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Pretty all in the question, but to add a bit of detail, I'm asking this regarding to the reflectance curves of refractive.info website wich have P/S/non polarized curves.

I'm using Cycles fresnel to a color input of an emission shader.

Edit:

  1. In fact I must be more precise, I'm looking without any transformation in the color management, so pure data, and I know I will not have the exact reflectance curve with this scene.
  2. To be more precise, those are the curves I'm speaking about:

And to respond to @JtheNinja I assume the k fresnel coefficient extinction is the part of the curve that drop down before jumping to 1 on the P polarized.
If I'm right about the k, we have this hardcodded to the fresnel node:
My curve is x compressed at the end because it's hard to deal with grazing angles and making a linear progression. But you can see the curve drop down even with the default fresnel node.
At the begining I was choosing the non polarized but looking at the fresnel curves I have, they look closer to the P polarized, so the question remains. What are P/S and non plolarized and which one to choose if I what to make fresnel with RGB curves?

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  • $\begingroup$ I don't believe that an Emmision Shader has reflectance. As it is a source of light. $\endgroup$ – Rick Riggs Dec 16 '16 at 20:58
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    $\begingroup$ Ok I was thinking just saying "I'm using Cycles fresnel to a color input of an emission shader." will be enough but I'll add an image. $\endgroup$ – Mareck Dec 16 '16 at 21:14
  • $\begingroup$ I don't understand your question, but you can surely look at wikipedia if you want to understand what p and s polarizations are: en.wikipedia.org/wiki/Plane_of_incidence#S_and_p_polarizations $\endgroup$ – lbalazscs Dec 17 '16 at 15:11
  • $\begingroup$ Thanks for the link @lbalazscs it helps a bit but I've really understood after reading this one: physicsclassroom.com/class/light/Lesson-1/Polarization I think I'll start an answer with what I understand for now and I'll complet/correct when new infos come. $\endgroup$ – Mareck Dec 18 '16 at 20:47
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I'm answering my own question because I start understanding a bit more about the topic of polarized light. I'm niether a scientist nor a mathematician so I will simplify things but don't hesitate to correct me or give more information in the comment so I can update/complete the answer.

What I know :

After reading more about Polarization :

  1. A ray light is a wave moving multiple directions, 2 plans x and y (non polarized)
  2. A non polarized light is P polarized after bounced on a dielectric material (water/glass/plastic...).
  3. Bounces on metals are non polarized.
  4. We haven't have spectral rendering in Cycles, so the light isn't acting as a wave, no polarization of rgb light.

What I assume :

  1. Light in Cycles must be think like non polarized

  2. Defaut fresnel considerate that most light is polarized because it seems to apply a P polarized curve on fresnel

  3. Direct ray need non polarized fresnel curve, and after one bounce on dielectric surface, it need P polarized fresnel curve.

Edit :

The second assumption isn't right because I was assuming that the drop down of the curve was a sign of polarization, in fact it's a sign that default fresnel node isn't suitable for metallic shaders (dielectrics are ok).
The third assumption isn't right either, first because if a ray hasn't bounced on something there is no fresnel curve, second because polarized light is way more complicated than that, and we haven't got any way to know which ray is polarized or not. So the best choice is to use unpolarized curve.

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Just n, the simple IOR. Cycles, like most raytracers, does not implement the other stuff as it's usually not important for artistic use. k is sometimes used for metallic shaders, but Cycles does not currently support these.

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  • $\begingroup$ In fact I'm searching for more in depth information. Perhaps you can explain what correspond to the 3 curves. If I'm right it's the light that is polarized, so when in real life the light is polarized S/P or not. And what is the k you speak about? $\endgroup$ – Mareck Dec 17 '16 at 1:54
  • $\begingroup$ k represents the extinction coefficient. It describes how light is absorbed, but I confess i don't understand it well. It is not a significant value for typical dielectric substances such as glass, plastic, tile. It's mainly has a significant effect in metallic reflections. Not sure which 3 curves you're talking about? The graphs on that site just show how the selected values change over wavelength. $\endgroup$ – JtheNinja Dec 17 '16 at 3:08
  • $\begingroup$ Ok I should have known that the question need more in depth infos to have more in depth answer. The curves wich I'm talking about are a bit a the bottom, you need to scroll down in the site. I'll add an image of the curves and what I think is the k coefficient. $\endgroup$ – Mareck Dec 17 '16 at 13:48

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