Quote - BB, would you post your calculus equations?  Unfortunately mathematical notations are not part of our standard keyboard although they really should be.  Please use whatever notations you wish to indicate their corresponding math symbols, e.g., sum for summation, S for integration (the elongated S), {lim x->0} for limit as x approaches zero, etc.
 

 
There are actually two ways to approach this, one with calculus, the other as an example of compound interest.

The compound interest approach:

We are trying to find a formula for a scenario in which a volume is modeled with n slices each with thickness x, and there is an infinite number of them, and each has continuous opacity k. Each slice takes away some light via opacity, and the rest goes to the next slice.

Transmittance (or transparency) is 1 - opacity.

So if we let the unknown continuous opacity be represented by k, we are asking for:

lim as n->infinity of (1 - k/n)^nx

The answer is e^(-kx)

This is exponential decay.

We can get the same answer via differential equations. Starting with the premise that the change in the amount of light transmitted is a decrease that is directly proportional to the amount passing reaching each slice:

dy/dx = -ky

The answer is again y = e^-kx.

Since k is some arbitrary constant, the negation can be absorbed without loss of generality, so the function is e^kx.

Now the question becomes how to figure out what value to use for k.

This is pretty easy.

Assuming you measured the ratio of light entering and leaving a sample of known thickness, X,  and you find the effective transparency is T, then

e^kX = T

Take the log of both sides:

kX = ln(T)

Then divide by X

k = ln(T) / X

An example. Suppose you are simulating a material that is 90% transparent at 2 mm.

k = ln(.9)/2

Now the interesting thing about e^kx is it is the same as (e^k)^x.

Which means that e^(ln(.9)/2) can be used as a new base, T (note I'm changing what I mean by "T" now), and you can just use T^x.

In this case, T is .948683298.

If you calculate T^2, i.e. the transmittance for 2 mm, you should get .9, and you do. Furthermore, T^4 (double thickness) will give you .81 (,9 * .9) which makes sense.

You can more directly get to the base T by simply taking the measured transparency and raise it to the power 1/D, where D is the depth of the sample.

.9 ^ (1/2) gives the same answer directly.

Another example:

Suppose you want 15% transparency at 3 inches of thickness.

.15 ^ (1/3) gives T = .793700526

What is the transparency of one inch then?

.793700526 ^ 1 = .793700526

Oh! The base T is the unit transparency. How about 2 inches?

.793700526 ^ 2 = .62996

The final opacity is just the complement of the transmittance, so 1 - T^x is the formula.

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