Diffuse Reflection

  • #1
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I'm confused as to how the diffuse reflection is computed as :

[tex]\frac{1} {\pi} * LightRadiance * cos(\theta)[/tex]

Where [tex]cos(\theta)[/tex] 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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  • #2
Andy Resnick
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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.
 
  • #3
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
 
  • #5
sophiecentaur
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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.
 
  • #6
Andy Resnick
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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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  • #7
sophiecentaur
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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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