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The inverse square law for point light sources inside an opaque medium

  1. May 18, 2012 #1
    Hello all,

    I already know that the radiant intensity of a point light source falls off with the inverse square of the distance to the source. This, however, only happens in a vacuum. My question is, what is the more general law for a point source inside an opaque medium with a known absorption coefficient σ(x) that may vary across space. From symmetry considerations alone, I would expect that the result will still be a function only of the distance to the light source, as before, just not an inverse square power anymore. The actual function will, of course, depend on σ(x) and should be the outcome of some 1D differential equation whose control variable is the distance to the source - it is the form of this 1D equation that I am looking for.

    To clarify a bit further, the above absorption coefficient occurs in the light transport equation when stating that the derivative of radiance L(t) along a light ray parameterised by t is:

    dL(t)/dt = -σ(t) L(t)

    or stating the same in 3D space:

    (ω.∇) L(x, ω) = -σ(x) L(x, ω)

    where L(x, ω) is the radiance at point x in the direction ω.

    Thank you,
  2. jcsd
  3. May 18, 2012 #2


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    Both factors apply. The inverse square plus (or, rather, multiplied by) the absorption per metre travelled.
  4. May 21, 2012 #3
    Hi Sophie, you mean that there is an effective absorption coefficient given by σ(t)/t^2 ? I don't see how that would work because in an optically transparent medium (σ = 0), the inverse square term would vanish together with the σ, which we know is not true.

  5. May 21, 2012 #4


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    Absorption PER METRE. It's exponential with distance (in addition to the spreading loss).
  6. May 21, 2012 #5


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    Better way to look at it, perhaps, is that in vacuum, the total power of the radiation passing through a spherical shell around a point source is a constant regardless of shell radius. This gives you inverse square per unit area. In opaque medium, the total power drops of as exp(-λr). Therefore, power per unit area drops as exp(-λr)/r²
  7. May 21, 2012 #6
    Thank you both - I understand it now. I didn't imagine the solution was so simple as to just multiply the inverse square with the absorption (after application of the exponential to the latter).
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