{j k_1 z}$ (75)

and on the right hand side of the interface the electric field is given by

$\displaystyle E_{2}=E^{+}_{2} e^{-j k_2 z}+E^{-}_{2} e^{j k_2 z}$ (76)

Maxwel's equations give us the relationship between the electric and magnetic fields for a plane wave.

$\displaystyle \nabla \times E=-j\omega \mu H$ (77)

which simplifies to:

$\displaystyle \frac{\partial E} {\partial z}=-j\omega \mu H$ (78)

Applying equation 78 to equations 75-76, we can get the magnetic field on the left of the interface

$\displaystyle -j \mu \omega H^{y}_{1}=-j k_1 E^{+}_{1} e^{-j k_1 z}+j k_1 E^{-}_{1} e^{j k_1 z}$ (79)

and on the right of the interface

$\displaystyle -j \mu \omega H^{y}_{2}=-j k_2 E^{+}_{2} e^{-j k_2 z}+j k_2 E^{-}_{2} e^{j k_2 z}.$ (80)

Tidying up gives,

$\displaystyle H^{y}_{1}=\frac{k}{\omega \mu}E^{+}_{1} e^{-j k_1 z}-\frac{k}{\omega \mu} E^{-}_{1} e^{j k_1 z}$ (81)

$\displaystyle H^{y}_{2}=\frac{k}{\omega \mu}E^{+}_{2} e^{-j k_2 z}-\frac{k}{\omega \mu} E^{-}_{2} e^{j k_2 z}$ (82)



Subsections
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Next: Boundary conditions Up: The physical model Previous: Energy balance - hydrodynamic
rod 2017-12-08