Peeter Joot's (OLD) Blog.

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Posts Tagged ‘Polarization’

Polarization angles for normal transmission and reflection

Posted by peeterjoot on January 22, 2014

[Click here for a PDF of this post with nicer formatting]

Question: Polarization angles for normal transmission and reflection ([1] pr 9.14)

For normal incidence, without assuming that the reflected and transmitted waves have the same polarization as the incident wave, prove that this must be so.

Answer

Working with coordinates as illustrated in fig. 1.1, the incident wave can be assumed to have the form

fig 1.1: Normal incidence coordinates

 

\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{I}} = E_{\mathrm{I}} e^{i (k z - \omega t)} \hat{\mathbf{x}}\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{I}} = \frac{1}{{v}} \hat{\mathbf{z}} \times \tilde{\mathbf{E}}_{\mathrm{I}} = \frac{1}{{v}} E_{\mathrm{I}} e^{i (k z - \omega t)} \hat{\mathbf{y}}.\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Assuming a polarization \hat{\mathbf{n}} = \cos\theta \hat{\mathbf{x}} + \sin\theta \hat{\mathbf{y}} for the reflected wave, we have

\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{R}} = E_{\mathrm{R}} e^{i (-k z - \omega t)} (\hat{\mathbf{x}} \cos\theta + \hat{\mathbf{y}} \sin\theta)\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{R}} = \frac{1}{{v}} (-\hat{\mathbf{z}}) \times \tilde{\mathbf{E}}_{\mathrm{R}} = \frac{1}{{v}} E_{\mathrm{R}} e^{i (-k z - \omega t)} (\hat{\mathbf{x}} \sin\theta - \hat{\mathbf{y}} \cos\theta).\end{aligned} \hspace{\stretch{1}}(1.0.2b)

And finally assuming a polarization \hat{\mathbf{n}} = \cos\phi \hat{\mathbf{x}} + \sin\phi \hat{\mathbf{y}} for the transmitted wave, we have

\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{T}} = E_{\mathrm{T}} e^{i (k' z - \omega t)} (\hat{\mathbf{x}} \cos\phi + \hat{\mathbf{y}} \sin\phi)\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{T}} = \frac{1}{{v}} \hat{\mathbf{z}} \times \tilde{\mathbf{E}}_{\mathrm{T}} = \frac{1}{{v'}} E_{\mathrm{T}} e^{i (k' z - \omega t)} (-\hat{\mathbf{x}} \sin\phi + \hat{\mathbf{y}} \cos\phi).\end{aligned} \hspace{\stretch{1}}(1.0.3b)

With no components of any of the \tilde{\mathbf{E}} or \tilde{\mathbf{B}} waves in the \hat{\mathbf{z}} directions the boundary value conditions at z = 0 require the equality of the \hat{\mathbf{x}} and \hat{\mathbf{y}} components of

\begin{aligned}\left( \tilde{\mathbf{E}}_{\mathrm{I}} + \tilde{\mathbf{E}}_{\mathrm{R}} \right)_{x,y} = \left( \tilde{\mathbf{E}}_{\mathrm{T}} \right)_{x,y}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned} \left( \frac{1}{\mu} \left( \tilde{\mathbf{B}}_{\mathrm{I}} + \tilde{\mathbf{B}}_{\mathrm{R}} \right) \right)_{x,y} = \left( \frac{1}{\mu'} \tilde{\mathbf{B}}_{\mathrm{T}} \right)_{x,y}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

With \beta = \mu v/\mu' v', those components are

\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} \cos\theta = E_{\mathrm{T}} \cos\phi \end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}E_{\mathrm{R}} \sin\theta = E_{\mathrm{T}} \sin\phi\end{aligned} \hspace{\stretch{1}}(1.0.5b)

\begin{aligned}E_{\mathrm{R}} \sin\theta = - \beta E_{\mathrm{T}} \sin\phi\end{aligned} \hspace{\stretch{1}}(1.0.5c)

\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} \cos\theta = \beta E_{\mathrm{T}} \cos\phi\end{aligned} \hspace{\stretch{1}}(1.0.5d)

Equality of eq. 1.0.5b, and eq. 1.0.5c require

\begin{aligned}- \beta E_{\mathrm{T}} \sin\phi = E_{\mathrm{T}} \sin\phi,\end{aligned} \hspace{\stretch{1}}(1.0.6)

or (\theta, \phi) \in \{(0, 0), (0, \pi), (\pi, 0), (\pi, \pi)\}. It turns out that all of these solutions correspond to the same physical waves. Let’s look at each in turn

  • (\theta, \phi) = (0, 0). The system eq. 1.0.5.5 is reduced to

    \begin{aligned}\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} &= E_{\mathrm{T}} \\ E_{\mathrm{I}} - E_{\mathrm{R}} &= \beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.7)

    with solution

    \begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= \frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= \frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.8)

  • (\theta, \phi) = (\pi, \pi). The system eq. 1.0.5.5 is reduced to

    \begin{aligned}\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} &= -E_{\mathrm{T}} \\ E_{\mathrm{I}} + E_{\mathrm{R}} &= -\beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.9)

    with solution

    \begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= -\frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= -\frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.10)

    Effectively the sign for the magnitude of the transmitted and reflected phasors is toggled, but the polarization vectors are also negated, with \hat{\mathbf{n}} = -\hat{\mathbf{x}}, and \hat{\mathbf{n}}' = -\hat{\mathbf{x}}. The resulting \tilde{\mathbf{E}}_{\mathrm{R}} and \tilde{\mathbf{E}}_{\mathrm{T}} are unchanged relative to those of the (0,0) solution above.

