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# Light through translucent cube

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Imagine there's a tranclucent cube and a lightray goes through it:
     ________
|      |
100% |      | 50% intensity
*****|* * * |*****
|      |
|______|

After going through the cube, the ray has 50% intensity left. Now, if the ray would go through a cube made of the same material, but twice as thick:
     ______________
|            |
100% |            | ???% intensity
*****|* * * * * * |*****
|            |
|____________|

Then how much intensity should the ray have after going through the material? How much intensity will the lightray have after going through the material for a given length? Thanks. P.S. No this is not a homework question, but me who wants to try to make voxels in a voxel engine translucent. [edited by - lode on November 18, 2002 9:06:25 PM]

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My best guess is that it''s probably an inverse-squared relationship. In other words, given the same opacity of the material, for every factor of a base thickness, the opacity decreases by that factor squared. So if you double the thickness, you''d cut the intensity by a fourth; multiply the width by 6, and you decrease the intensity by a factor of 36, etc.

However that''s just an educated guess. I guess it would depend on the material really.

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Think of it as photons and statistics. Consider your original cube. A particular photon has a 50%, or 0.5, chance of being absorbed, and a 0.5 chance of being allowed through.

Now consider the double-width cube. Essentially, it''s just two of the cubes stuck together. So in order for a photon to travel through, it needs to be allowed through both. The chance of this happening is 0.5*0.5 = 0.25, or 25%. In other words, 25% of the photons will get through, on average.

You can extend this to a formal equation. If q is the percentage of light allowed through a translucent surface of width 1, then the percentage of light Q let through a translucent surface of width w, as a function of w, is Q(w)=qw.

Don''t listen to me. I''ve had too much coffee.

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> ... If q is the percentage of light allowed through a
> translucent surface of width 1, then the percentage of light Q
> let through a translucent surface of width w, as a function of > w, is Q(w)=qw.
>---------------------------------------------------------------->Don''t listen to me. I''ve had too much coffee.

Sneftel stop drink that coffee! You had a perfect reply up to the cited text above. Your function increases with depth -surely an error.

The whole thing is a standard differetial equation:
(I beeing intensity, x just a lenght coordinate)
(f is a damping factor)

dI = -f*I*dx (intensity changes proportinally to intensity)

dI/I = -f*dx

integrate over x:

ln(I) = -f*x

I = exp(-f*x) (witch is the solution) (falls of nicely with depth)

/Ola

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quote:
Original post by Anonymous Poster
> ... If q is the percentage of light allowed through a
> translucent surface of width 1, then the percentage of light Q
> let through a translucent surface of width w, as a function of > w, is Q(w)=qw.
>---------------------------------------------------------------->Don''t listen to me. I''ve had too much coffee.

Sneftel stop drink that coffee! You had a perfect reply up to the cited text above. Your function increases with depth -surely an error.

His reply makes perfect sense, since as he says you can treat a block of length N as 2 adjacent blocks of length N/2. Also, his function doesn''t increase with distance because q < 1, so q > q^2 > q^3 > ...

Your solution might work, but I see no proof that your assertion dI=-f*I*dx is true.

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quote:
Original post by Anonymous Poster
Sneftel stop drink that coffee! You had a perfect reply up to the cited text above. Your function increases with depth -surely an error.

What the heck are you talking about? q is a fractional value, between 0 and 1. remember third grade math? when using percentages in math, you divide them by 100. For instance, if a 1-meter width transmits 25% (that''s 0.25, ya see) then the following widths transmit as such:
0 meters: 0.250 = 1.0
1 meter: 0.251 = 0.25
2 meters: 0.252 = 0.0625
3 meters: 0.253 = 0.015625

Does that look like it increases with depth?

quote:

The whole thing is a standard differetial equation:
(I beeing intensity, x just a lenght coordinate)
(f is a damping factor)

Your function does indeed give a smooth tapering off. But it isn''t the RIGHT ONE. if you don''t believe me, try it out yourself, with a lamp, pieces of paper, and a light meter.

Don''t listen to me. I''ve had too much coffee.

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Gentlemen, relax. You are both saying the same thing.

K^x = exp(-log(1/K)*x)

It''s only a question of how which expression you understand better. I would go with exp(-f*x) myself, because it''s easier to compute and makes more obvious that it''s the solution of the above-mentioned differential equation. But that''s just a matter of taste.

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Despite the fact that Zipster''s guess isn''t right, I think it is important to note that the intensity of light moving through space (something where the optical opaqueness is close to zero) is an inverse square relationship. This is just like every other inverse field (gravity, electrostatics, magnetism), and is an inherent property of the geometry of the space we live in. Anyone see an application of Gauss''s law for finding the amount of light hitting a plane perhaps? Would this be useful? I don''t know. And for the people who have posted about the exponentials: All exponentials will be correct. The difference between a confusing one and a concise one is whether it can incorporate quantities that we give names to. I was at a lecture a few months ago and there is a name given to the constant in the exponent before the depth of the material. This constant has been measured for various mediums, and because of that, you will want to use the formula that incorporates the constant. At any rate, the question has been answered. Sorry for being so picky.

Brendan

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I'm not sure that your answers are correct, even though I don't have a better one myself.

We're talking about translucency, here. The light is diffused through the material; photons are absorbed and reemitted (I'm not sure that this assertion is correct, but for the purpose of this discussion, it will suffice). Consider:
      |      |A     |    B      |      |  vs  A     |    B

A is a light source separated from point B by a block of negligible width. B will obviously receive more light from A in the first case, since the photons emitted from A that are going at 30 degrees of the horizontal might be reemitted in the direction of B when they hit the block.

My point is that, in the original post, height is not modified, but the ratio height/width is. It might have an impact on the solution. It's a pretty complex problem.

Cédric

[edited by - cedricl on November 19, 2002 8:34:08 PM]

[edited by - cedricl on November 19, 2002 8:35:57 PM]

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We''re talking about translucency, here. The light is diffused through the material; photons are absorbed and reemitted (I''m not sure that this assertion is correct, but for the purpose of this discussion, it will suffice).

Not all translucent materials scatter the incident light. Moreover when they do the emitted photons often have longer wavelengths, in the infrared or microwave.

Sometimes electrons exited by incoming light might jump into a conduction band and no photons are re-emmitted at all. I think this is what happens in tinted glass where metal powders are added to the glass in the manufacturing process.

There are also some processes involving acoustic waves in the translucent material, but I am a little hazy about these.

For just about everything the exponential dropoff with thickness method is appropriate.

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