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What is the absorption coefficient of black-colored water?

  1. Aug 30, 2014 #1
  2. jcsd
  3. Aug 30, 2014 #2

    Simon Bridge

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    "m^-1" reads "per meter". What dimensions were you expecting?

    It's a good point: how can you possibly absorb more than 100% of the incident radiation?

    But that ##\alpha_w## which is on the vertical axis is the "attenuation coefficient", is this the same as "absorption coefficient"? There is a link to the definition under the top graph in your link, did you check it out?
  4. Aug 30, 2014 #3
    well obviously but I was expecting it to have no units .. there are three types of coefficients you take into consideration when a beam of light hits an object: the reflection co ,the transmission co, and the absorption coefficient and they're all equal to 1
    so the absorption co α has to be less than 1


    Attenuation versus absorption

    The terms "attenuation coefficient" and "absorption coefficient" are generally used interchangeably. However, in certain situations they are distinguished, as follows.[4]

    When a narrow (collimated) beam of light passes through a substance, the beam will lose intensity due to two processes: The light can be absorbed by the substance, or the light can be scattered (i.e., the photons can change direction) by the substance. Just looking at the narrow beam itself, the two processes cannot be distinguished. However, if a detector is set up to measure light leaving in different directions, or conversely using a non-narrow beam, one can measure how much of the lost intensity was scattered, and how much was absorbed.

    In this context, the "absorption coefficient" measures how quickly the beam would lose intensity due to the absorption alone, while "attenuation coefficient" measures the total loss of narrow-beam intensity, including scattering as well. "Narrow-beam attenuation coefficient" always unambiguously refers to the latter. The attenuation coefficient is always larger than the absorption coefficient, although they are equal in the idealized case of no scattering.

    and this is where i got the chart, I clicked on that link but I could not understand s***
  5. Aug 30, 2014 #4

    Simon Bridge

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    That page gives you an equation: $$I(z)=I_0e^{-\alpha z}$$ ... where z is the distance through the material. I0 is the incoming flux, and I(z) is the flux after distance z.

    This means that $$\alpha_z = -\frac{1}{z}\ln\left|\frac{I(z)}{I_0}\right|$$ ... so if a particular distance through the material removes roughly 35% or less of the incoming light, then the coefficient will be bigger than 1. A dimensional analysis will tell you what units ##\alpha## should have.
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