  • (\theta, \phi) = (0, \pi). The system eq. 1.0.5.5 is reduced to

    \begin{aligned}\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} &= -E_{\mathrm{T}} \\ E_{\mathrm{I}} - E_{\mathrm{R}} &= -\beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.11)

    with solution

    \begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= -\frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= \frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.12)

    Effectively the sign for the magnitude of the transmitted phasor is toggled. The polarization vectors in this case are \hat{\mathbf{n}} = \hat{\mathbf{x}}, and \hat{\mathbf{n}}' = -\hat{\mathbf{x}}, so the transmitted phasor magnitude change of sign does not change \tilde{\mathbf{E}}_{\mathrm{T}} relative to that of the (0,0) solution above.

  • (\theta, \phi) = (\pi, 0). The system eq. 1.0.5.5 is reduced to

    \begin{aligned}\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} &= E_{\mathrm{T}} \\ E_{\mathrm{I}} + E_{\mathrm{R}} &= \beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.13)

    with solution

    \begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= \frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= -\frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.14)

    This time, the sign for the magnitude of the reflected phasor is toggled. The polarization vectors in this case are \hat{\mathbf{n}} = -\hat{\mathbf{x}}, and \hat{\mathbf{n}}' = \hat{\mathbf{x}}. In this final variation the reflected phasor magnitude change of sign does not change \tilde{\mathbf{E}}_{\mathrm{R}} relative to that of the (0,0) solution.

We see that there is only one solution for the polarization angle of the transmitted and reflected waves relative to the incident wave. Although we fixed the incident polarization with \mathbf{E} along \hat{\mathbf{x}}, the polarization of the incident wave is maintained regardless of TE or TM labeling in this example, since our system is symmetric with respect to rotation.

References

[1] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

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A final pre-exam update of my notes compilation for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

Here’s my third update of my notes compilation for this course, including all of the following:

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

plus everything detailed in the description of my previous update and before.

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PHY452H1S Basic Statistical Mechanics. Lecture 20: Bosons. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 2, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Bosons

In order to maintain a conservation of particles in a Bose condensate as we decrease temperature, we are forced to change the chemical potential to compensate. This is illustrated in fig. 1.1.

Fig 1.1: Chemical potential in Bose condensation region

 

Bose condensatation occurs for T < T_{\mathrm{BEC}}. At this point our number density becomes (except at \mathbf{k} = 0)

\begin{aligned}n(\mathbf{k}) = \frac{1}{{e^{\beta \epsilon_\mathbf{k}} - 1}}.\end{aligned} \hspace{\stretch{1}}(1.2.1)

Except for \mathbf{k} = 0, n(\mathbf{k}) is well defined, and not described by this distribution. We are forced to say that

\begin{aligned}N = N_0 + \sum_{\mathbf{k} \ne 0} n(\mathbf{k}) = N_0 + V\int \frac{d^3 \mathbf{k}}{(2 \pi)^3} \frac{1}{{ e^{\beta \epsilon_\mathbf{k}} - 1 }}.\end{aligned} \hspace{\stretch{1}}(1.2.1)

Introducing the density of states, our density is

\begin{aligned}\rho = \rho_0 + \int_0^\infty d\epsilon \frac{N(\epsilon)}{e^{\beta \epsilon} - 1 },\end{aligned} \hspace{\stretch{1}}(1.2.3)

where

\begin{aligned}N(\epsilon) = \frac{1}{{4 \pi^2}} \left( \frac{2m}{\hbar} \right)^{3/2} \epsilon^{1/2}.\end{aligned} \hspace{\stretch{1}}(1.2.4)

We worked out last time that

\begin{aligned}\rho = \rho_0 + \rho \left( \frac{T}{T_{\mathrm{BEC}}} \right)^{3/2},\end{aligned} \hspace{\stretch{1}}(1.2.4)

or

\begin{aligned}\rho_0 = \rho \left( 1 - \left( \frac{T}{T_{\mathrm{BEC}}} \right) ^{3/2} \right).\end{aligned} \hspace{\stretch{1}}(1.2.6)

This is plotted in fig. 1.2.

Fig 1.2: Density variation with temperature for Bosons

 

\begin{aligned}\rho_0 = \frac{N_{\mathbf{k} = 0}}{V}.\end{aligned} \hspace{\stretch{1}}(1.7)

For T \ge T_{\mathrm{BEC}}, we have \rho_0 = 0. This condensation temperature is

\begin{aligned}T_{\mathrm{BEC}} \propto \rho^{2/3}.\end{aligned} \hspace{\stretch{1}}(1.8)

This is plotted in fig. 1.3.

