How does one calculate the polarization state of random light after having been totally reflected by a single dielectric interface? Please consider pure specular reflexions from a plane interface between two dielectric mediums of indexes $n_1,\,n_2$ when the angle of incidence $\theta_1$ is greater than the critical angle $\arcsin(n_2/n_1)$.

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    $\begingroup$ I think you wrote it quite well. By the way, you are like the optics lord here! $\endgroup$ – Yossarian Jan 9 '14 at 0:08

I'll stick to pure total internal, specular reflexion in this answer.

When TIR happens, both linear polarisation components are fully relfected, but the phase change (the Goos-Hänchen shift) is different for the two states. In scalar theory the Goos-Hänchen shift (see my answer here) for the two states, but the full vector theory shows a subtle difference. What this means practically is that the two polarisation states seem to reflect from ever so slightly different depths into the denser medium beyond the totally internally reflecting interface.

The Fresnel equations still apply in this situation. Now, of course, we get $\sin\theta_t>1$ so that $\cos\theta_t = \sqrt{1-\sin\theta_t^2}$ is imaginary. We interpret the trigonometric functions exactly as they are interpreted in the derivation of the Fresnel equations (e.g. in Reference [1]), to wit, the sine and cosine are the ratios $k_x/k$ and $k_z/k$ of the wavevector components tangential and normal to the interface, respectively. The cosine is imaginary beyond the interface, so the wave is evanescent and exponentially decaying with depth as in my answer cited above. So, from the Fresnel equations:

$$r_s = \frac{n_1 \cos \theta_i - n_2 \cos \theta_t}{n_1 \cos \theta_i + n_2 \cos \theta_t} = \frac{n_1 c_1 - i\, n_2 c_2}{n_1 c_1 + i\,n_2 c_2}$$

$$t_s = \frac{2 n_1 \cos \theta_i}{n_1 \cos \theta_i + n_2 \cos \theta_t}= \frac{2 n_1 c_1}{n_1 c_1 + i\,n_2 c_2}$$

$$r_p = \frac{n_2 \cos \theta_i - n_1 \cos \theta_\text{t}}{n_1 \cos \theta_t + n_2 \cos \theta_i} = \frac{n_2 c_1 -i\, n_1 c_2}{n_1 c_2 + i\, n_2 c_1} = -i$$

$$t_p = \frac{2 n_1\cos \theta_i}{n_1 \cos \theta_t + n_2 \cos \theta_i}= \frac{2 n_1 c_1}{n_1 c_2 + i\,n_2 c_1}\tag{1}$$

where $c_1 = \cos\theta_1$ and $c_2 = \text{Im}(\cos\theta_2)$ are both real numbers. Take heed that the reflexion co-efficients are either a complex numbers of the form $r_s=z^*/z$ or $r_p = -i$ and thus have unity magnitude, so all the power is reflected. The Goos-Hänchen shifts for the two orthogonal linear polarisations are:

$$\phi_s = -2\,\arg(n_1 c_1 + i\,n_2 c_2) = -2\,\arctan\left(\frac{n_2 c_2}{n_1 c_1}\right)$$

$$\phi_p = -\frac{\pi}{2}\tag{2}$$

which are almost equal in most cases: they deviate more significantly from one another for highly glancing angles $\theta_1\approx\pi/2$ (which condition invalidates the scalar theory of my other answer).

In particular, as the incidence angle approaches $\pi/2$ and the reflexion becomes highly grazing, $\phi_s\to-\pi$ and $\phi_p\to-\pi/2$. That is, the TIR mechanism mimics a quarter wave plate.

So, for your random polarisation, you are going to either represent it as a known, pure polarisation with a $2\times1$ Jones vector $X = \left(\begin{array}{c}x_p\\x_s\end{array}\right)$ and transform it by:

$$X\mapsto U\,X;\;\text{where}\;U = \left(\begin{array}{cc}e^{i\,\phi_p}&0\\0&e^{i\,\phi_s}\end{array}\right)$$

or if the light is depolarised (a mixed state) you will use the Mueller calculus / density matrix formalism:

$$\rho\mapsto U\,\rho\,U^\dagger;\;\text{where}\;\rho = \sum\limits_{j=0}^3 s_j \sigma_j$$

$\sigma_0 = {\rm id}$ is the $2\times 2$ identity matrix, $\sigma_j$ the Pauli spin matrice and the co-efficients $s_j$ are the four Stokes parameters as described in my answer on dealing with calculations with depolarised light.


[1] §1.5 "Reflexion and Refraction of a Plane Wave" in the seventh edition of Born and Wolf, "Principles of Optics".


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