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Radiation Intensity Thermodynamics

  1. Sep 9, 2014 #1
    So I understand that a black body is a diffuse emitter, and that it radiates energy at a rate of σT^4.
    I also understand the correspondence between a solid angle and the superficial area of a sphere.

    [itex]
    dS=r^2sin(θ)dθd\phi=\frac{ds}{r^2}=d\omega=\frac{dAcos(\alpha)}{r^2}
    [/itex]

    What I don't understand is how they get to the formulation for intensity of radiation.

    Why is the intensity of radiation proportional to the cos of the angle between the two areas? Isn't the intensity the same for every part of the half sphere since it's a black body? Why is it proportional to the solid angle? The area is proportional to the solid angle right?

    The formula for Radiation Intensity is:

    [itex]
    I_e(θ,\phi)=\frac{\frac{dQ}{dt}}{dA*cos(θ)sin(θ)dθd\phi}
    [/itex]
     
  2. jcsd
  3. Sep 9, 2014 #2

    Mech_Engineer

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  4. Sep 9, 2014 #3
    Alright! Thank you! Now I know.
     
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