# If the oceans were distilled water... **URL:** \ **Category:** Factual Questions\ **Created:** [March 28, 2007, 2:37am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779 "2007-03-28T02:37:36Z")\ **Posts on this page:** 18\ **Page:** 1
**Author:** ![Nars\_Glinley](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/nars_glinley/32/8177_2.png) [@Nars\_Glinley](https://boards.straightdope.com/u/Nars_Glinley)\ **Post date:** [March 28, 2007, 2:37am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/1 "2007-03-28T02:37:36Z")
could we see the bottom from the surface or would the water itself absorb the light rays? ---
**Author:** ![Nars\_Glinley](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/nars_glinley/32/8177_2.png) [@Nars\_Glinley](https://boards.straightdope.com/u/Nars_Glinley)\ **Post date:** [March 28, 2007, 2:40am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/2 "2007-03-28T02:40:09Z")
:smack: I thought I was in GQ. Could a mod move for me please? ---
**Author:** ![Squink](https://avatars.discourse-cdn.com/v4/letter/s/b5e925/32.png) [@Squink](https://boards.straightdope.com/u/Squink)\ **Post date:** [March 28, 2007, 3:31am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/3 "2007-03-28T03:31:27Z")
This plot of the [Absorption Coefficient of pure water](http://www.lsbu.ac.uk/water/vibrat.html) (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. ---
**Author:** ![Sage\_Rat](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/sage_rat/32/399_2.png) [@Sage\_Rat](https://boards.straightdope.com/u/Sage_Rat)\ **Post date:** [March 28, 2007, 3:49am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/4 "2007-03-28T03:49:11Z")
Is there anything, outside of a vacuum, that won’t filter out all light by at least some distance? ---
**Author:** ![Gaudere](https://avatars.discourse-cdn.com/v4/letter/g/e495f1/32.png) [@Gaudere](https://boards.straightdope.com/u/Gaudere)\ **Post date:** [March 28, 2007, 4:01am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/5 "2007-03-28T04:01:13Z")
[Moderator Hat ON] To GQ. [Moderator Hat OFF] ---
**Author:** ![threemae](https://avatars.discourse-cdn.com/v4/letter/t/5f9b8f/32.png) [@threemae](https://boards.straightdope.com/u/threemae)\ **Post date:** [March 28, 2007, 4:07am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/6 "2007-03-28T04:07:02Z")
**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](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](http://nobelprize.org/nobel_prizes/physics/laureates/1995/illpres/neutrino.html) ---
**Author:** ![Squink](https://avatars.discourse-cdn.com/v4/letter/s/b5e925/32.png) [@Squink](https://boards.straightdope.com/u/Squink)\ **Post date:** [March 28, 2007, 4:20am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/7 "2007-03-28T04:20:52Z")
[QUOTE=threemae] So, one absorption coefficient doesn’t mean that all light through that sample is blocked. [/QUOTE] 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. ---
**Author:** ![Triskadecamus](https://avatars.discourse-cdn.com/v4/letter/t/b19c9b/32.png) [@Triskadecamus](https://boards.straightdope.com/u/Triskadecamus)\ **Post date:** [March 28, 2007, 5:11am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/8 "2007-03-28T05:11:56Z")
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 ---
**Author:** ![Napier](https://avatars.discourse-cdn.com/v4/letter/n/ce73a5/32.png) [@Napier](https://boards.straightdope.com/u/Napier)\ **Post date:** [March 28, 2007, 4:12pm UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/9 "2007-03-28T16:12:52Z")
\>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. ---
**Author:** ![WarmNPrickly](https://avatars.discourse-cdn.com/v4/letter/w/d78d45/32.png) [@WarmNPrickly](https://boards.straightdope.com/u/WarmNPrickly)\ **Post date:** [March 28, 2007, 4:32pm UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/10 "2007-03-28T16:32:13Z")
> [@](#): > > you have to remember that absorption coefficients are arbitrary units. You can call me confused. I’m pretty sure the units I always used were liters/(mole_centemeter). I don’t know if there is a name for that unit, I just derived it from the Beer-Lambert Law: A = epsilon_l\*c Anyone know how to do greek lettering? ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [March 28, 2007, 4:43pm UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/11 "2007-03-28T16:43:27Z")
