Diffuse Reflection

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Diffuse Reflection

Consider a surface that is lit by ?Point Light Sources, each emanating rays uniformly in all directions from a single point. The surface's brightness varies from one part to another, depending on the positions of the light sources.

Dull, matte surfaces - such as chalk - exhibit diffuse reflection. In diffuse reflection, a fraction of incident light penetrates the surface, and the rest is reflected uniformly in all directions. Thus, these surfaces appear equally bright from all viewing angles.

TODO Insert Figure Here (You've made it so hard guys. Uploading images should be supported)
Light beam of infinitesimal cross-sectional area $dA$ at angle of incidence $\theta$ intercepts area of $dA/cos\theta$

Figure# shows that a light beam that intercepts a surface covers an area whose size is ?Inversely Proportional to the cosine of the angle that the beam makes with the ?surface normal at that point. If the beam has an infinitesimally small cross-sectional differential area $dA$ , then the beam intercepts an area $dA/cos\theta$ on the surface. Thus, for an incident light beam, the amount of light energy that falls on $dA$ is proportional to $\theta$.

#Figure#
Diffuse reflection

Given the point, surface and ?light source configuration in #Figure#, the ?illumination equation for this model is

$I_{d\lambda}} = k_d * O_{d\lambda} \sigma{ I_{p\lambda} * max( cos\theta, 0 ) }
I_{d\lambda} = k_d * O_{d\lambda} \sigma{ I_{p\lambda} * max( \hat N dot \hat L, 0 ) }$

Where

k_d is the material's diffuse-reflection coefficient, which defines the material color.
O_{d\lambda} is the object's diffuse color
I_{p\lambda} is the intensity of the ?light source
\hat N is the ?unit-length ?normal to the surface at the given point
\hat L is the ?unit-length vector from the point to the ?light source

If the surface is pointing away from the light-source, $cos\theta$ becomes negative. Given that we take the maximum of the cosine, and zero, the outcome becomes zero. So, the surface color intensity then becomes 0, which translates to a pure black color – which is very unrealistic. This is due to the fact that local illumination models don't take into account that light is reflected off - and is thus received from - all objects, not only ?light sources. To account for this, diffuse lighting is often combined with an ambient light term to approximate the ?global illumination effect, giving the following ?illumination equation

$I_{d\lambda} = k_a * O_{d\lambda} + k_d * O_{d\lambda} \sigma{ I_{p\lambda} * max( \hat N dot \hat L, 0 ) }$

Next: ?Light Attenuation

last edited (July 4, 2004) by MHaggag, Number of views: 2492, Current Rev: 1

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