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We assume that a homogeneous plane wave coming from a certain
direction
k0 = (kx ky k0, z)T
strikes onto a planar homogeneous layer l.
Within the layer the electric field consists of two plane waves
traveling in opposite directions
kl+ and
kl-.
This situation is schematically illustrated in Figure C.1.
The electric phasor
El(x) is thus written as
![$\displaystyle \mathbf{E}_l(\mathbf{x}) = \left(\mathbf{E}_l^+ e^{+jk_{l,z}z} + \mathbf{E}_l^- e^{-jk_{l,z}z}\right) e^{j(k_xx+k_yy)},$](img1524.gif) |
(C.1) |
whereby
El+ and
El- are the wave amplitudes
traveling downwards and upwards the layer. The two wavevectors are
given by
![$\displaystyle \mathbf{k}_l^+ = (k_x \;\, k_y \, +k_{l,z})^{\mathrm{T}}{}\qquad{\text{and}}\qquad{}\mathbf{k}_l^- = (k_x \;\, k_y \, -k_{l,z})^{\mathrm{T}},$](img1525.gif) |
(C.2) |
and the vertical wavevector component
![$\displaystyle k_{l,z} = \sqrt{k_l^2 - k_x^2 - k_y^2}$](img1526.gif) |
(C.3) |
depends on the wavenumber
kl = k0nl of the layer material with
refractive index nl.
Figure C.1:
In a homogeneous planar layer
with refractive index nl the electric field consists of two plane waves
El+ and
El- traveling downwards and upwards,
respectively.
|
Due to the transverseness of the plane waves the
vertical amplitude components can be expressed by the lateral ones, i.e.,
![$\displaystyle \left.\begin{aligned}k_xE^+_{l,x} + k_yE^+_{l,y} + k_{l,z}E^+_{l,...
...yle+\frac{k_x}{k_{l,z}} E^-_{l,x} + \frac{k_y}{k_{l,z}} E^-_{l,y}.\end{aligned}$](img1528.gif) |
(C.4) |
Hence it suffices to study only the lateral field components of
(C.1) given by
In these two equations
El, x(z) and
El, y(z) refer to the z-dependent
part of the electric field phasor. As can be seen from (C.1)
the complete phasor is obtained by multiplication with the exponential factor
exp(j(kxx + kyy)), which describes the lateral dependence.
Next: Magnetic Field
Up: C.1 One Homogeneous Planar
Previous: C.1 One Homogeneous Planar
Heinrich Kirchauer, Institute for Microelectronics, TU Vienna
1998-04-17