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Backwards propagating wave

Rearrange equation, 88 to give,

$\displaystyle E_{1}^{+}=E_{1}^{-} +\frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})$ (93)

Inserting in equation 85, gives

$\displaystyle E_{2}^{+}+E_{2}^{-}=E_{1}^{-} +\frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})+E_{1}^{-}$ (94)

$\displaystyle 2E_{1}^{-}=E_{2}^{+}+E_{2}^{-}- \frac{n_2}{n_1}(E_{2}^{+}-E_{2}^{-})$ (95)

$\displaystyle 2E_{1}^{-}\frac{n_1}{n_1+n_2}=E_{2}^{+}\frac{n_1-n_2}{n_1+n_2}+E_{2}^{-}$ (96)

Which is the same result as obtained in [2].

These equations become:

$\displaystyle E_{1}^{-}t_{12}=E_{2}^{+}r_{12}+E_{2}^{-}$ (97)

and

$\displaystyle E_{1}^{+}t_{12}=E_{2}^{+}+E_{2}^{-}r_{12}$ (98)

Accounting for propagation we can write. Note the change in sign between [2] and this work, this is because of how I have defined my wave equation.

$\displaystyle E_{1}^{+}t_{12}=E_{2}^{+}e^{\zeta_2 d_1}+E_{2}^{-}r_{12}e^{-\zeta_2 d_1}$ (99)

and

$\displaystyle E_{1}^{-}t_{12}=E_{2}^{+}r_{12}e^{\zeta_2 d_1}+E_{2}^{-}e^{-\zeta_2 d_1}$ (100)

where

$\displaystyle \zeta=\frac{2\pi}{\lambda} \bar{n}$ (101)

The boundary condition on the right hand side is given by:

$\displaystyle E_{4}^{-}t_{45}=E_{5}^{+}r_{45}e^{\zeta_1 d_1}+E_{5}^{-}e^{-\zeta_1 d_1}$ (102)

where we say that $ E_{5}^{-}$ is zero and $ r_{12}$ is zero. Therefore we end up with

$\displaystyle E_{4}^{-}t_{45}=0$ (103)

as the boundary condition.


next up previous
Next: Refractive index and absorption Up: Optical model Previous: Forward propagating wave
rod 2017-12-08