# Tag Info

## Hot answers tagged fresnel

12

Using two Fresnel terms is correct in the sense that any given diffuse path will pass through the surface twice. If you're solving diffusion by tracing a path through the medium until it bounces out again then that you will get two (or more) Fresnel terms for that path as it interacts with the surface. However, that's not what you're doing with a diffuse ...

11

Warning: I am not a physicist. As Dan Hulme already explained, light can't travel through metals, so dealing with IOR is a lot more... complex. I will answer why that happens and how to calculate the reflection coefficient. Explanation: Metals are filled with free electrons. Those electrons react to external fields and reposition until electrostatic ...

7

The Fresnel coefficient should be evaluated using $H$, not $N$. You wrote, I have trouble seeing why we can still use that formula in a BRDF, which is supposed to approximate the integral over all the hemisphere. It's not. The BRDF in itself does not approximate the integral over all the hemisphere. The rendering equation does that: you integrate over ...

7

In Schlick's 1994 paper, "An Inexpensive Model for Physically-Based Rendering", where they derive the approximation, the formula is: $$F_{\lambda}(u) = f_{\lambda} + (1 - f_{\lambda})(1 - u)^{5}$$ Where So, to answer your first question, $\theta$ refers to the angle between the view vector and the half vector. Consider for a minute that the surface is a ...

7

While browsing to properly write my question, I actually found the answer, which happens to be very simple. Another Fresnel term is also going to weight in as the photons make their way out of the material (so being refracted into the air) and become the diffuse term. Thus the correct factor for the diffuse term would be: $$(1 - F_{in}) * (1 - F_{out})$$

7

Yes, because refractive index can vary with wavelength. This is the origin of colored specular reflection in metals such as gold and copper; most other materials have essentially uncolored specular. At each wavelength, the specular reflectance at normal incidence is related to the refractive index at that wavelength according to the formula you mentioned. ...

6

Look at the refractive index of several metals. They are all complex numbers and the math does work out when you put this into the fresnel equation: you get the expected high reflectivity at all angles. There are also subtle color shifts because the index depends on wavelength. This is actually used in rendering but it is not common. The function is ...

5

I finally figured out a flaw in my argumentation to use the half vector for the diffuse part. tl;dr version: $\alpha_{hi}$ and $\alpha_{ho}$ are not equal, this assumption only works for the specular part. Therefore the energyconservation is not given. More correct: Per definition $\alpha_{hi} = \alpha_{ho}$, but you are not allowed to use them in the ...

4

This is the complex number version of refraction, were K is the extinction coefficient. This is commonly used for metals. You can check the Wikipedia on refraction: Complex refractive index | Wikipedia

4

I wasn't really expecting that, no. The formula in the paper is not the most elegant - there's quite a few parentheses in there. In this case I think it's just a matter of shuffling the parentheses around a little in the get_k implementation - both terms should be divided by (1-r): float get_k(float r, float n) { return std::sqrt( 1.0/(1.0-r) * (...

3

Unfortunately, the iridescence model is not made to be applied to a diffuse term. Pascal and I made it for microfacet models only (that is the specular term). One way to understand how to include it to a game engine might be to look at Unity's HDRP implementation. In the Lit.hlsl to see how to incorporate the iridescence Fresnel into a specular + diffuse ...

3

@PaulHK's answer is correct I'm sure, here's a bit of a check to show that the IOR() function is calculating the reflection coefficients for $s$ and $p$ polarizations then averaging the two assuming unpolarized incident light. The reflection is for a single interface, and at least at normal incidence the results reduce to the simpler $(n_2-n_1)^2/(n_2+n_1)^... 2 Physically, the origin of diffuse light is subsurface scattering, which happens continuously as light travels through a material. So, the proportion of transmitted light depends on the thickness of the object. There's no precise equivalent to the Fresnel law, but maybe the closest thing is the Beer–Lambert law. It states that the transmitted light falls off ... 1 We have$t_{12} = t_{21}$, by the law of reciprocity. Also, $$r_{12} + t_{12} = 1\\r_{21} + t_{21} = 1$$ It follows that$r_{12} = r_{21}\$ as well.

1

We're actually going through that paper for our own GGX BRDF metallic & edge_tint model, and we've spotted one crucial problem. Hammon is still using the Fresnel Schlick approximation for his microfacet field ray-trace which in reality only works for dielectrics (and conductors but with a hack). But we went all out on actual Fresnel (discarding ...

1

The refractive index is related to the speed at which light travels through the medium, and only applies to materials which are at least partially transparent. Metals are electrically conductive, so they are opaque, so light can't travel through them at any speed, so they don't have a refractive index. This is why Fresnel's law doesn't apply: it's for ...

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