Reef Chemistry Question of the Day #227 Math of Light Absdorption

[Randy Holmes-Farley]

Reef Chemistry Question of the Day #227

You are interested in knowing how much UV is penetrating your reef aquarium. You do an experiment and determine that at a particular wavelength, 31.767% of the UV is removed after the light passes through 6 inches of your aquarium water.

If that same wavelength passed through 18 inches of the same water, what fraction of the UV would remain?

A. 0.000%
B. 4.699%
C. 15.239%
D. 31.767%
E. 68.231%

Good luck!


























.

 

[Alfrareef]

3 times more will leave 4.699 which should be B. The only doubt I have it’s the linearity of the salt water because I don’t remember wave propagation classes at the university... if it’s non-linear than it must be A.

 

[Randy Holmes-Farley]

It is a surprisingly tricky question. lol

 

[jmerideth1]

D seems closest since the light would only reduce 31% of the already reduced by every 6 inches?

 

[rkpetersen]

D. ((100-31.767)/100) x ((100-31.767)/100) x ((100-31.767)/100) = .317675

 

[S-t-r-e-t-c-h]

D. ((100-31.767)/100) x ((100-31.767)/100) x ((100-31.767)/100) = .317675

Yup. Result is:

D. 31.767%

 

[zsuzsu]

d n stuff

 

[CNDReef]

B

 

[tgionet]

D

 

[Alfrareef]

After some investigation and since we talk about UV I’ll go with E

 

[JimWelsh]

D. ((100-31.767)/100) x ((100-31.767)/100) x ((100-31.767)/100) = .317675

What a verbose way of saying, "0.68233 ^ 3 = 0.31767"!

I'm more interested in how Randy came up with that "31.767%" value in order to get this mathematical funny business to happen in the first place.

Transmittance is defined as that portion of the light recieved by the sample that is transmitted by the sample. If 31.767% of the recieved light is NOT transmitted ("removed"), then 68.233% of the light is transmitted, for a transmittance "T" = 0.68233.

Absorbance "A" is related to transmittance "T" by:

A = -log10(T), and
T = 10^(-A),

so for T = 0.68233, A = -log10(0.68233) = 0.16601. Since absorbance is directly related to the path length, in Randy's question, increasing the path length by a factor of 3 will increase the absorbance also by a factor of 3, so for 18 inches of water, A will equal 0.16601 * 3 = 0.49802. Randy's question is about the amount of light transmitted ("would remain") in the 18 inch scenario, so we can now calculate transmittance T = 10 ^ -A = 10^-0.49802 = 0.31767.

What I want to do is to see how we could set up this problem to solve for "X", where "X" is that 31.767% value, but we don't know exactly what that value is; all we know is that we want the initial amount of light removed to equal the same amount of light remaining after multiplying the path length by a factor of three. So, we will replace "31.767%" by "X", and then solve for "X".

The initial 6" absorbance:
"A6" = -log10(1-X).

The absorbance for 18" will be three times the initial A6 value:
"A18" = 3 * -log10(1-X).

The transmittance "T18" for the 18" scenario will be:
"T18"= 10^(-3*-log10(1-X)) = 10^(3*log10(1-X)).

Since the initial "X" and the final answer 10^(3*log10(1-X)) are the same number (our goal), we have:
X = 10^(3*log10(1-X)).

Taking the logarithm of both sides:
log10(X) = 3*log10(1-X).

Using the properties of logarithms:
log10(X) = log10((1-X)^3).

Taking the antilog of both sides:
X = (1-X)^3. (EDIT: This is what @rkpetersen said.)

Rearranging:
-X^3 + 3X^2 - 4X + 1 = 0

And it becomes a problem of solving the cubic equation.

 

[Randy Holmes-Farley]

What a verbose way of saying, "0.68233 ^ 3 = 0.31767"!

I'm more interested in how Randy came up with that "31.767%" value in order to get this mathematical funny business to happen in the first place.
.

lol

I was wondering how many people would reject the right answer because it matches a wrong answer.

