---
title: "Short note: Intensity attenuation, transparency"
subject: "Computer Graphics"
chapter: 8
key: "semester_5/computer_graphics:short-note-intensity-attenuation-transparency"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/short-note-intensity-attenuation-transparency
site: BCT Engineering
---

# Short note: Intensity attenuation, transparency

Computer Graphics · Chapter 8: Illumination and Surface Rendering methods

- Asked in: 2069 Asar, 4 marks
- Video: [Intensity Attenuation in Computer Graphics](https://www.youtube.com/watch?v=zrd0bN1Hlz8) — Jamie King
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/short-note-intensity-attenuation-transparency
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-8

## Question

Write a short note on intensity attenuation and 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.

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the function f(d) = min(1/(a0 + a1d + a2d^2), 1) used instead of 1/d^2?
   A. 1/d^2 does not decrease with distance
   B. 1/d^2 changes too strongly in small scenes and gives too large a value for small d
   C. 1/d^2 only works for ambient light
   D. f(d) increases with distance
   Answer: B. The constants are tuned for a good visual effect and the min caps f at 1. (Key point: Why not 1/d^2)

2. What is f(d) for a source at infinity?
   A. 0
   B. Infinity
   C. 1
   D. 1/d^2
   Answer: C. Distance has no effect for a very distant source, so there is no attenuation. (Key point: Source at infinity)

3. In the simple transparency model I = (1 - kt)*I_surface + kt*I_background, what does kt = 1 mean?
   A. The surface is opaque
   B. The surface is a perfect mirror
   C. The surface has no ambient light
   D. The surface is fully transparent
   Answer: D. kt is the transmission coefficient; 0 means opaque and 1 means fully transparent. (Key point: Transparency formula)

4. What visual effect does intensity attenuation give?
   A. Farther objects look dimmer, a depth cue
   B. Farther objects look brighter
   C. Objects look transparent
   D. Highlights disappear
   Answer: A. Weaker light from distance makes distant objects darker. (Key point: Depth cue)

## Related in Computer Graphics

- [Distance consideration in intensity](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aintensity-attenuation-distance) — Computer Graphics, chapter 8, asked 2×