Fig 1.3: Temperature vs pressure demarkation by T_BEC curve

 

There is a line for each density that marks the boundary temperature for which we have or do not have this condensation phenomina where \mathbf{k} = 0 states start filling up.

Specific heat: T < T_{\mathrm{BEC}}

\begin{aligned}\frac{E}{V} &= \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} \frac{1}{{ e^{\beta \hbar^2 k^2/2m} - 1}}\frac{\hbar^2 k^2}{2m} \\ &= \int_0^\infty d\epsilon N(\epsilon) \frac{1}{{ e^{\beta \epsilon} - 1 }} \epsilon \\ &\propto \int_0^\infty d\epsilon \frac{\epsilon^{3/2}}{ e^{\beta \epsilon} - 1 } \\ &\propto \left( k_{\mathrm{B}} T \right)^{5/2},\end{aligned} \hspace{\stretch{1}}(1.9)

so that

\begin{aligned}\frac{C}{V} \propto \left( k_{\mathrm{B}} T \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.10)

Compare this to the classical and Fermionic specific heat as plotted in fig. 1.4.

Fig 1.4: Specific heat for Bosons, Fermions, and classical ideal gases

 

One can measure the specific heat in this Bose condensation phenomina for materials such as Helium-4 (spin 0). However, it turns out that Helium-4 is actually quite far from an ideal Bose gas.

Photon gas

A system that is much closer to an ideal Bose gas is that of a gas of photons. To a large extent, photons do not interact with each other. This allows us to calculate black body phenomina and the low temperature (cosmic) background radiation in the universe.

An important distinction between a photon sea and some of these other systems is that the photon number is actually not fixed.

Photon numbers are not “conserved”.

If a photon interacts with an atom, it can impart energy and disappear. An excited atom can emit a photon and change its energy level. In a thermodynamic system we can generally expect that introducing heat will generate more photons, whereas a cold sink will tend to generate fewer photons.

We have a few special details that distinguish photons that we’ll have to consider.

  1. spin 1.
  2. massless, moving at the speed of light.
  3. have two polarization states.

Because we do not have a constraint on the number of particles, we essentially have no chemical potential, even in the grand canonical scheme.

Writing

\begin{aligned}\lambda = \left\{\begin{array}{l l}+1 & \quad \mbox{Right circular polarization} \\ -1 & \quad \mbox{Left circular polarization}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.11)

Our number density, since we have no chemical potential, is of the form

\begin{aligned}n_{\mathbf{k}, \lambda}= \frac{1}{{e^{\beta \epsilon_{\mathbf{k}, \lambda}} - 1 }},\end{aligned} \hspace{\stretch{1}}(1.12)

Observe that the average number of photons in this system is temperature dependent. Because this chemical potential is not there, it can be quite easy to work out a number of the thermodynamic results.

Photon average energy density

We’ll now calculate the average energy density of the photons. The energy of a single photon is

\begin{aligned}\epsilon_{\mathbf{k}, \lambda} = \hbar c k = \hbar \omega,\end{aligned} \hspace{\stretch{1}}(1.2.13)

so that the average energy density is

\begin{aligned}\frac{E}{V} &= \sum_{\mathbf{k}, \lambda} \frac{1}{{ e^{ \beta \epsilon_\mathbf{k}} - 1}} \epsilon_\mathbf{k}\rightarrow\underbrace{2}_{\text{number of polarizations}}\int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\frac{ \hbar c k}{ e^{ \beta \epsilon_\mathbf{k}} - 1} \\ &= 2 \int_0^\infty d\epsilon \underbrace{\frac{1}{{(2 \pi)^3}} 4 \pi \frac{\epsilon^2}{(\hbar c)^3} }_{\text{Photon density of states}}\frac{\epsilon}{e^{\beta \epsilon} - 1} \\ &= \frac{1}{{\pi^2}} \frac{1}{{ (\hbar c)^3 }} \int_0^\infty d\epsilon \frac{\epsilon^3}{e^{\beta \epsilon} - 1}\end{aligned} \hspace{\stretch{1}}(1.2.13)

Mathematica tells us that this integral is

\begin{aligned}\int_0^\infty d\epsilon \frac{\epsilon^3}{e^{\beta \epsilon} - 1} =\frac{\pi ^4}{15 \beta ^4},\end{aligned} \hspace{\stretch{1}}(1.2.13)

for an end result of

\begin{aligned}\frac{E}{V} =\frac{\pi^2}{15} \frac{1}{{(\hbar c)^3}} \left( k_{\mathrm{B}} T \right)^4.\end{aligned} \hspace{\stretch{1}}(1.2.13)

Phonons and other systems

There is a very similar phenomina in matter. We can discuss lattice vibrations in a solid. These are called phonon modes, and will have the same distribution function where the only difference is that the speed of light is replaced by the speed of the sound wave in the solid. Once we understand the photon system, we are able to look at other Bose distributions such as these phonon systems. We’ll touch on this very briefly next time.

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Derivation of Fresnel equations for mixed polarization (using Geometric Algebra)

Posted by peeterjoot on September 25, 2012

[Click here for a PDF of this post with nicer formatting and figures]

Motivation

In [2] we have a derivation of the Fresnel equations for the TE and TM polarization modes. Can we do this for an arbitrary polarization angles?

