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Boundary conditions

We now apply the electric and magnetic boundary conditions[1]

$\displaystyle \mathbf{n} \times (\mathbf{E_2}-\mathbf{E_1})=0$ (83)

$\displaystyle \mathbf{n} \times (\mathbf{H_2}-\mathbf{H_1})=0$ (84)

We let the interface be at z=0, which gives,

$\displaystyle (E_{2}^{+}+E_{2}^{-})-(E_{1}^{+}+E_{1}^{-})=0$ (85)


$\displaystyle \frac{k_1}{\omega \mu}(E_{2}^{+}-E_{2}^{-})-(E_{1}^{+}-E_{1}^{-})\frac{k_2}{\omega \mu}=0$ (86)

. The wavevector is given by

$\displaystyle k=\frac{2 \omega }{\lambda}=\frac{\omega n}{c}$ (87)

. We can therefore write the magnetic boundary condition as

$\displaystyle n_2 (E_{2}^{+}-E_{2}^{-}) - n_1 (E_{1}^{+}-E_{1}^{-})=0$ (88)

rod 2017-12-08