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Reflection and transmission (Fresnel equation) of polarized light are treated in many optics or electromagnetism books.

If $E_s$ and $E_p$ is incident electric field with s-polarization and p-polarization, respectively, reflected electric field would be $E'_s = r_s E_s$ or $E'_p = r_p E_p$, where $r$ is reflection coefficient and $'$ means reflected field.

Then electric field of any polarized light can be calculated by decomposition.

For instance, since $E_{45^{\circ}} = \frac 1 {\sqrt{2}} ({ E_p + E_s })$, $E'_{45}$ would be $\frac{1}{\sqrt 2} (E'_p + E'_s )$

However, how can I express electric field of reflected unpolarized light?

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    $\begingroup$ If you have unpolarized light, will the mirror change that? $\endgroup$ – yankeefan11 Aug 13 '13 at 12:44
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Suppose the Fresnel equations give us complex reflexion co-effcients $R_p$ and $R_s$ for $p$- and $s$-polarized light, respectively. Then the intensity reflexion co-efficient (power reflexion coefficient) for depolarized light is (in most cases):

$\frac{1}{2} (|R_s|^2 + |R_p|^2)$

You do likewise for the transmission co-efficients, so that the transmitted power ratio is:

$\frac{1}{2} (|T_s|^2 + |T_p|^2) = 1- \frac{1}{2} (|R_s|^2 + |R_p|^2)$

where $T_p$ and $T_s$ are the Fresnel equation-derived complex transmission co-efficients for $p$- and $s$-polarized light. Forming average square magnitudes like this is often called "incoherent summing".

To understand fully how to do your calculation, you need to understand exactly what depolarized light is, and it has quite a complicated description: it is bound up with decoherence and partially coherent light, a topic which Born and Wolf in "Principles of Optics" give a whole chapter to. A classical description, roughly analogous to Born and Wolf's is as follows: if the transverse (normal to propagation) plane is the $x,y$ plane, then we represent the electric field at a point as:

$\mathbf{E} = \left(\begin{array}{cc}E_x(t) \cos(\omega t + \phi_x(t))\\E_y(t) \cos(\omega t + \phi_y(t))\end{array}\right)$

where $\omega$ is the centre frequency and the phases $\phi_x(t)$, $\phi_y(t)$ and envelopes $E_x(t)$, $E_y(t)$ are stochastic processes, which can be as complicated as you like. The formulas I cite above just assume that:

  1. $E_x$, $E_y$ and $\phi$ behave like independent random variables, and
  2. They vary with time swiftly compared to your observation interval (the time interval whereover you gather light in a sensor to come up with an "intensity" measurement) but not so swiftly that the light's spectrum broadened so much that we cannot still think of the light as roughly monochromatic.

A simple quantum description is actually conceptually clearer than Born and Wolf's classical one, as long as light states do not become entangled. Each photon can be thought of as a perfectly coherent wave propagating following Maxwell's equations. The Fresnel equations thus apply to each photon as they would to a perfectly coherent wave. For each photon, therefore, you calculate the intensity of reflexion and transmission, and then average this intensity over all photon polarization states - we assume the source is creating "random" pure states. Thus, suppose the Fresnel equations give us complex reflexion co-effcients $R_p$ and $R_s$ for $p$- and $s$-polarized light: the complex amplitude reflexion co-efficient for a general polarization state is then:

$R(\alpha, \phi) = \alpha R_p e^{i \frac{\phi}{2}} + \sqrt{1-\alpha^2} R_s e^{-i \frac{\phi}{2}}$

where $\alpha \in [0, 1]$ and $\phi \in [0, 2\,\pi)$. Summing intensities over all values of $\phi$ (assuming all phases equally likely) yields:

$\frac{1}{2\pi}\int\limits_0^{2\pi} \left(\alpha^2 |R_p|^2 + (1-\alpha^2) |R_s|^2 + 2 \alpha\sqrt{1-\alpha^2} |R_p| |R_s| \cos\phi\right)\mathrm{d}\phi = \alpha^2 |R_p|^2 + (1-\alpha^2) |R_s|^2$

and then summing intensities over all values of $\alpha^2$ (assuming the $\alpha^2$ is uniformly distributed in $[0,1]$) leaves the formulas above.

This will not give a full picture for general entangled polarization states, when you have to resort to more general coherent and cross correlation functions to describe what is going on. Likewise for Born and Wolf's classical description. But it is an excellent first approximation and it is probably true to say that it is hard to arrange for it not to hold in the laboratory. Deviations from it are likely to be seen if you sample the light intensities over very short sampling intervals, when you will see complicated, extremely swift fluctuations in scattered and transmitted intensities, often following white noise processes.

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  • $\begingroup$ Would you say that unpolarized has 50% of s-polarization or none, since it's not polarized at all? $\endgroup$ – jinawee Dec 26 '14 at 10:06
  • $\begingroup$ Thanks, I was asking this to get things clear about a question I asked: physics.stackexchange.com/questions/154828. $\endgroup$ – jinawee Dec 28 '14 at 20:32
  • $\begingroup$ Since you talk about single photons here, and all polarization, including partial polarization, is the result of time (or ensemble) averaging some coherent field, what is the phase of a single photon? Does the phase of a single photon even make sense? $\endgroup$ – daaxix Aug 17 '15 at 6:44
  • $\begingroup$ @daaxix Yes it does. If you look at my answer here and, further, at the details of the argument in the Scully and Zubairy reference I cite there, there is a vector valued function of space and time that (1) uniquely defines, and is uniquely defined by, a one-photon state of the EM field and (2) fulfils Maxwell's equations. Local phases are most certainly meaningful here, for they govern diffraction. Of course global phase changes have no effect. I think you are thinking of a photon as a particle on an empty background, whereas ... $\endgroup$ – WetSavannaAnimal Aug 17 '15 at 7:08
  • $\begingroup$ @daaxix ... one does better to think of one entity permeating spacetime -the quantum EM field.Here we're thinking about this beast in a one photon Fock state.One doesn't of course need to bring up photons here: a completely classical description is more than enough. It's simply that I find the classical description messy and unintuitive for partial polarization. When you get to full analysis, both paradigms are equally difficult, but to think of the (unentangled) photons as the units of coherence, then think of depolarized light as classical mixtures of these pure states seems the easiest way. $\endgroup$ – WetSavannaAnimal Aug 17 '15 at 7:13

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