Setup

The task at hand is to find evaluate the boundary value constraints. Following the interface plane conventions of [1], and his notation that is

\begin{aligned}\epsilon_1 ( \mathbf{E}_i + \mathbf{E}_r )_z = \epsilon_2 ( \mathbf{E}_t )_z\end{aligned} \hspace{\stretch{1}}(1.2.1a)

\begin{aligned}( \mathbf{B}_i + \mathbf{B}_r )_z = ( \mathbf{B}_t )_z\end{aligned} \hspace{\stretch{1}}(1.2.1b)

\begin{aligned}( \mathbf{E}_i + \mathbf{E}_r )_{x,y} = ( \mathbf{E}_t )_{x,y}\end{aligned} \hspace{\stretch{1}}(1.2.1c)

\begin{aligned}\frac{1}{{\mu_1}} ( \mathbf{B}_i + \mathbf{B}_r )_{x,y} = \frac{1}{{\mu_2}} ( \mathbf{B}_t )_{x,y}\end{aligned} \hspace{\stretch{1}}(1.2.1d)

I’ll work here with a phasor representation directly and not bother with taking real parts, or using tilde notation to mark the vectors as complex.

Our complex magnetic field phasors are related to the electric fields with

\begin{aligned}\mathbf{B} = \frac{1}{{v}} \hat{\mathbf{k}} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.2.2)

Referring to figure (see pdf) shows the geometrical task to tackle, since we’ve got to express all the various unit vectors algebraically. I’ll use Geometric Algebra here to do that for its compact expression of rotations. With

Figure: See pdf: Reflection and transmission of light at an interface

\begin{aligned}j = \mathbf{e}_3 \mathbf{e}_1,\end{aligned} \hspace{\stretch{1}}(1.2.3)

we can express each of the k vector directions by inspection. Those are

\begin{aligned}\hat{\mathbf{k}}_i = \mathbf{e}_3 e^{j \theta_i} = \mathbf{e}_3 \cos\theta_i + \mathbf{e}_1 \sin\theta_i\end{aligned} \hspace{\stretch{1}}(1.2.4a)

\begin{aligned}\hat{\mathbf{k}}_r = -\mathbf{e}_3 e^{-j \theta_r} = -\mathbf{e}_3 \cos\theta_r +\mathbf{e}_1 \sin\theta_r\end{aligned} \hspace{\stretch{1}}(1.2.4b)

\begin{aligned}\hat{\mathbf{k}}_t = \mathbf{e}_3 e^{j \theta_t} = \mathbf{e}_3 \cos\theta_t + \mathbf{e}_1 \sin\theta_t.\end{aligned} \hspace{\stretch{1}}(1.2.4c)

Similarly, the perpendiculars \hat{\mathbf{m}}_p = \mathbf{e}_2 \times \hat{\mathbf{k}}_p are

\begin{aligned}\hat{\mathbf{m}}_i = \mathbf{e}_{1} e^{j \theta_i}= \mathbf{e}_1 \cos\theta_i - \mathbf{e}_3 \sin\theta_i= \mathbf{e}_3 j e^{j \theta_i}\end{aligned} \hspace{\stretch{1}}(1.2.5a)

\begin{aligned}\hat{\mathbf{m}}_r = -\mathbf{e}_{1} e^{-j \theta_r}= -\mathbf{e}_1 \cos\theta_r - \mathbf{e}_3 \sin\theta_r= -\mathbf{e}_3 j e^{-j \theta_r}\end{aligned} \hspace{\stretch{1}}(1.2.5b)

\begin{aligned}\hat{\mathbf{m}}_t = \mathbf{e}_{1 } e^{j \theta_t} = \mathbf{e}_1 \cos\theta_t - \mathbf{e}_3 \sin\theta_t= \mathbf{e}_3 j e^{j \theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.5c)

In [1] problem 9.14 we had to show that the polarization angles for normal incident (\mathbf{E} \parallel \mathbf{e}_1) must be the same due to the boundary constraints. Can we also tackle that problem for both this more general angle of incidence and a general polarization? Let’s try so, allowing temporarily for different polarizations of the reflected and transmitted components of the light, calling those polarization angles \phi_i, \phi_r, and \phi_t respectively. Let’s set the \phi_i = 0 polarization aligned such that \mathbf{E}_i, \mathbf{B}_i are aligned with the \mathbf{e}_2 and -\hat{\mathbf{m}}_i directions respectively, so that the generally polarized phasors are

\begin{aligned}\begin{bmatrix}\mathbf{E}_p \\ \mathbf{B}_p \\ \end{bmatrix}=\begin{bmatrix}\mathbf{e}_2 \\ -\hat{\mathbf{m}}_p \\ \end{bmatrix}e^{ \hat{\mathbf{m}}_p \mathbf{e}_2 \phi_p }\end{aligned} \hspace{\stretch{1}}(1.2.6)