If X= fraction removed in 6", then the fraction removed in 18" = X^3 and the fraction remaining in 18" is 1-X^3

To satisfy the oddity of the fraction removed in 6" (X) equaling the fraction remaining after 18", then 1-X^3=x (or X^3 = 1-x)

I did it an cheater way.
Three columns in excel.
Column One was numbers from 0.9 down to 0 by 0.01,
column two was column one^3
column three was 1-column one

I narrowed in on the two rows with the closest column 2 and 3 values (X^3 = 1-x), and kept expanding column one to more and more digits to get a more precise answer.

It's a lot easier to verify the correct answer to the problem than it is to find the right values to set up the problem with this oddity.

 

[Randy Holmes-Farley]

After some investigation and since we talk about UV I’ll go with E

Because E rhymes with UV?

 

[rkpetersen]

What a verbose way of saying, "0.68233 ^ 3 = 0.31767"!

I'm verbose?!? ;Joyful Seriously, you just took an EASY problem and made it profoundly more complex.

The reason I posted it that way was to to provide a bit of 'showing my work', rather than just saying D. I probably should have given a text explanation right there.

Anyway, you don't need logs, or a calculator that does cubes, to solve this. 31.767% absorbed equals 68.233% transmitted through each six inches of water depth. After the first six inches, 68.233% remains. After the next six inches, 68.233 x .68233 = 46.557%. After the final six inches, 46.557 x .68233 = 31.767% of the initial UV illumination remains.

I didn't even notice the numerical trickery until some time after I posted my answer.

 

[Randy Holmes-Farley]

I didn't even notice the numerical trickery until some time after I posted my answer.

Oh my wasted work...

 

[tgionet]

I'm verbose?!? ;Joyful Seriously, you just took an EASY problem and made it profoundly more complex.

The reason I posted it that way was to to provide a bit of 'showing my work', rather than just saying D. I probably should have given a text explanation right there.

Anyway, you don't need logs, or a calculator that does cubes, to solve this. 31.767% absorbed equals 68.233% transmitted through each six inches of water depth. After the first six inches, 68.233% remains. After the next six inches, 68.233 x .68233 = 46.557%. After the final six inches, 46.557 x .68233 = 31.767% of the initial UV illumination remains.

I didn't even notice the numerical trickery until some time after I posted my answer.

Yeah... I just did a rough cubing of .7 in my head. Definitely didn't notice the trickery.

 

[Alfrareef]

Because E rhymes with UV?

Not really... but since at wavelength (UVA or B) you’ll loose 8% over 1 meter of pure water and I don’t feel like finding the absorption ratio that you used over the initial equation Jim as posted, and find the correct value I choose the not much loss option.
What I really like with your challenges it’s taking people to non comfort areas... Love it!!!

 

[JimWelsh]

I'm verbose?!? ;Joyful Seriously, you just took an EASY problem and made it profoundly more complex.

I'm glad my irony wasn't lost on you!

 

[JimWelsh]

I was wondering how many people would reject the right answer because it matches a wrong answer.

Hunh?

If X= fraction removed in 6", then the fraction removed in 18" = X^3

Umm, no. For example: 0.31767 ^3 = 0.03206. Not even close to the correct answer of 0.68233.

and the fraction remaining in 18" is 1-X^3

Umm, no. 1 - 0.31767 ^ 3 = 0.96794. Not even close. It is actually (1-X) ^ 3. The parenthesis matter.

Your Excel spreadsheet cheat was not solving for "X", where X= the fraction removed in 6". You were solving for X^3 = 1-X where X= the fraction removed in 18". And you took the answer from the other two columns, not the "X" column. The "X" column would have had the value 0.68233 for the row where X^3 = 1-X. That's different.

Obligatory smileys: (BTW, I use that kind of Excel cheat all the time to avoid having to sort through the hard math on things.)

 

[JimWelsh]

To clarify:

To satisfy the oddity of the fraction removed in 6" (X) equaling the fraction remaining after 18", then 1-X^3=x (or X^3 = 1-x)

I think you are confusing the removed and remaining parts. To satisfy the oddity of the fraction removed in 6" (X) equaling the fraction remaining after 18" (which we know to be 0.31767), then X = (1-X)^3. However, to satisfy the oddity of the fraction remaining in 6" (X) equaling the fraction removed after 18" (which we know to be 0.68233), then 1-X^3 = X (or X^3 = 1-X).