For a transparent or translucent object there may be light sources behind the object so there is transmitted or refracted light. There can be both diffuse refracted light, as in frosted glass, and specular refracted light, as with clear glass or water. The modeling of translucent objects with diffuse refracted light is difficult so we will only do specular refracted light. In scan line rendering we usually ignore refraction or the "shifting" effect and only use transmitted light. Note that Ray tracing easily handles the refraction of light (see the next section).
In scan-line graphics the intensity of the background object is added to the intensity of the transmitting surface:
I = r * It + (1 - r) * Ib
Ib = intensity background, It = intensity surface, r = weighting factor
For a perfectly transparent surface r = 0 and I would be only the background object intensity. For an opaque surface r = 1 and I would be only the front surface intensity.
When light passes through different types of media, it has different speed.This causes refraction or the apparent bending of light. Example of a ruler in water
The Index of refraction for a substance i is defined as:
hi = (speed of light in vacuum) / (speed of light in substance i)
Substances have h > 1.0
h air = 1.0003
h water = 1.33
h vacuum = 1.00
h is a function of the wavelength of light.
Example: Fused quartz at 180 C, h = 1.46 (589 nm)
Snell's law : sin q1 / sin q2 = h2 /h1 = h21
Different h gives different bending angles, which is how a quartz prism separates the light. One possible behavior of light at boundaries is total internal reflection (TIR) when light passes from a dense to a less dense medium.
If the reflection angle is beyond critical angle, then there is no transmission, only reflection. This is the basis for using optical fiber for high speed computer networking. When the reflection angle exceeds the critical angle for TIR, just use specular reflection and ignore transmission. So, we need to find the critical angle.
Look at 2 physical laws.
Let Sin qt = St and Sin qi = Si
then Snell's law: St / Si = hti , or Si / St = hit = 1/hti
Now I, N, T are co-planar, i.e., T
= aI + bN
cos qi = Ci = N · (-I)
cos qt = Ct = (-N) · T
We will look at both algebraic and geometric solutions.
St / Si = hit = h => St^2 = Si^2 * h^2
S0^2 + C0^2 = 1 => (1 - Ci^2) * h^2 = (1 - Ct^2)
(1 - Ci^2) * h^2 - 1 = Ct^2 = (-N · T)^2
= [-N · (aI + bN)]^2
= [a(-N · I) + b(-N · N)]^2
(1 - Ci^2)h^2 - 1 = [aCi - b]^2
Now: want T · T = 1
or 1 = (aI + bN) · (aI + bN)
= a^2(I · I) + 2 ab (I · N) + b^2 (N · N)
= a^ - 2ab Ci + b^2
Solve two equations for a, b, and get four solutions:1 for each quadrant. From physics the only valid real solution is:
T = hit I + [hitCi - (1 + hit^2(Ci^2 - 1))^.5] N
when (1 + hit^2(Ci^2 - 1))^ .5 is negative then there will be no real solution, and we have TIR and no transmitted light.
The above gives us: T = hitI + [hitCi - (1 + hit^2(Ci^2 - 1)) ^.5] N
Example: Compute the critical angle for crown glass with hit = 1.52
0 = 1 + h^2(Ci^2 - 2) = 1 + h^2*Ci^2 - h^2
h^2 = 1 + h^2*Ci^2
h^2 - 1 (h^2 - 1)^.5 (1.52 - 1)^.5 --------- = Ci^2 => Ci = -------------- = ---------------- h^2 h 1.52
= .753 = cos q => q = 41 degrees
T = 1/n L - (cos q2 - 1/n
cosq1) N where n = n1/n2 cos q2 = [1 - 1/n^2 ( 1
Note: n = h
After the substitutions, this gives the previous result, i.e., T = hitI + [hitCi - (1 + hit^2(Ci^2 - 1)) ^.5] N
For perfect specular transmission, light passes through with no interference. For perfect diffuse transmission, light may hit many small particles that bounce and scatter photons around the substance. This colors the light, but can't see anything from the other side. So all the light arriving on other side can contribute.
Specular transmission adds the following term to the Phong
local illumination model:
Ip * Kt * (T · V)^ m.
Image Refererences: Images in this section were taken from
Andrew S. Glassner, "Surface Physics for Ray Tracing" in An Introduction to Ray Tracing, Andrew Glassner, ed., Academic Press Limited, 1989.
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