could we see the bottom from the surface or would the water itself absorb the light rays?
:smack: I thought I was in GQ. Could a mod move for me please?
This plot of the Absorption Coefficient of pure water (2/3 down page) reaches a minimum of about 2 X 10[sup]-4[/sup] cm[sup]-1[/sup] at the blue end of the visible range. Blue light will be significantly absorbed by 50 meters, *i.e.*1/(2 X 10[sup]-4[/sup]cm[sup]-1[/sup]), of pure water. Other wavelengths will be filtered out by lesser depths.
Is there anything, outside of a vacuum, that won’t filter out all light by at least some distance?
[Moderator Hat ON]
To GQ.
[Moderator Hat OFF]
Squink, I’m not sure about your numbers.
Essentially you’ve solved for a wavelength of light that would have an absorption coefficient of 1 at 50 meters, but you have to remember that absorption coefficients are arbitrary units. So, one absorption coefficient doesn’t mean that all light through that sample is blocked. I know that I’ve measured samples in the lab where they had absorption coefficients of greater than 1/cm that I could still see through in a 1 cm cuvette.
Also, here’s a clear example of light penetrating further:
http://www.waterencyclopedia.com/La-Mi/Light-Transmission-in-the-Ocean.html
Basically, they say the oceans are “dimly lit” even to 330 meters in the “disphotic zone.” And that’s even with all of the organic and inorganic solutes and mixtures and crap we’ve got in there.
“The uppermost, sunlit layer of the ocean where 70 percent of the entire amount of photosynthesis in the world takes place is called the euphotic zone. It generally extends to a depth of 100 meters (330 feet). Below this is the disphotic zone, between 100 and 1,000 meters (330 and 3,300 feet) deep, which is dimly lit. Some animals are able to survive here, but no plants.”
Finally, here’s a giant tube of distilled water where I’d say that you can see clear distilled water is, well, very clear. I’ve tried to find a top-bottom photo with dimensions, but I haven’t had any luck via Google.
http://nobelprize.org/nobel_prizes/physics/laureates/1995/illpres/neutrino.html
That’s right, which is why I used the weasel term “significantly absorbed.” 50 meters will knock out about 90% of the blue, but sunlight is pretty bright to start with.
I would like to point out that the OP says, “could we see the bottom from the surface” not “could we see the sun from the bottom”. The difference is significant.
No, we could not see the bottom of any of the abyssal deeps even if the sea was chemically pure water without suspended material.
The amount of light theoretically reflected off the abyssal plains and subsequently traveling back to the surface would be well below the threshold of human visual perception.
Tris
>you have to remember that absorption coefficients are arbitrary units
I think they’re in units like nepers per meter or (as in the cite) nepers per cm.
You can call me confused. I’m pretty sure the units I always used were liters/(molecentemeter). I don’t know if there is a name for that unit, I just derived it from the Beer-Lambert Law: A = epsilonl*c
Anyone know how to do greek lettering?
A Bose-Einstein condensate can be tuned to be totally transparent (no absorption) to a very narrow frequency band of light. The is a quantum electrodynamic effect which obviously doesn’t apply to natural light across a wide spectrum, and good luck trying to make a BEC of any significant practical depth, but that technically applies. Ditto for a field of neutrinos; while they do have a very small electromagnetic cross-section, you could fill the galaxy with them and see an otherwise visible light source from one side to the other without noticable attenuation.
Stranger
The absorption units are in: (arbitrary units/length) and absorbitivity (an intrinsic property of a substance) is in (arbitrary units/(lengthconcentration)) or (arbitrary unitsvolume/(length*quantity)) like you put it. So, it isn’t entirely arbitrary units, but the actual absorption units are. Now, how this relates to the actual proportion of absorbed light? I don’t know.
Damnit, just outside of the edit window. Anyway, here’s actually something that explains it all, and I wish I had tried Wikipedia a lot earlier this week. It would have made me look a lot smarter.
Also, Squink was right, A=log(I[sub]1[/sub]/I[sub]0[/sub]), so when the A is equal to 1, the transmitted light is 10% of the initial light.
I suppose once you consider the trip back up where light will need to go from the surface, hit the object at the bottom, and then travel back to the surface, 50 meters of water ought to knock out 99% of the light (at just the right blue wavelength). Would that still be visible to an observer looking at the bottom from a shaded viewpoint? I would guess so if the observer’s vantage point was very well shaded but I don’t really know that much about the dynamic range of light sensitivity for human eyes.
Just looking over the rail of a boat? No.
There is a dimensionless quantity called “optical depth” (denoted by the greek letter tau) which is the product of the absorption coefficient and the length you are interested in. Generally, you can see into an opaque medium only up to only about one optical depth, since any light on its way to you from farther into the material has loses about 60% of its intensity (falls by a factor 1/e) on its way out. Squink has exactly the right idea, but didn’t quantify “significant.”
Assuming a sea of distilled water with a perfectly mirrored bottom and depth of 50 meters, you’ll lose 60% of the intensity of blue light on its way to the bottom and another 60% of that on the way up. That still leaves you with a bit more than 10% of the incident light coming back to your eyes, which should (I think) be visible, under the right conditions. To get only 1% of the incident light from a perfectly reflecting sea floor, you’d need a depth of about 115 meters.
I was up at Crater Lake in Oregon last fall, and they advertise it to be one of the clearest lakes on the planet. According to this web site:
The water is damn clear and I wouldn’t be surprised if it approaches the clarity of distilled water.
No, I think it’s 90% as suggested by Squink in the first place.
Again, the original cite provided by Squink has a chart comparing wavelength to absorption coefficient from the Beer-Lambert law for absorption of light through a solution.
That law is:
A=log[sub]10/sub
So, if A=1, then I[sub]1[/sub]/I[sub]0[/sub]=10.
So, the emitted light is 10% of the original light, and 90%, not 60% of the light has been absorbed.
I’ve kept out of this, because you seem to have gotten the right results (and I didn’t openm it until they’d all been given), but I have to point out a couple of things here:
1.) I’ve usually seen Beer’s Law given in terms of "e’ raised to the power of minus the product of the absorption coefficient and the length, not ten to that power. Conversely, the inverse expressions involve the natural logarithm, not the (base 10) common log. Of course, the functional form of the law is the same, but the value of your absorption coefficient will be different by a factor of ln(10), so watch out. One time I had this issue bite me – when I calcu;lated the coefficient on my calculator I got a different result than I did plotting it on semilog paper. but semilog paper gives you the result you’d get from common, not natural logs.
2.) The units of the absorption coefficient aren’t aritrary. They have very definite values, but it depends upon whether they’re used with natural or common logs (see above) and upon the concentration of absorbers, and the wavelength of light. But there are certainly "correct’ values, which aren’t arbitrary. (And “dimensionless” is not a synonym for “arbitrary”)
Thanks, CalMeacham, for the clarification. I had never heard of absorption coefficients modified for use with base 10.
Footnote d of Squink’s link to the absorption coefficient gives the expression for the transmitted intensity as I = I[sub]0[/sub]e[sup]kl[/sup], with *k *the absorption coefficient in cm [sup]-1[/sup] and *l *the length in cm. The optical depth, tau = kl, is dimensionless, and when tau = 1, I/I[sub]0[/sub]=0.36.
The natural-log absorption also seems to correspond better with your statement above that the oceans are dimly lit down to 330 m, where it gives I/I[sub]0[/sub]=0.1% and base 10 gives I/I[sub]0[/sub]=2x10[sup]-7[/sup].