[QUOTE=Sage Rat] Is there anything, outside of a vacuum, that won’t filter out all light by at least some distance? [/QUOTE] 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 ---
**Author:** ![threemae](https://avatars.discourse-cdn.com/v4/letter/t/5f9b8f/32.png) [@threemae](https://boards.straightdope.com/u/threemae)\ **Post date:** [March 28, 2007, 7:36pm UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/12 "2007-03-28T19:36:41Z")
[QUOTE=Christopher] You can call me confused. I’m pretty sure the units I always used were liters/(mole_centemeter). I don’t know if there is a name for that unit, I just derived it from the Beer-Lambert Law: A = epsilon_l\*c Anyone know how to do greek lettering? [/QUOTE] The absorption units are in: (arbitrary units/length) and absorbitivity (an intrinsic property of a substance) is in (arbitrary units/(length_concentration)) or (arbitrary units_volume/(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. ---
**Author:** ![threemae](https://avatars.discourse-cdn.com/v4/letter/t/5f9b8f/32.png) [@threemae](https://boards.straightdope.com/u/threemae)\ **Post date:** [March 28, 2007, 7:49pm UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/13 "2007-03-28T19:49:10Z")
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. > **[Beer–Lambert law](https://en.wikipedia.org/wiki/Beer-Lambert_law)** > > The Beer-Lambert law is commonly applied to chemical analysis measurements to determine the concentration of chemical species that absorb light. It is often referred to as Beer's law. In physics, the Bouguer–Lambert law is an empirical law which relates the extinction or attenuation of light to the properties of the material through which the light is travelling. It had its first use in astronomical extinction. The fundamental law of extinction (the process is linear in the intensity of radiati... 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. ---
**Author:** ![illoe](https://avatars.discourse-cdn.com/v4/letter/i/a3d4f5/32.png) [@illoe](https://boards.straightdope.com/u/illoe)\ **Post date:** [March 29, 2007, 2:36am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/14 "2007-03-29T02:36:14Z")
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. ---
**Author:** ![Darryl\_Lict](https://avatars.discourse-cdn.com/v4/letter/d/7cd45c/32.png) [@Darryl\_Lict](https://boards.straightdope.com/u/Darryl_Lict)\ **Post date:** [March 29, 2007, 3:18am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/15 "2007-03-29T03:18:35Z")
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](http://craterlake.wr.usgs.gov/facts.html): > [@](#): > > The record clarity of Crater Lake was measured at a depth of 41 m (134 ft) in August 1994. The lake clarity is measured with a secchi disk, a black and white disk lowered into the water with a cable. The water is damn clear and I wouldn’t be surprised if it approaches the clarity of distilled water. ---
**Author:** ![threemae](https://avatars.discourse-cdn.com/v4/letter/t/5f9b8f/32.png) [@threemae](https://boards.straightdope.com/u/threemae)\ **Post date:** [March 29, 2007, 6:43pm UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/16 "2007-03-29T18:43:58Z")
[QUOTE=illoe] 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.” [/QUOTE] 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. ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [March 29, 2007, 7:41pm UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/17 "2007-03-29T19:41:01Z")
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”) ---
**Author:** ![illoe](https://avatars.discourse-cdn.com/v4/letter/i/a3d4f5/32.png) [@illoe](https://boards.straightdope.com/u/illoe)\ **Post date:** [March 30, 2007, 3:22am UTC](https://boards.straightdope.com/t/if-the-oceans-were-distilled-water/397779/18 "2007-03-30T03:22:56Z")
Thanks, **CalMeacham** , for the clarification. I had never heard of absorption coefficients modified for use with base 10. [QUOTE=threemae] No, I think it’s 90% as suggested by **Squink** in the first place… [/QUOTE] Footnote d of **Squink** ’s [link](http://www.lsbu.ac.uk/water/vibrat.html) 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].