# Fresnel and specular colour

I'm trying to implement Cook-Torrance BRDF and had previously bookmarked this question as it looked very well thought out. Going through it, I'm confused about the Fresnel equation. The author has the equation:

$$F = c_{spec} + (1 - c_{spec}) (1 - \mathbf{w_i} \cdot \mathbf{h})^5 \tag{1}$$

and says "$c_{spec}$ is the specular color"

Looking elsewhere, I can see this equation is Schlick's approximation, but in other references, $c_{spec}$ is called $R_0$ and is $\left(\frac{n_1-n_2}{n_1+n_2}\right)^2$ which is clearly a number based upon refractive indices and not a colour.

Does equation $(1)$ make sense with colour?

• Inspired by your question, I've posted a related one regarding the use of more accurate Fresnel approximations in a RGB-based path tracer. I am just commenting it here in the case you are interested. – Christian Pagot Mar 19 '17 at 21:06
• Brilliant. I have tried to implement gold appearance but don't think it looks right and have been reading more about Fresnel, extinction coefficients, etc. I'm not on this full time so it's going slow. Thanks. – PeteUK Mar 19 '17 at 22:32 • Thanks. So to come up with the specular colour for gold, I'd need to lookup the wavelengths for red, green, and blue components. With those wavelengths I lookup the corresponding refractive indices for gold. Then plug them in the formula for $R_0$ and those 3 values are the components of the specular colour? – PeteUK Feb 25 '17 at 0:56
• @PeteUK Yep, as an easy way, you can use representative wavelengths for R, G, B like that. Probably a more accurate way is to integrate $R_0(\lambda)$, multiplied by one of the CIE color matching functions, over wavelength. – Nathan Reed Feb 25 '17 at 0:59