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

[MnFish1]

Well if you used the initial experiment to calculate the extinction coefficient(k) and then plug that back into the equation Id = I0 e^-kD (the extinction of Light in water)

Where Id is the intensity at the depth (D), Io Intensity at the surface (100), you get about 31% (depending on the number of places you plug into the equation. Which seems easier than all of the other methods.

 

[JimWelsh]

Well if you used the initial experiment to calculate the extinction coefficient(k)

OK, sure. Do tell how to perform that calculation, please.

 

[MnFish1]

The interaction of light and water can be described mathematically with the following equation:

ID = I0e^-kD
Where;

ID = Light Intensity at depth (measured in einsteins)
I0 = Light Intensity at the Surface
k = Extinction Coefficient
D = Depth

If we know the light intensity at the surface (IO) and at depth (ID), we can rearrange the equation to determine the extinction coefficient of the water we are dealing with.

k = (ln I0 - ln Id) * D^-1

 

[JimWelsh]

How does your choice of a different base for the logarithmic function make the math any "easier than all the other methods"?

 

[Martin Kuhn]

as absorption is like "filtering out certain wavelenghts of light" and this is done by particles in the saltwater, the chance to filter out a certain wavelength is highest in the first inches

Excel is my friend for ocasions where i don't like to stretch my brain making complex formulas
depht [inch] | depht [cm] | initial. % light |percentage remaing | % light at this depth
6 | 15,24 |100 % |31,77% | 68,233%
12 | 30,48| 68,233% | 31,77% |46,557%
18 | 45,72 |46,557% | 31,77% |31,767%

i expect the correct answer should be D
should be a red light source with something about 700..800 nm....this is what we divers hate

 

[MnFish1]

How does your choice of a different base for the logarithmic function make the math any "easier than all the other methods"?

Hm. Well Firstly, I guess it relates to the number of words needed to describe the process. Second, It is the equation used to calculate light penetration into seawater. Rather than deriving it myself or making a spreadsheet I just googled it (to me that makes it easier)

 

[Alfrareef]

ln it’s exponential. The curve it’s quite different from Log10.
My assumption was that the initial absorption was due to the surface effect since Randy posted it was UV wavelength which we all know go thru water with very little absorption unlike the red has was already mentioned.

 

[Hans-Werner]

My pocket calculator says 31.76752636 % which I would round up to 31.768 %.

 

[Kent12456]

I'll add this to the list of things I'll never understand.

 

[Randy Holmes-Farley]

To clarify:

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).

I did mess up the explanation of what I did as I quickly tried to reply to you.

 

[Randy Holmes-Farley]

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

Hunh?

All I meant by that was that the normally unexpected result that percent removed in 6"(31.767%) matched the amount remaining after 18".

 

[Randy Holmes-Farley]

So just to make it totally clear...

The answer is D...

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?

D. 31.767%

After the first 6 inches, 31.767% is removed and 68.233% remains.

When that light passes through another 6", the same fraction of the light is removed (31.767%) and the amount remaining is 68.233% of the 68.233%, which is 0.68233 x 0.68233 = 0.46557, or 46.557%.

When that light passes through another 6", the same fraction of the light is removed (31.767%) and the amount remaining is 68.233% of the 46.557%, which is 0.68233 x 0.46557 = 0.31767, or 31.767%. Answer D.

The unusual fact that the answer D matches the amount removed in the first 6 inches is just an artifact of the exact numbers I chose. Any other number elected for the percent removed in the first six inches would not have matched the final answer for the UV remaining after 18".

Happy reefing.

 

[rkpetersen]

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

I did miss your irony! To be honest, I skimmed the first bit pretty quickly, it seemed very bizarre, so I stopped.

Hey I just read details about the Trident! I'd heard about it of course, but wow, that's going to be excellent. The relatively small sample volume is important, it will allow use of a container rather than a drain for the waste. Congratulations on developing and helping get this piece of technology out to hobbyists. Well, us Apex owners, anyway.

 

[JimWelsh]

Thank you for your kind words about Trident! And I think many people find my mathematical ramblings to be "very bizarre"; I'm sure you're not alone!

 

[BluewaterLa]

Wow I started thinking that I stumbled upon a Rocket science exam with all the math that appeared to be as alien as a transformer haha.
I give props to the persons who understand all the letters/number equations or advanced math thinking minds out there.
Lost is an understatement for this old fool. Time, money, fish and corals I can count very well