We are now set to at least express our boundary value constraints

\begin{aligned}\epsilon_1 \left( \mathbf{e}_2 E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + \mathbf{e}_2 E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \cdot \mathbf{e}_3 = \epsilon_2 \left( \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7a)

\begin{aligned}\frac{1}{{v_1}} \left( -\hat{\mathbf{m}}_i E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - \hat{\mathbf{m}}_r E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \cdot \mathbf{e}_3 = \frac{1}{{v_2}} \left( -\hat{\mathbf{m}}_t E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7b)

\begin{aligned}\left( \mathbf{e}_2 E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + \mathbf{e}_2 E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \wedge \mathbf{e}_3 = \left( \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( -\hat{\mathbf{m}}_i E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - \hat{\mathbf{m}}_r E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \wedge \mathbf{e}_3 = \frac{1}{{\mu_2 v_2}} \left( -\hat{\mathbf{m}}_t E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7d)

Let’s try this in a couple of steps. First with polarization angles set so that one of the fields lies in the plane of the interface (with both variations), and then attempt the general case, first posing the problem in the tranditional way to see what equations fall out, and then using superposition.

Before doing so, let’s introduce a bit of notation to be used throughout. When we wish to refer to all the fields or angles, for example, \mathbf{E}_i, \mathbf{E}_r, \mathbf{E}_t then we’ll write \mathbf{E}_p where p \in \{i, r, t\}. Similarily, to refer to just the incident and transmitted components (or angles) we’ll use \mathbf{E}_q where q \in \{i, t\}. Following [1] we’ll also write

\begin{aligned}\beta = \frac{\mu_1 v_1} {\mu_2 v_2} \end{aligned} \hspace{\stretch{1}}(1.2.8)

\begin{aligned}\alpha = \frac{\cos\theta_t}{\cos\theta_i},\end{aligned} \hspace{\stretch{1}}(1.2.8)

Question: Sanity check. Verify for \mathbf{E} parallel to the interface.

Answer

For the \mathbf{E}_p \parallel \mathbf{e}_2 polarization (\phi_i = \phi_r = \phi_t) our phasors are

\begin{aligned}\mathbf{E}_p = \mathbf{e}_2 E_p\end{aligned} \hspace{\stretch{1}}(1.2.9)

\begin{aligned}\mathbf{B}_p = -\frac{1}{{v_p}} \hat{\mathbf{m}}_p E_p\end{aligned} \hspace{\stretch{1}}(1.2.9)

Our boundary value constraints then become

\begin{aligned}\epsilon_1 \left( \mathbf{e}_2 E_i  + \mathbf{e}_2 E_r  \right) \cdot \mathbf{e}_3 = \epsilon_2 \left( \mathbf{e}_2 E_t  \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10a)

\begin{aligned}\frac{1}{{v_1}} \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r  \right) \cdot \mathbf{e}_3 = \frac{1}{{v_2}} \left( \hat{\mathbf{m}}_t E_t  \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10b)

\begin{aligned}\left( \mathbf{e}_2 E_i  + \mathbf{e}_2 E_r  \right) \wedge \mathbf{e}_3 = \left( \mathbf{e}_2 E_t  \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( \hat{\mathbf{m}}_i E_i  + \hat{\mathbf{m}}_r E_r  \right) \wedge \mathbf{e}_3 = \frac{1}{{\mu_2 v_2}} \left( \hat{\mathbf{m}}_t E_t  \right) \wedge \mathbf{e}_3.\end{aligned} \hspace{\stretch{1}}(1.2.10d)

With \hat{\mathbf{m}}_p substitution this is

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i  + \mathbf{e}_2 E_r  \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t  \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.11a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_i} E_i  - \mathbf{e}_1 e^{-j \theta_r} E_r  \right) }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_t} E_t  \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.11b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i  + \mathbf{e}_2 E_r  \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t  \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.11c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_i} E_i  -\mathbf{e}_1 e^{-j \theta_r} E_r  \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_t} E_t  \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.11d)

Evaluating the grade selections we have a separation into an analogue of real and imaginary parts for

\begin{aligned}0 = 0\end{aligned} \hspace{\stretch{1}}(1.2.12a)

\begin{aligned}\frac{1}{{v_1}} \left( -\sin\theta_i E_i  - \sin\theta_r E_r  \right) = \frac{1}{{v_2}} \left( -\sin\theta_t E_t  \right)\end{aligned} \hspace{\stretch{1}}(1.2.12b)

\begin{aligned}E_i + E_r = E_t\end{aligned} \hspace{\stretch{1}}(1.2.12c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( \cos{\theta_i} E_i  - \cos{\theta_r} E_r  \right) = \frac{1}{{\mu_2 v_2}} \left( \cos{ \theta_t} E_t  \right).\end{aligned} \hspace{\stretch{1}}(1.2.12d)

With \theta_i = \theta_r and \sin\theta_t/\sin\theta_i = n_1/n_2 1.2.12b becomes

\begin{aligned}E_i + E_r = \frac{n_1 v_1}{n_2 v_2} E_t = \frac{v_2 v_1}{v_1 v_2} E_t = E_t,\end{aligned} \hspace{\stretch{1}}(1.2.12d)

so that we find 1.2.12b and 1.2.12c are dependent. We are left with a pair of equations

\begin{aligned}E_i + E_r = E_t\end{aligned} \hspace{\stretch{1}}(1.2.14)

\begin{aligned}E_i - E_r = \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} E_t,\end{aligned} \hspace{\stretch{1}}(1.2.14)

