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I am interested in ray-tracing polarized photons. I have code that works very well for unpolarized light. When a ray hits a dielectric interface the photon is either reflected or refracted by comparing the Fresnel coefficients with a random number.

The calculation of the reflected or refracted direction vector is relatively simple. Say we have a direction vector $\boldsymbol{v}$ and surface normal to the dielectric interface $\boldsymbol{n}$.

Reflected direction vector

$$ \boldsymbol{v} - 2d\boldsymbol{n} $$

Refracted direction vector

$$ N\boldsymbol{v} + (Nd - c) \boldsymbol{n} $$

In the above $d=\left(\boldsymbol{n}\cdot\boldsymbol{v}\right)$, $N=\frac{n_1}{n_2}$ which is the ratio of the refraction index at the interface, and $c =\left(1 - N^2\left(1 - d^2\right)\right)^{1/2}$. These transformation are easily applied within the ray tracing loop because call values are known.

Question

My question is what transformation needs to be applied to calculate the correct polarisation vector of a reflected or refracted ray? Before hitting the interface the polaristaion vector of the ray $\boldsymbol{k}$ is known (I assume linear polarised states only). Clearly $\boldsymbol{k}\cdot\boldsymbol{v}=0$, but where in the plane perpendicular to the direction vector $\boldsymbol{v}$ does the polaristaion vector $\boldsymbol{k}$ lie? Can I apply a simple transformation, similar to the above, to find the new polaristaion vector?

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  • $\begingroup$ I actually do this somewhere in a code I wrote about ten years ago. I shall have a look at it and get back to you if no one else answers, but I'm pretty sure that I just brute force it: my ray objects carry also polarization vectors, one resolves into $s$ and $p$, applies the Fresnel equations and reassembles the new rays. I don't think I found any particular elegant transformations for this. The Fresnel equations, being so nonlinear, would seem to rule out a simple solution like you're susggesting. What's the grey square BTW? $\endgroup$
    – Selene Routley
    Commented Sep 3, 2013 at 14:08
  • $\begingroup$ Well it is grey only on stackoverflow :) Thanks for the offer. So rather than have one vector transformation, you recommend transforming the components. But actually I still don't understand exactly how to do that, how do you determine the reflection polarisation vector for TE and TM components? $\endgroup$
    – boyfarrell
    Commented Sep 3, 2013 at 14:25

2 Answers 2

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You have separate Fresnel equations for s- and p-polarized light. The two polarizations reflect/refract separately. You can reconstitute them on the other side to recover the new polarization vector if you want.

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  • $\begingroup$ I can decompose into s and p components by calculating the plane of incidence for the ray. In the above is $\bf{k}$ the polarisation vector (normally $\bf{k}$ mean direction). What is the meaning the $\bf{n}$ terms? $\endgroup$
    – boyfarrell
    Commented Dec 5, 2013 at 0:07
  • $\begingroup$ Yes, that's what you are supposed to do. What I wrote was crap :) $\endgroup$
    – lionelbrits
    Commented Dec 5, 2013 at 1:31
  • $\begingroup$ I realize now this is essentially WetSavannaAnimal's comment. $\endgroup$
    – lionelbrits
    Commented Dec 5, 2013 at 1:33
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The only part that needs to change is "the photon is either reflected or refracted by comparing the Fresnel coefficients with a random number."

If you know both the polarization and the phase, you can calculate whether to reflect or refract. No longer does the number need to be random.

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  • $\begingroup$ Isn't reflection from an interface was inherently probabilistic? In any case, the Monte-Carlo method requires that one can calculate a probability (this is where the reflection coefficients are used, and I can do this easily for unpolarised or polarised photons) and compared with a random number. The result determine the next path of the photon and the direction vector gets transformed accordingly. Where I am stuck is on how to transform the polarisation vector. $\endgroup$
    – boyfarrell
    Commented Sep 4, 2013 at 2:10
  • $\begingroup$ Probabilistic if you dont know the phase. Depending of the phase of the photon, it has a higher or lower chance of begin reflected or refracted. $\endgroup$
    – AnimatedPhysics
    Commented Sep 5, 2013 at 3:01
  • $\begingroup$ I didn't know that, do you have a reference I can follow? $\endgroup$
    – boyfarrell
    Commented Sep 5, 2013 at 7:43
  • $\begingroup$ I like the description about 7:45 of Feynman_youtube lecture_on_light where Feynman talks about something that 'follows the particle along changing its disposition to do things' - ie, it reflects or refracts depending on the phase. $\endgroup$
    – AnimatedPhysics
    Commented Sep 7, 2013 at 2:06
  • $\begingroup$ "Probabilistic if you dont know the phase. Depending of the phase of the photon, it has a higher or lower chance of begin reflected or refracted" -- it is still probabilistic then, isn't it? $\endgroup$
    – lionelbrits
    Commented Dec 5, 2013 at 1:33

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