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Radiance of a Lambertian emitter

  1. Jan 19, 2012 #1

    I am considering an infinitesimal surface dA that receives a total radiant flux [itex]d\Phi[/itex], so basically we know the irradiance, that is given by [itex]dE = \frac{d\Phi}{dA}[/itex].

    If I assume that this surface reflects 100% of the light, and it is Lambertian, how am I supposed to calculate the radiance coming out from it, in some direction?

    -- My attempt was the following: --

    The fact that dA is Lambertian means that the radiance emitted by it must be constant for any direction, moreover dA reflects all the radiation, so the radiant exitance is equal to its irradiance. However the radiance is defined as:

    [tex]L=\frac{d^2\Phi}{dAcos\theta\cdot d\omega}[/tex]

    where θ is the angle between the direction we are considering, and the normal of dA.
    Since we know the radiant exitance is dE we can write:

    [tex]L=\frac{dE}{cos\theta\cdot d\omega}[/tex]

    and here something weird has happened, because L is supposed to be constant, dE is constant too, but we have a cosθ term, so the right term is dependent on the direction. Where is the mistake?
  2. jcsd
  3. Jan 19, 2012 #2

    Andy Resnick

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    According to my book (Introduction to Radiometry, by Wolfe), the radiance is defined as:

    [tex]L = \frac{∂^{2}\Phi}{∂(Acosθ)∂ω}[/tex]

    with units of W/(m^2*sr). For a Lambertian emitter (or reflecting surface), L ≠ L(θ) and the emitted flux density (or excitance, or incidance, or irradiance) with units W/m^2 is the radiance integrated over the projected solid angle:

    [tex] E =\frac{∂\Phi}{∂A} = L\int^{2\pi}_{0}\int^{\pi/2}_{0}sinθcosθdθd\phi= πL[/tex]

    I admit the geometry is highly confusing. Does this help?
  4. Jan 20, 2012 #3
    Hi Andy,

    thanks for the info. I am a bit puzzled about that L≠L(θ.) I think that for a Lambertian emitter radiance is independent of direction, so shouldn't it be L=L(θ) instead?

    In any case, I understand the equations you wrote, but my real question was slightly different.
    Basically all we know is that the total outgoing flux [itex]d\Phi[/itex] from the surface dA is given by [itex]d\Phi/dA[/itex], and that dA is a Lambertian emitter.

    I wanted to see how the flux must be distributed in every direction, in such a way that L remains constant.
  5. Jan 20, 2012 #4

    Andy Resnick

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    If L is independent of direction θ, then L does not depend on θ- am I defining θ differently than you?

    Hang on- the exitance/incidance/irradiance is given by [itex]d\Phi/dA[/itex], but those quantities do not specify the direction the flux [itex]\Phi[/itex] is propagating- similarly, the intensity [itex]d\Phi/d\Omega[/itex] specifies how much flux is emitted in a particular direction but does not specify how the flux is emitted from the surface- to combine the two (incidance and intensity), you use the radiance.

    I've got a 'magic slide' somewhere that nicely shows the different radiometric quantities... let me see if I can find it and I'll post it.
  6. Jan 20, 2012 #5

    Andy Resnick

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    here it is... does this help?

    http://img821.imageshack.us/img821/8241/physicsoflight.jpg [Broken]
    Last edited by a moderator: May 5, 2017
  7. Jan 20, 2012 #6
    Hi Andy,

    thanks for the reply and for the slide!
    we agree now on L≠L(θ). I just got confused with the notation.

    I understand that if we know the radiant exitance (or irradiance) we don't know anything about the flux in specific directions. However, I thought that the fact that our surface is Lambertian might impose a constraint on how the flux propagates in each direction. In other words, the flux is not allowed to go wherever it wants and however it wants. It must be carefully scattered in such a way that the radiance originating from dA at each direction is constant. Or am I wrong?

    ---- What follows next, is what I tried to do. I don't know if it makes sense ------

    If we call [itex]d\Phi_i[/itex] the total incoming flux onto dA, and [itex]d\Phi_L = d\Phi_L(\theta,\phi)[/itex] the reflected radiant flux at an arbitrary direction from our Lambertian ideal diffuser (I parametrized the direction with two spherical coordinates angles), we can verify that by imposing:

    [tex]d\Phi_L(\theta,\phi) = d\Phi_i \cdot \cos(\theta) \cdot sin(\theta)/\pi[/tex]

    the radiance becomes constant at any direction, because we have:

    [tex]L = \frac{d^2\Phi_L}{dA \cos(\theta) \cdot d\omega} = \frac{d^2\Phi_i \cdot \cos(\theta)\sin(\theta)}{\pi \; dA \cos(\theta) \cdot \sin(\theta)d\theta d\phi} = \frac{d^2\Phi_i}{\pi \; dA \; d\theta d\phi}=const[/tex]

    Moreover notice that if we calculate the total flux exiting from each differential patch on the unit-hemisphere we obtain:

    [tex]\int_{0}^{\pi/2} \int_{0}^{2\pi} d\Phi_L(\theta,\phi) \; d\phi d\theta = d\Phi_i[/tex]

    and that is needed in order to be a perfect reflecting surface (i.e. incoming flux=outgoing flux).
    Is this correct?
    Does it make sense?
    Last edited: Jan 20, 2012
  8. Jan 20, 2012 #7

    Andy Resnick

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    I think that's all correct- at least it roughly matches up with my other reference:


    In Appendix C a Lambertian reflector is defined as a 'perfectly diffuse' reflector with a bidirectional reflectance distribution function (BRDF) = 1/π (for a lossless reflector). Your sine and cosine term may refer to the *projected* solid angle =∫cosθsinθdθd[itex]\phi[/itex] and should be retained, but as long as you understand what is going on.... :)
  9. Jan 20, 2012 #8
    Alright, thanks a lot once again also for the useful reference that you pointed out in the last post. I am glad to see that whenever I have a problems with these topics you manage to help to me.
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