Adding and subtracting we have

\begin{aligned}2 E_i = \left( 1 + \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} \right) E_t\end{aligned} \hspace{\stretch{1}}(1.2.15)

\begin{aligned}2 E_r = \left( 1 - \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} \right) E_t,\end{aligned} \hspace{\stretch{1}}(1.2.15)

with a final rearrangement to yield

\begin{aligned}\frac{E_t}{E_i}=\frac{2 \mu_2 v_2 \cos\theta_i}{\mu_2 v_2 \cos\theta_i+\mu_1 v_1 \cos\theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.16)

\begin{aligned}\frac{E_r}{E_i}=\frac{\mu_2 v_2 \cos\theta_i-\mu_1 v_1 \cos\theta_t}{\mu_2 v_2 \cos\theta_i+\mu_1 v_1 \cos\theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.16)

Using the \alpha and \beta notation above we have

\begin{aligned}\frac{E_t}{E_i}=\frac{2 }{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.17)

\begin{aligned}\frac{E_r}{E_i}=\frac{1 - \alpha \beta}{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.17)

Question: Sanity check. Verify for \mathbf{B} parallel to the interface.

Answer

As a second sanity check let’s rotate our field polarizations by applying a rotation e^{\mathbf{e}_2 \hat{\mathbf{m}}_p \pi/2} = \mathbf{e}_2 \hat{\mathbf{m}}_p (\phi_i = \phi_r = \phi_t = -\pi/2) so that

\begin{aligned}-\hat{\mathbf{m}}_p \rightarrow -\hat{\mathbf{m}}_p \mathbf{e}_2 \hat{\mathbf{m}}_p = \mathbf{e}_2\end{aligned} \hspace{\stretch{1}}(1.2.18)

\begin{aligned}\mathbf{e}_2 \rightarrow \mathbf{e}_2 \mathbf{e}_2 \hat{\mathbf{m}}_p = \hat{\mathbf{m}}_p\end{aligned} \hspace{\stretch{1}}(1.2.18)

This time we have \mathbf{E}_p \parallel \hat{\mathbf{m}}_p and \mathbf{B}_p \parallel \mathbf{e}_2. Our boundary value equations become

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_i E_i  + \hat{\mathbf{m}}_r E_r  \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_t E_t  \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.19a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r  \right) }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t  \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.19b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_i E_i  + \hat{\mathbf{m}}_r E_r  \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_t E_t  \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.19c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i  + \mathbf{e}_2 E_r  \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t  \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.19d)

This second equation 1.2.19b is a 0 = 0 identity, and the remaining after \hat{\mathbf{m}}_p substitution are

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_i} E_i  + (-\mathbf{e}_3) j e^{-j \theta_r} E_r  \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_t} E_t  \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.20a)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_i} E_i  + (-\mathbf{e}_3) j e^{-j \theta_r} E_r  \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_t} E_t  \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.20b)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i  + \mathbf{e}_2 E_r  \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t  \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.20c)

Simplifying we have

\begin{aligned}\epsilon_1 \left(  -\sin \theta_i E_i  - \sin{\theta_r} E_r  \right) = - \epsilon_2 \sin{\theta_t} E_t  \end{aligned} \hspace{\stretch{1}}(1.2.21a)

\begin{aligned} \cos{ \theta_i} E_i  - \cos{ \theta_r} E_r = \cos{ \theta_t} E_t\end{aligned} \hspace{\stretch{1}}(1.2.21b)

\begin{aligned}E_i  + E_r = \frac{\mu_1 v_1} {\mu_2 v_2} E_t\end{aligned} \hspace{\stretch{1}}(1.2.21c)

We expect an equality

\begin{aligned}\frac{\epsilon_2 \sin\theta_t}{\epsilon_1 \sin\theta_i} =  \frac{\mu_1 v_1} {\mu_2 v_2},\end{aligned} \hspace{\stretch{1}}(1.2.21c)

Noting that \epsilon_p v_p = 1/(v_p \mu_p) we find that to be true

\begin{aligned}\frac{\epsilon_2 \sin\theta_t}{\epsilon_1 \sin\theta_i} = \frac{\epsilon_2 n_1}{\epsilon_1 n_2} = \frac{\epsilon_2 v_2}{\epsilon_1 v_1} = \frac{\mu_1 v_1}{\mu_2 v_2} \end{aligned} \hspace{\stretch{1}}(1.2.21c)

we see that 1.2.21a and 1.2.21c are dependent. We are left with the system

\begin{aligned}E_i - E_r = \alpha E_t\end{aligned} \hspace{\stretch{1}}(1.2.24a)

\begin{aligned}E_i + E_r = \beta E_t,\end{aligned} \hspace{\stretch{1}}(1.2.24b)

with solution

\begin{aligned}\frac{E_t}{E_i} = \frac{2 }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.25)

\begin{aligned}\frac{E_r}{E_i} = \frac{\beta - \alpha}{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.25)

Question: General case. Arbitrary polarization angle.

Determine the set of simulaneous equations that would have to be solved for if the incident polarization angle was allowed to be neither TE nor TM mode.

