Does anyone know how light paths “see” emission materials in Cycles? By that I mean, do they pass straight through as though nothing is there? Is a bounce calculated?
Strictly emissive surfaces strictly emit light, that’s it. The rays stop !
They absolutely do not use a bounce? I ask because many renderers will not display null or transparent materials, but when the light ray passes through them, a bounce is used up. This can result in “ghosts” of shapes appearing in fog, mist, or other volumes. I haven’t encountered this in Cycles but I have created a light source where a significant amount of light comes from a bounce inside the light fixture. If the light path uses up a bounce as the light passes back through the emitting light bulb, I want to make sure to compensate for that with additional bounces in the render.
You mean transmissive ? Depends, transparent BSDF counts as one ray depth but no bounce afaik, while refraction, glass, etc. count as one bounce. This is explained in the wiki.
Thanks for the input. I think I’ve got my answer from the various information online. Emission material is completely invisible to light rays. I thought that this was the answer but wanted to make sure because a number of other renderers treat the emission material as a thing, arguing that this is more physically accurate since any substance in real life would both emit and absorb light.
An emissive object casts shadows, thus it is not invisible to rays - it absorbs 100% of incoming light, reflecting nothing. In Cycles closures are pretty “strict”, as in if you want to have a material that’s mostly emissive, partly absorbing, you’ll have to mix the corresponding shaders to your liking.