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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]


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?


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.


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.


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.


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.


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)


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

[2] E. Hecht. Optics. 1998.

5 Responses to “Derivation of Fresnel equations for mixed polarization (using Geometric Algebra)”

  1. Antonio said

    Peeter, how can we demonstrate that when an incident TE wave hits a surface, both reflected and transmitted wave maintain the original linear polarization? Similarly for a TM it is assumed (in all Fresnel law derivations) that the initial linear polarization is preserved. Are the boundary conditions sufficient to prove that?
    In fact this happens only in these two very special initial conditions, in every other case, an initial linear polarization is modified. Antonio

    • peeterjoot said

      I believe that this is what I was referring to when I wrote:

      In ‘[1] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.’, problem 9.14 we had to show that the polarization angles for normal incident (E k e 1 ) must be the same due to the boundary constraints

      I showed using 1.2.32 above that an input that is either a pure TE or pure TM mode, has a solution where both the reflected and transmitted portions are of the same mode. However, I don’t think I really showed the opposite: that it must be so (i.e. what was desired in the Griffiths problem). To tackle that problem I think you would have to set the input phase angle \phi_i to one of 0, or \pi/2 and then show from 1.2.32 that \phi_r and $\phi_t$ must match.

      • Antonio said

        Thanks for your quick reply. Indeed I was wandering if showing valid coherent solutions was sufficient, specially because in this problem we provide actually two valid solutions which are linearly independent. I looks like we are addressing every case, providing two particular solutions. I think we are complete in the space, I am not sure however if this intuitive reasoning is mathematically correct. I have looked a lot in the literature, everyone seems satisfied providing these two solutions, so I suspect that just finding them complete the demonstration. Feynman instead argues that because the dipoles in the dielectric are moved by the by a linearly polarized field Ei, they can only produce linearly polarized fields, however that would not work in polarization different from TE or TM, I don’t think the inference is solid, he also states though that the results can be derived mathematically and not physically, using the boundary conditions.

      • peeterjoot said

        Here’s the Griffith’s problem:

    • peeterjoot said

      I attempted the Griffiths problem 9.14 again (it was for a normal incident wave). There end up being exactly four solutions that satisfy the boundary value constraints, but they are all physically the same (example: one toggles the sign of \mathbf{E}_{\mathrm{R}}, but it also toggles the sign of that reflected waves’ polarization vector in a way that compensates. I’ll post that problem’s solution later. Short story, for normal incidence the polarization is maintained. I suspect the same thing can be shown for non-normal incidence, but that the math will get messier.

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