Answer

Substituting our \hat{\mathbf{m}}_p vector expressions into the boundary value constraints we have

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \mathbf{e}_2 \left( E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle \end{aligned} \hspace{\stretch{1}}(1.2.26a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ j e^{j \theta_i} E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - j e^{-j \theta_r} E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ j e^{j \theta_t} E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.26b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \mathbf{e}_2 \left( E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.26c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ j e^{j \theta_i} E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - j e^{-j \theta_r} E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ j e^{j \theta_t} E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

With \alpha \in \{i,r\} we want to expand some intermediate multivector products

\begin{aligned}\mathbf{e}_{32} e^{\hat{\mathbf{m}}_q \mathbf{e}_2 \phi_q}=\mathbf{e}_{32} \cos \phi_q+\mathbf{e}_{32} \hat{\mathbf{m}}_q \mathbf{e}_2 \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q+\mathbf{e}_{32} \mathbf{e}_3 j e^{j \theta_q} \mathbf{e}_2 \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q-j e^{j \theta_q} \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q - \mathbf{e}_{31} \cos\theta_q \sin\phi_q+ \sin\theta_q \sin{\phi_q}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}\mathbf{e}_{32} e^{\hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r}=\mathbf{e}_{32} \cos \phi_r+\mathbf{e}_{32} \hat{\mathbf{m}}_r \mathbf{e}_2 \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r+\mathbf{e}_{32} (-\mathbf{e}_3) j e^{-j \theta_r} \mathbf{e}_2 \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r+ j e^{-j \theta_r} \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r + \mathbf{e}_{31} \cos\theta_r \sin{\phi_r}+ \sin \theta_r \sin{\phi_r}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}j e^{j \theta_q} e^{\hat{\mathbf{m}}_q \mathbf{e}_2 \phi_q}=j e^{j \theta_q} \left( \cos \phi_q+ \hat{\mathbf{m}}_q \mathbf{e}_2 \sin{\phi_q}\right)=j e^{j \theta_q} \left( \cos \phi_q+ \mathbf{e}_3 j e^{j \theta_q} \mathbf{e}_2 \sin{\phi_q}\right)=j e^{j \theta_q} \left( \cos \phi_q- j e^{-j \theta_q} \mathbf{e}_{32} \sin{\phi_q}\right)=j e^{j \theta_q} \cos \phi_q+ \mathbf{e}_{32} \sin{\phi_q}=\mathbf{e}_{31} \cos {j \theta_q} \cos \phi_q+ \mathbf{e}_{32} \sin{\phi_q}- \sin{ \theta_q} \cos \phi_q\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}-j e^{-j \theta_r} e^{\hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r}=-j e^{-j \theta_r} \left(\cos\phi_r + \hat{\mathbf{m}}_r \mathbf{e}_2 \sin\phi_r\right)=-j e^{-j \theta_r} \left(\cos\phi_r - \mathbf{e}_3 j e^{-j \theta_r} \mathbf{e}_2 \sin\phi_r\right)=-j e^{-j \theta_r} \left(\cos\phi_r + j e^{j \theta_r} \mathbf{e}_{32} \sin\phi_r\right)=-j e^{-j \theta_r} \cos\phi_r + \mathbf{e}_{32} \sin\phi_r=-\mathbf{e}_{31} \cos{\theta_r} \cos\phi_r + \mathbf{e}_{32} \sin\phi_r- \sin{ \theta_r} \cos\phi_r \end{aligned} \hspace{\stretch{1}}(1.2.26d)

Our boundary value conditions are then

\begin{aligned}\epsilon_1 \left( E_i \sin\theta_i \sin{\phi_i}+ E_r \sin \theta_r \sin{\phi_r}\right) = \epsilon_2 E_t \sin\theta_t \sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{v_1}}\left(E_i \sin{ \theta_i} \cos \phi_i+E_r \sin{ \theta_r} \cos\phi_r \right)=\frac{1}{{v_2}}E_t \sin{ \theta_t} \cos \phi_t\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}E_i \cos \phi_i + E_r \cos \phi_r =E_t \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}-E_i \cos\theta_i \sin\phi_i+ E_r \cos\theta_r \sin{\phi_r}=-E_t \cos\theta_t \sin\phi_t\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{\mu_1 v_1}}\left(E_i\cos { \theta_i} \cos \phi_i -E_r\cos{\theta_r} \cos\phi_r \right)=\frac{1}{{\mu_2 v_2}}E_t \cos { \theta_t} \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{\mu_1 v_1}}\left(E_i\sin{\phi_i}+E_r\sin\phi_r\right)=\frac{1}{{\mu_2 v_2}}E_t\sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.31)

Note that the wedge product equations above have been separated into \mathbf{e}_3 \mathbf{e}_1 and \mathbf{e}_3 \mathbf{e}_2 components, yielding two equations each. Because of 1.2.21c, we see that 1.2.31 and 1.2.31 are dependent. Also, as demonstrated in 1.2.12d we see that 1.2.31 and 1.2.31 are also dependent. We can therefore consider only the last four equations (and still have additional linear dependencies to be discovered.)

