So, from what I understand, different wavelengths of visible light, when shone through water, will have different attenuation coefficients due to scattering and absorption from the water molecules. This means that blue light would have a higher attenuation coefficient (since it gets scattered most) right? So, does this mean that white light would have a higher attenuation coefficient than light composed of only one wavelength? Would it mean that white light's attenuation coefficient is just like the weighted average of the attenuation coefficients of each wavelength? Please answer if you can.
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$\begingroup$ The attenuation of white light is not a well defined concept. Dispersion affects the spectral composition of light. If the ingoing light is defined as white, then the outgoing light no longer is exactly white. $\endgroup$– my2ctsCommented Mar 27, 2020 at 23:12
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$\begingroup$ It is why sunsets are red and why the sky is blue. $\endgroup$– user137289Commented Mar 27, 2020 at 23:44
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You are right: White light’s total attenuation coefficient is an average of that for the individual frequency components, weighted by the spectral power in each component.
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$\begingroup$ So if you have two light sources with the same total power, one blue and the other white, the white light would be less attenuated. $\endgroup$ Commented Mar 28, 2020 at 14:44
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$\begingroup$ @Not_Einstein Not necessarily! If blue is the least attenuated color in the spectrum, then the white light would on average be attenuated more. $\endgroup$– GilbertCommented Mar 28, 2020 at 15:18
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$\begingroup$ But your assumption was that blue light is more highly attenuated. $\endgroup$ Commented Mar 29, 2020 at 17:02
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$\begingroup$ @Not_Einstein Sometimes blue light is more attenuated than the rest of the spectrum, sometimes less! $\endgroup$– GilbertCommented Mar 29, 2020 at 17:34