If we assume the ocean is sufficiently deep so that the blue light transmitted inside the water gets absorbed completely before it reaches the ocean floor and be scattered back towards the surface, and subtract the contribution of specular reflection of the blue sky, is there any blue color left? How strong is the Rayleigh scattering in water?

For example, the blue color of the oceans far away from the coasts in satellite images should be due to specular reflection of the blue sky, right?

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    $\begingroup$ If you have a white bathtub and twenty minutes to fill it all the way up, you should be able to convince yourself that water is blue. You may also discover that your local blue-colored swimming pool has white paint on its walls and bottom. $\endgroup$
    – rob
    Commented Jan 2 at 0:57
  • $\begingroup$ @rob I'm talking about deep ocean where the blue light inside the water cannot be scattered back. Of course a bathtub is blue because water is blue. Not related to my question. $\endgroup$ Commented Jan 2 at 0:59
  • $\begingroup$ If we put lots and lots of red koolade in the ocean and replace your question with "red" and "koolade" for blue and water, is it the same question? $\endgroup$
    – g s
    Commented Jan 2 at 1:29
  • $\begingroup$ @gs my question would be why the coolade ocean is not black. Note you're not looking at a bottle of coolade, but an infinite depth coolade ocean. The red light goes down but not up. $\endgroup$ Commented Jan 2 at 2:26
  • $\begingroup$ Water has a higher refractive index at lower wavelengths (bluish colors), making it slightly more reflective there. Water is more absorptive at higher wavelengths (reddish colors), which means light scattered back up from impurities beneath the surface is bluish. I suspect the latter effect is primarily responsible for the color in the case of deep ocean. See also en.wikipedia.org/wiki/Color_of_water. $\endgroup$
    – Puk
    Commented Jan 2 at 2:54

1 Answer 1


The ocean is non-black because fluids capable of scattering a given wavelength glow when illuminated by that wavelength. Particles (including tiny particles like molecules of H2O, but also larger particles like suspended minerals etc) scatter light, and multiple scatterings result in a diffuse glow. Imagine giving a million people randomized directions that read "turn $x_1$ degrees, then go $y_1$ meters, then turn $x_2$ degrees, then go $y_2$ meters, then turn...". If most $x_n$ are small, most will keep going more or less the direction they started in, but a few of them will end up going in any given direction, including a few coming back towards you. Each instruction to turn is like a scattering event, and each instruction to go a certain distance is a path in which there were no scattering events.

If water was colorless, one might expect wavelength-dependent Rayleigh scattering to give the ocean the same range of colors as the sky, and for deep sea scenes to be lit in sunset hues.

However, water is itself blue/cyan in color: water molecules transmit all colors of visible light pretty well, which makes small amounts of water appear perfectly colorless, but they absorb a little less blue and a little more of the other visible colors. The larger-wavelength red, orange, and yellow light is all absorbed by the time it has traveled a few tens of meters through water in whatever path. The blue and green wavelengths that can penetrate farther provide the azure or cyan glow that is characteristic of especially underwater photographs. I'm not sure why it's sometimes azure and sometimes cyan, although I imagine it has to do with scattering profiles for different mixtures of suspended particulates.

Of course, the surface of water is reflective, so from the surface most of the color is reflected color.


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