# Diffuse Reflection I'm confused as to how the diffuse reflection is computed as :

$$\frac{1} {\pi} * LightRadiance * cos(\theta)$$

Where $$cos(\theta)$$ is the angle between the light and the surface normal, and accounts for the light projected onto the surface

When considering light reflecting at a point from some direction wi (on a diffuse surface), which is scattered in an infinite amount of directions, how can the BRDF function only mulitply the incoming by 1/pi?

Wouldnt the outgoing light in any direction be drastically smaller than the incoming light?

I dont understand why dividing it by ~3.14 works.

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Andy Resnick
IIRC, the factor 1/pi is used to normalize the total reflectance- integrating the BRDF over the 2*pi hemisphere should give '1', for non-absorbing materials.

Where did you read about that? I've been looking for an explanation on google but can't seem to find one, all I find is the actual formular itself.

Do you know of any books that exlain it in depth?

Thanks

sophiecentaur
Gold Member
2020 Award
That diagram seems correct and so does the 1/pi factor but, on its own, that information is counter intuitive and needs to be put in context. When you have a diffuse reflection, light hits the observer from a wide area on the reflector , whereas with a specular reflection, only the light from a single point reaches the observer - it is only for an infinite / large reflector that the 1/pi factor tells you just how much light, in a particular direction, you will get from a diffuse reflection. If a rough bit of dust is the reflector then you will only get a tiny amount to the observer.

Andy Resnick
That diagram seems correct and so does the 1/pi factor but, on its own, that information is counter intuitive and needs to be put in context. When you have a diffuse reflection, light hits the observer from a wide area on the reflector , whereas with a specular reflection, only the light from a single point reaches the observer - it is only for an infinite / large reflector that the 1/pi factor tells you just how much light, in a particular direction, you will get from a diffuse reflection. If a rough bit of dust is the reflector then you will only get a tiny amount to the observer.
The BRDF doesn't relate to detection- it simply computes how the scattered power is distributed in solid angle.

A Lambertian surface is a model of a perfect diffuse reflector (plain white paper is approximately Lambertian, so is the lunar surface). More complex models (Kubelka-Munk, Brown and Miller) exist as well:

http://webstaff.itn.liu.se/~freco/Publications/Courses/Paper_optics_presentation.pdf [Broken]

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sophiecentaur
Gold Member
2020 Award
The BRDF doesn't relate to detection- it simply computes how the scattered power is distributed in solid angle.

A Lambertian surface is a model of a perfect diffuse reflector (plain white paper is approximately Lambertian, so is the lunar surface). More complex models (Kubelka-Munk, Brown and Miller) exist as well:

http://webstaff.itn.liu.se/~freco/Publications/Courses/Paper_optics_presentation.pdf [Broken]
Yes; fair enough. The point I was making is that the actual fraction of light emerging in a particular direction from a perfect diffuser (an infinitessimal) must be infinitessimal. Compare what would hit you in the eye from a laser beam, reflected in a clean mirror, and what you would end up in your direction if the beam hit a diffuse reflector. The 1/pi factor could not apply there. The integral over a hemisphere must be the same as the incident power in a 'single' direction.

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