Because the math may be somewhat dense, I outline the procedure below:
- Assume we have an input field with a known spatio-temporal distribution.
- Fourier transform that input field in space and time to get the
plane wave components.
- Apply Fresnel analysis to each component and obtain the plane wave
spectrum for the evanescent side of the interface.
- Superpose all the evanescent components back together (if possible).
- The superposition should result in a single penetration depth, which
may in fact be infinite if all components of the input field are not
above the critical angle.
- Find the resulting exponential decay term and take 1e
of that value to be the penetration depth.
Suppose that we have some incident electric field incident on an interface completely polarized in the p-plane, given by f(r,t)ˆp, where ˆp is the unit vector in the direction perpendicular to the Poynting vector for the propagating fields. We also assume that the incident field contains no evanescent wave components. Any vector field whose components are elements of the Schwartz class (of functions) can be represented as a superposition of plane waves \cite{Lax02}, so f(r,t) can represented as
f(r,t)=∫∞−∞F(k,ω)ei(k⋅r−ωt)d3kdω
where
F(k,ω) is the 4-dimensional Fourier transform of
f(r,t), (3 spatial dimensions and 1 temporal dimension). In general,
k can be a complex valued vector, but here we assume that the
k is propagating, (i.e., homogeneous) so it is real valued. Additionally, because of the symmetries of a planar interface, we can assume
ky=0, and
kz≥0, when we define the interface to lie in the
xy-plane, with the Poynting vector of the incoming beam point somewhere in the
+z direction, and when, for simplicity, we assume that the beam is only varying spatially in the
xz-plane. For this case, the general Fresnel amplitude coefficient for the p-polarization is given by :
τp(kx,ω)=2ϵa(ω)√ϵb(ω)μb(ω)−(ckxω)2ϵa(ω)√ϵb(ω)μb(ω)−(ckxω)2+ϵb(ω)√ϵa(ω)μa(ω)−(ckxω)2=2ϵa(ω)√n2b(ω)−(ckxω)2ϵa(ω)√n2b(ω)−(ckxω)2+ϵb(ω)√n2a(ω)−(ckxω)2
where
n(ω)=√ϵ(ω)μ(ω) is the complex index of refraction,
ϵ(ω) is the dielectric permittivity,
μ(ω) is the magnetic permeability, and
ω is angular frequency. Is this case we assume that we are at optical frequencies, i.e., that
μ(ω)≈1, then
n2(ω)→ϵ(ω)
which implies that the Fresnel coefficient becomes:
τp(kx,ω)=2n2a(ω)√n2b(ω)−(ckxω)2n2a(ω)√n2b(ω)−(ckxω)2+n2b(ω)√n2a(ω)−(ckxω)2
From Maxwell's equations we have the following relationship for each plane wave component:
k⋅k=(ωc)2ϵ(ω)μ(ω)=(ω⋅n(ω)c)2
where
c is the speed of light. In our case this results in:
k2x+k2z=(ω⋅n(ω)c)2
We now need to answer the question, what is the resultant
k′ on the other side of the interface? From the generalized Snell's law we know that, for our case,
kx=k′x. So we must compute
k′z using the generalized Snell's law:
na(ω)sin(θa)=nb(ω)sin(θb)⟹na(ω)nb(ω)sin(θa)=sin(θb)
where
θ is the angle of the
ks with respect to the
z-axis, and the sine functions can be complex valued. If the left hand side of the above is real and
na(ω)>nb(ω) the we have the following equation where
C is real:
eiθb−e−iθb2i=C⟹eiθb−e−iθb=2Ci
In this case we have that
ℜ(θb)=π/2, and the imaginary part varies, to meet the above equation. Note that in whatever latex interpreter StackExchange is using
ℜ(⋅) is the real part and
ℑ(⋅) is the imaginary part. Then
ie−ℑ(θb)+ieℑ(θb)=2Ci⟹e−ℑ(θb)+eℑ(θb)=2C⟹cosh(ℑ(θb))=na(ω)nb(ω)sin(θa)⟹ℑ(θb)=cosh−1(na(ω)nb(ω)sin(θa))
since
eiπ/2=i,e−iπ/2=−1. Now since
θa is defined to the
z-axis, we have the following definition for
k:
k=ω⋅na(ω)c(sin(θa)0cos(θa)).
Note, that we have:
k′⋅k′=∥k′re∥−∥k′im∥+2ik′re⋅k′im=(ωc)2n2b(ω)
for the complex valued
k′=k′re+k′im. Since we are assuming that
nb(ω) is real valued, then the imaginary part of the dot product must be zero, and hence the real and imaginary parts of
k′ are perpendicular. Therefore, since
k′x=kx and
kx is real,
k′z is either real or purely imaginary. When
k′z is purely imaginary then we have the definition of an evanescent wave. Those components in the input beam which have
k′z purely real will not be evanescent. Now we just need to compute
k′z. We know that
k′=ω⋅nb(ω)c(sin(θb)0cos(θb))=ω⋅nb(ω)c(kx0cos(θb)).
It is easy to show that for
θb=π/2+ℑ(θb) and the above solution for
ℑ(θb) that
cos(θb)=isinh[ℑ(θb)]=isinh[cosh−1(na(ω)nb(ω)sin(θa))]=isinh[cosh−1(cωnb(ω)kx)]
Now, we have each
k and associated
k′, and the
τp. The field on the
nb side of the interface is then
τp(kx,ω)F(k,ω)ei(k′⋅r−ωt)
for each
k and each
ω. We must superpose these to obtain the field. First, we shall rewrite the above in terms of
kx.
k′=ω⋅nb(ω)c(kx0isinh[cosh−1(cωnb(ω)kx)])
Note that the above is only for the evanescent components! We have the following relations (a restatement of above):
k′x=kxk′y=ky=0k′z=isinh[cosh−1(cωnb(ω)kx)]
The superposition then becomes:
∫∞−∞∫Eτp(kx,ω)F(kx,kz,ω)e−sinh[cosh−1(cωnb(ω)kx)]zei(kxx−ωt)dkxdkzdt+[propagating components]
where
E is the domain for the evanescent components. Then by analyzing the inner integral (wrt
kx) we get
∫∞−∞rect((kx−kx,0)D)τp(kx,ω)F(kx,kz,ω)e−sinh[cosh−1(cωnb(ω)kx)]zeikxxdkx=F−1{rect((kx−kx,0)D)τp(kx,ω)F(kx,kz,ω)e−sinh[cosh−1(cωnb(ω)kx)]z},
i.e., the inverse Fourier transform of a somewhat nasty function for the evanescent part where
rect((kx−kx,0)D) is the frequency window for the evanescent components. Evaluating the above inverse Fourier transform and computing the
1e point of it would yield the depth (there is an integral of an exponentially decaying term here), I'm still working on this part...I'll update as it is completed.
This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user daaxix