Let’s write these as

\begin{aligned}E_i \cos \phi_i + E_r \cos \phi_r =E_t \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}-E_i \sin\phi_i+ E_r \sin{\phi_r}=-E_t \alpha\sin\phi_t\end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}E_i\cos \phi_i -E_r\cos\phi_r =\alpha \beta E_t\cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}E_i\sin{\phi_i}+E_r\sin\phi_r=\beta E_t\sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.32)

Observe that if \phi_i = \phi_r = \phi_t = 0 (killing all the sine terms) we recover 1.2.14, and with \phi_i = \phi_r = \phi_t = \pi/2 (killing all the cosines) we recover 1.2.24.

Now, if \phi_i n \pi/2 we’ve got a different story. Specifically it appears that should we wish to solve for the reflected and transmitted magnitudes, we also have to simulaneously solve for the polarization angles in the reflected and transmitted directions. This is now a problem of solving four simulaneous equations in two linear and two non-linear variables.

Does it make sense that we would have polarization rotation should our initial polarization angle be rotated? I think so. In dicusssing this problem with Prof Thywissen, he strongly suggested treating the problem as a superposition of two light waves. If we consider that, even without attempting to solve the problem, we see that we must have different reflected and transmitted magnitudes associated with the pair of incident waves since we have to calculate each of these with different Fresnel equations. This would have an effect of scaling and rotating the superimposed reflected and transmitted waves.

Question: General case using using superposition

Using superposition determine the Fresnel equations for an arbitrary incident polarization angle. This should involve solving for both the magnitude and the polarization angle of the reflected and transmitted rays.

Answer

For a polarization of \phi = 0 and \phi = \pi/2 respectively, we have from problems \ref{fresnelAlternatePolarization:pr1-Answer} and \ref{fresnelAlternatePolarization:pr2-Answer}, or from 1.2.32 we have

\begin{aligned}\frac{E_{r \parallel}}{E_{i \parallel}} = \frac{1 - \alpha \beta}{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{t \parallel}}{E_{i \parallel}} = \frac{2 }{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{r \perp}}{E_{i \perp}} = \frac{ \beta - \alpha }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{t \perp}}{E_{i \perp}} = \frac{ 2 }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.33)

We can use these results to consider a polarization of \phi < \pi/2 as illustrated in figure (see pdf)

Figure: see pdf: Polarization of incident field to be considered

Our incident, reflected, and transmitted fields are

\begin{aligned}\mathbf{E}_i = E_{i} \mathbf{e}_2 e^{\mathbf{e}_2 \hat{\mathbf{m}}_i \phi}\end{aligned} \hspace{\stretch{1}}(1.2.34)

\begin{aligned}\mathbf{E}_r = E_{i \parallel}\frac{1 - \alpha\beta}{1 + \alpha\beta} \mathbf{e}_2 + E_{i \perp}\frac{\beta - \alpha}{\beta + \alpha} \hat{\mathbf{m}}_r\end{aligned} \hspace{\stretch{1}}(1.2.34)

\begin{aligned}\mathbf{E}_t = E_{i \parallel}\frac{2}{1 + \alpha\beta}\mathbf{e}_2+ E_{i \perp}\frac{2}{\beta + \alpha}\hat{\mathbf{m}}_i\end{aligned} \hspace{\stretch{1}}(1.2.34)

However, E_{i \parallel} = E_i \cos \phi and E_{i \perp} = E_i \sin\phi leaving us with

\begin{aligned}\mathbf{E}_i = E_{i} \left( \mathbf{e}_2 \cos\phi + \mathbf{e}_1 e^{j \theta_i} \sin\phi \right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}\mathbf{E}_r = E_i \left(\cos\phi\frac{1 - \alpha\beta}{1 + \alpha\beta} \mathbf{e}_2 - \sin\phi\frac{\beta - \alpha}{\beta + \alpha} \mathbf{e}_1 e^{-j \theta_r}\right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}\mathbf{E}_t = E_i\left(\cos\phi\frac{2}{1 + \alpha\beta}\mathbf{e}_2+ \sin\phi\frac{2}{\beta + \alpha}\mathbf{e}_1 e^{j \theta_t}\right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

We find that the reflected and transmitted polarization angles are respectively

\begin{aligned}\tan \phi_r = \tan \phi\frac{\beta - \alpha}{\beta + \alpha} \frac{1 + \alpha \beta}{1 - \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.36)

\begin{aligned}\tan \phi_t = \tan \phi \frac{ 1 + \alpha \beta}{ \beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.36)

where the associated magnitudes are

\begin{aligned}\frac{E_r}{E_i}= \sqrt{\left(\cos\phi\frac{1 - \alpha\beta}{1 + \alpha\beta} \right)^2+ \left( \sin\phi\frac{\beta - \alpha}{\beta + \alpha} \right)^2}\end{aligned} \hspace{\stretch{1}}(1.2.37)

\begin{aligned}\frac{E_t}{E_i}=\sqrt{\left(\cos\phi\frac{2}{1 + \alpha\beta}\right)^2+\left(\sin\phi\frac{2}{\beta + \alpha}\right)^2}\end{aligned} \hspace{\stretch{1}}(1.2.37)

References

[1] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

[2] E. Hecht. Optics. 1998.

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