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I have a spherical light source of radius R. There is a surface patch of area $\delta A$ located at a distance $d$ from the spherical source. How do I calculate the scene radiance due to surface patch $\delta A$? I want to find the scene radiance at an arbitrary point, which is at angle of $\theta_e$ and $\phi_e$ with respect to the surface patch. The BRDF of the surface patch is given to be $f(\theta_e, \phi_e, \theta_i, \phi_i)$, and the center of the spherical light source can be assume to be in the direction of $\theta_i$,$\phi_i$. Light source has power $E_0$.

Edit: Clarified the question.

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  • $\begingroup$ Is your patch a specular reflector? If it isn't, then I believe the intensity of light reaching every point on the patch is the same as if the light source was a point at the center of the sphere (inverse square law). $\endgroup$
    – Floris
    Mar 1 '17 at 15:53
  • $\begingroup$ @Floris I think you misinterpret the question. The OP's patch is a patch on the source. Also the question is about radiance, not irradiance. $\endgroup$
    – garyp
    Mar 1 '17 at 16:14
  • $\begingroup$ I think the question needs clarification. Is it "What is the radiance at an arbitrary point in space due to the patch?" And are those angles the location of the patch on the sphere? $\endgroup$
    – garyp
    Mar 1 '17 at 16:14
  • $\begingroup$ @Floris I have edited the question a bit for more clarity. I want to find the radiance of the surface patch. The light source will illuminate the patch, and there will some irradiance on the patch, which will be converted to scene radiance using the BRDF. I want to find the scene radiance. As far as I know, radiance L, will be a function of emitting angles, theta_e and phi_e. So yes, I want to find the radiance at an arbitrary point in space due to the patch, and I am assuming the arbitrary point considered in a direction of $theta_e$ and $phi_e$. $\endgroup$
    – Arka Sadhu
    Mar 1 '17 at 16:27
  • $\begingroup$ Do you know the radiation pattern of your spherical source? Is it Lambertian? Can we assume that $d \gg R$? $\endgroup$
    – Floris
    Mar 1 '17 at 20:08
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The BRDF, Bidirectional Reflectance Distribution Function, tells you exactly what intensity of light you will get in a direction $\theta_r, \phi_r$ for a given ray incident at $\theta_i, \phi_i$.

Now to a reasonable approximation, a spherical light source of radius $R$ that is "far" (distance $d\gg R$) from a patch will look like a uniform circular light source (not quite true when you get close).

To calculate the scene radiance, you would have to compute the integral of all the possible rays from the light source to your patch; where for each ray, you compute the reflectance for the given angle of incidence and each possible angle of reflection.

The intensity of the incident light from a given patch on the light source is calculated by assuming the source is Lambertian; then an area $dA$ on the surface that makes an angle $\alpha$ to the observer will appear equally bright regardless of the viewing angle, and the integral over all possible patches and all possible angles should equal $E_0$. Again, if we can consider the reflector to be "far" from the source, we can say that the total power $E_0$ is spread over an area $4\pi d^2$, and that therefore the fraction incident on the patch is $$E_{patch}=\frac{\delta A}{4\pi d^2}E_0$$

Treating all that energy as incident at the same angle, your BRDF will immediately give you the distribution of lighting. If you have to treat the light source as "large" relative to the variations in the BRDF, you have to divide it into smaller areas for which this assumption does hold.

It is very likely that you can make all kinds of simplifying assumptions - that really depends on the degree of fidelity that you need, and how well-behaved the BRDF is.

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  • $\begingroup$ I agree with all your arguments and got the same expression with the simplifying assumptions, though I really wanted to derive the exact expression with r involved, and use d>>r as a sanity check. But then I am unable to come up with integration limits. $\endgroup$
    – Arka Sadhu
    Mar 2 '17 at 20:03

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