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Consider a diffuse (Lambertian) emitter

  1. Jul 11, 2010 #1
    a simple question about radiometry.

    I consider a diffuse (Lambertian) emitter which has the shape of a square planar surface, with area A, parallel to the horizontal xy-plane, and centered at the point (0,0,h).

    I know the radiant flux at each point of the emitter [itex]\Phi(x,y,z,\Theta)=\mathrm{const}[/itex]

    What should I do to compute the irradiance at the location (0,0,0) of the surface represented by the xy-plane?

  2. jcsd
  3. Jul 16, 2010 #2
    Re: irradiance

    no reply at all(!)
    ...I thought this question to be really basic...I was probably wrong.
  4. Jul 16, 2010 #3

    Andy Resnick

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    Re: irradiance

    Thanks for the bump- I missed it the first time.

    Let me see if I understand your geometry: the source is a Lambertian emitter, which means the the radiance is independent of angle.

    The fundamental expression is E = L * Z / R^2, where E is the irradiance at the receiver, L the radiance of the source, R^2 the distance to the receiver, and Z the 'throughput'/entendue/lichtleitwet...

    [tex]dZ = \frac{dA_{1} cos(\theta_{1})dA_{2}cos( \theta_{2})}{\rho^{2}}[/tex]

    is the "fundamental equation of radiative transfer" involving two differential areas (source and reciever) separated by a distance 'rho', and their relative orientations with respect to the line of sight.

    So in order to calculate the irradiance, you have to integrate over the area of the source which will be cumbersome for a square, but easier for a disk. I'll step through the disk of radius R centered at {0,0,h}: dA_2 = r dr d[itex]\phi[/itex], [itex]\rho = h/ tan(\theta)[/itex], r = h tan([itex]\theta[/itex])

    [tex]dZ = \frac{dA_{1} cos(\theta_{1})dA_{2}cos( \theta_{2})}{\rho^{2}}[/tex]
    [tex]dZ = \frac{dA cos^{2}(\theta) r dr d\theta d\phi}{h^{2} sec^{2}(\theta)}[/tex]
    [tex]dZ = dA sin(\theta) cos(\theta)d\theta d\phi[/tex]
    [tex] E = \frac{LA sin^{2}(\theta)}{h^{2}} = \frac{I}{h^{2}}(1-(R/h)^{2}+(R/h^{4}-...)[/tex]

    It's all geometry- once you write down your expressions for dA_2 and rho, it's simply a matter of integrating over dA_2.

    Hope this helps. A great source is "Introduction to Radiometry" by Wolfe.
  5. Jul 16, 2010 #4
    Re: irradiance

    Uhm...I am trying to follow your steps, but there is something I don't understand.
    Let's try to spot my mistake:

    1) [itex]dA_1 = dA[/itex]
    2) [itex]cos(\theta_1)=cos(\theta_2)=cos(\theta)[/itex]
    3) [itex]dA_2=r\cdot dr d\phi[/itex]

    Let's substitute 1)2)3) into dZ:

    [tex]dZ = \frac{cos^2(\theta) \cdot dA \cdot r dr d\phi}{\rho^2}[/tex]

    where does that [tex]\mathbf{d\theta}[/tex] in your equations come from???
    I noticed it appears also in the book you mentioned, but I can't follow this step.

    -- Another question is: why you didn't consider the following:

    [tex]\rho^2 = h^2 + r^2[/tex]

    [tex]\cos^2(\theta) = \frac{h^2}{h^2+r^2}[/tex]
  6. Jul 16, 2010 #5

    Andy Resnick

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    Re: irradiance

    heh- you caught me, I cribbed the derivation directly from the book... :)

    Ok- I went from my Eq 1 through Eq. 3 and got the correct answer:

    [tex] Z = \frac{dA_{1}dA_{2}cos(\theta_{1})cos(\theta_{2})}{\rho^{2}} [/tex]

    [tex] dA_{1} = dA [/tex]
    [tex] dA_{2} = r dr d\phi [/tex]
    [tex] cos(\theta_{1}) = cos(\theta_{2}) = cos(\theta) [/tex]
    [tex] r = h tan(\theta) [/tex]

    I erred: [tex] \rho = h/cos(\theta) [/tex]

    We substitute:
    [tex] Z = \frac{dA cos^{2}(\theta) r dr d\phi}{h^{2} sec^2(\theta)} [/tex]

    now substitute for r and dr:
    [tex] dr = h sec^{2}(\theta) d\theta [/tex]

    [tex] Z = dA cos(\theta) sin(\theta) d\theta d\phi} [/tex]
  7. Jul 17, 2010 #6
    Re: irradiance

    Now I was finally able to follow all the steps ;)

    To conclude, I have a last and more general question related to the study of radiometry.
    When you deal with radiometric quantities, how do you usually picture them in your mind? For example when you think of radiant flux do you imagine a certain quantity of photons flowing in many directions through a differential area (in the unit time)?

    Also, all the radiometry books I opened surprisingly avoid introducing rigor in the definitions. This is giving me a lot of frustration. As a result, it is not yet clear to me whether:

    - radiometric flux is a function of position and direction ?
    - is radiometric flux a function at all? If so, what are the input variables?
    - does it make sense to talk about radiometric flux through a point or it can be only through a differential surface patch?
    Last edited: Jul 17, 2010
  8. Jul 17, 2010 #7

    Andy Resnick

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    Re: irradiance

    I agree, radiometric nomenclature is completely ridiculous, and photometry manages to be even worse.

    I tend to ignore all the words and instead focus on the *units*: W, W/sr, W/m^2, W/(m^2*sr).

    'Flux' is *usually* given as W/sr. Flux corresponds to the flow of energy, and W/sr tells you in what direction the power flows. 'Emittance', 'Excitance', 'Incidance', 'Irradiance', and a few others are given as W/m^2. That's the amount of energy passing through a surface- either emitted by an object (the emittance and excitance), hitting a detector (incidance), or some more generic concept (irradiance).

    'Radiance' is W/m^2*sr and is the fundamental radiometric quantity- it tells you how much power is emitted per unit area, and in what direction that energy flows. On the detector side, it's the amount of energy passing through a surface taking into account the acceptance angle.

    Wolfe calls the radiometric/photometric system of units "the 'chinese menu' system of units". That is, take one from column 'a' (Power? energy? etc.), one from column 'b' (direction? area? etc.), and one from column 'c' (MKS? CGS? Imperial? etc..?)
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