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Example: If one SCUBA diver looks past another SCUBA diver horizontally into negative space, how far away is the most distant emission of 'blue' light waves? Consider the sight angle in question to be that which offers the maximum perceived distance, which I speculate is horizontal or along the Mariana Trench to nullify the Earth's curvature or something else beyond my assumption.

Please disregard local water visibility conditions as in how far away an object can be discerned due to suspended matter. I am curious about the actual 'blue' region you see in an undersea photo. I assume this is scattered light from distant water molecules.

I suppose that, unlike the black of outer space between star light emission, the 'blue' of water is not the absence of emitted light from a source but a non-point diffusion of light.

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I only know the answer to this question from actual diving experience, but any "redness" is utterly swamped by the shorter wavelength end of the spectrum below about 7m depth. Your question would not have a well defined answer, because there will always be some red - just many orders of magnitude below blue. The mechanism I assume is the same as that which makes the sky blue - Rayleigh scattering, whose attenuation co-efficient $\propto\;1/\lambda^4$. However, I don't know how to do this calculation, because I don't know the right stochastic models of inhomogeneity - either for air or water. –  WetSavannaAnimal aka Rod Vance Sep 25 '13 at 6:56
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Have a look at this compilation of optical properties of water. If I take for example this recent measurement of the absorption coefficient, the minimum absorption coefficient is $0.0000442$ /cm at $417.5$ nm (which is blue light).

The next question is where you consider the cutoff for vision to be. The Wikipedia article on daylight gives the ratio between normal daylight and an overcast sky at night to be about $10^9$ ($100,000$ lux during the day and $0.0001$ lux at night). In an absorbing medium of absorption coefficient $a$ the light intensity at a distance $x$ is given by:

$$ \frac{I}{I_0} = e^{-ax} $$

so:

$$ x = \frac{-1}{a} \ln \frac{I}{I_0} $$

and feeding in $a = 0.0000442$/cm and $I/I_0 = 10^{-9}$ we get $x \approx 470,000$ cm or $4.7$ km.

This will seem ridiculously high to most of us since experience suggests we can hardly see from one end of a swimming pool to the other. Firstly these measurements are for pure water, and in practice seas (or indeed swimming pools) contains particulate matter that dominates the scattering. The other consideration is that in large bodies of water currents cause refractive index gradients and these scatter light. They don't reduce the overall light intensity, but they do blur vision.

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