The specular color in metalness/roughness workflow is usually defined as following:
float3 specColor = lerp(0.03f, albedoColor, metallic);
The Cook-Torrance BRDF is given with the following formula:
The final color will be:
Ci = (diffColor*(n.l)+Ks*specColor*cook);
My question is - won't specular be too dim for non-metals? Cook-Torrance term will be multiplied by 0.03 in case of dielectrics - this will make the specular component virtually non-existent. This doesn't seem realistic to me, because smooth plastic reflects a lot of light, almost as much as metals:
What am I missing?