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Is there an equation that describes the percentage of solar energy per wavelength that penetrates the ocean surface?

For example, this website states that only 44.5% of the surface light reaches a depth of 1 meter.

So, is there an equation that can be used to determine the percentage of solar energy per wavelength?

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    $\begingroup$ For more than you ever EVER want to know: "Light Absorption in Sea Water," Wozniak & Dera, Springer, 2010. $\endgroup$ – Carl Witthoft Jun 2 '14 at 11:34
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The concept you are looking for is the attenuation length of light in water. This is the quantitative measure of how strongly different wavelengths get absorbed in water. The absorption of light is fairly simple to describe: it mostly depends on the material, but also

  • if the light transverses a longer length of material, the attenuation will be stronger, and

  • if there is more light to begin with, then more will be absorbed. In fact, the amount of light absorbed is proportional to the amount of light present, so that the decay will be exponential.

To get rid of these factors and to get a constant which is purely a property of the material and not of how much of it there is or of the intensity of the light, we do two things:

  • we measure absorption per unit length of material transversed, and

  • we measure absorption as a percentage of the incident light.

This is what the absorption length is: the length $L$ it takes for a given light intensity to decrease to 36.7% of its initial value. Why this weird number? Since the decay of the intensity is exponential, it can be written as $$ I(l)=I(0)e^{-l/L}. $$ When $l=L$, the intensity has decreased to $e^{-1}=0.367\ldots$.

An alternative, more useful quantity is the inverse of the attenuation length, $\kappa=1/L$, which is called the attenuation coefficient of the medium, and this gets quoted somewhat more often as I understand it. Here is a sample absorption spectrum of liquid water, which as you can see depends strongly on the wavelength:

Absorption spectrum of water

To find out the incident light spectrum at a given depth, you need to apply this attenuation factor to the initial spectrum, which is the one at sea level.

enter image description here

To get the spectral irradiance (solar power per unit area per unit wavelength) at a wavelength $\lambda$ and at depth $l$, you start with the spectral irradiance at the sea surface, $\sigma(\lambda,0)$, and then you apply the Beer-Lambert law of exponential decay to it: $$ \sigma(\lambda,l)=\sigma(\lambda,0)\exp(\kappa(\lambda)l)). $$ This is the equation you were looking for. You might notice, though, that it doesn't get you that much closer to plotting a graph of this quantity, and that is because $\kappa(\lambda)$ is a complicated quantity with a complicated spectral dependence. To get this data, you can start e.g. here, though of course the amount of work you will need will depend on exactly how much detail you want in your final answer.

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    $\begingroup$ Important to note that if we are considering a light source incident on the ocean from the atmosphere then one must also factor in the variable reflection loss from the air/water interface. This will depend on the wavelength of the light, its polarization, and the angle of incidence with respect to the surface of the water. Waves or chop will obviously affect this and in all cases it can be a considerable percentage of the total incident light. $\endgroup$ – J... Jun 2 '14 at 13:18
  • $\begingroup$ So, is there no simple equation for the percentage of solar energy per wavelength i.e. is k(lambda) sometimes used as a constant? $\endgroup$ – KatyB Jun 2 '14 at 16:35
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The light intensity at a depth $d$ is given by the Beer-Lambert law:

$$ I = I_0 \exp\left(-\frac{d}{\ell}\right) $$

where $I_0$ is the intensity at the surface and $\ell$ is the depth at which the light intensity has fallen by 73%.

The depth $\ell$ is dependant on wavelength as seawater absorbs red light more strongly than blue light. I did a quick Google to see if I could find data on the absorptance of seawater, but while I found lots of articles I couldn't find any definitive figures. Presumably the absorptance of seawater is rather variable as different samples will contain different amounts of dissolved and suspended materials.

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  • $\begingroup$ Yeah, it's brutally variable (see my ref. in comment on the post). $\endgroup$ – Carl Witthoft Jun 2 '14 at 11:35

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