- 2069 Asar · 4 marks
Short note: Intensity attenuation, transparency
Short answer
Intensity attenuation reduces light as distance d increases, using f(d)=min(1/(a0+a1 d+a2 d^2),1) to avoid 1/d^2 harshness, and provides depth cues. Transparency lets light pass through: I=(1−kt)I_surface+kt I_background, with kt the transmission coefficient; advanced models include refraction via Snell's law.
Write a short note on intensity attenuation and transparency.
Intensity attenuation
Light from a point source becomes weaker as the distance d from the source increases. In theory it falls as 1/d², but this gives too strong a change for small scenes (and too large a value when d is small), so a function is used:
f(d) = min( 1 / (a0 + a1·d + a2·d²), 1 )
a0, a1, a2 are constants chosen for the best visual effect. The diffuse and specular terms are multiplied by f(d):
I = ka·Ia + f(d)·IL·[ kd (N·L) + ks (R·V)^n ]
For a source at infinity, f(d) = 1. Attenuation gives depth cues: farther objects look dimmer.
Transparency
A transparent object lets light pass through it, so objects behind it can be seen. A simple model (for a transparent surface in front of an opaque one):
I = (1 − kt)·I_surface + kt·I_background
where kt is the transmission coefficient (0 = opaque, 1 = fully transparent). Surfaces are processed back to front (or by using the A-buffer). More accurate models also consider refraction using Snell's law (n1 sinθi = n2 sinθr) and are used in ray tracing: I = I_local + kr·I_reflected + kt·I_transmitted.
✓ Answer checked by Claude Opus 5.5 (high effort)
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