Dear Radiance/Radiometry Experts,
I need help understanding the rtrace/rpict irradiance output for a sphere light source. The sphere has a radius of 1 mm. and it is projecting light onto a planar surface, 1 m. away. For this geometry, a point source approximation can be used to represent the light source.
The irradiance falling onto the planar surface is governed by the analytical relation:
E(theta) = I(theta=0)*cos(theta)*(1/d^2).
I(theta=0) is the radiant intensity falling perpendicular to the surface. I(theta=0) = L*A, where L=10, is the radiance of the light source and A=4*pi*r^2, is the surface area of the sphere light source. Plugging the numbers into the irradiance formula for sphere at theta=0 (normal projection) yields:
E_sphere(theta=0) = (L * A * 1)/(1) = 10 * 4 * pi * (0.01)^2 = 0.012566 W/m^2
I ran Radiance with this geometry, both rpict(-i) and rtrace (-i -I) produced the irradiance of 0.003141 W/m^2 for theta=0. This result is consistent with the analytical solution if a circle surface is used for the light source, instead of a sphere.
E_circle(theta=0) = (L * A * 1)/(1) = 10 * pi * (0.01)^2 = 0.003141 W/m^2
I changed my Radiance simulation to use a circle (ring with no inner radius) for the light source. For this case, rpict & rtrace both produced that same answer (0.003141), equal to the case of a sphere light source..
Why is the irradiance the same for both light source geometries? The sphere emanates light in all directions (4pi sr.), unlike the circular surface which emanates light into a hemisphere (2pi sr.). Since both light geometries were given the same radiance definition (Lr=10, Lg=10, Lb=10), why do they yield the same result in Radiance? While their analytical solutions are different, by a factor of 4.
I did not use the brighdata primitive to assign a beam pattern to the sphere or circle light sources. I'm assuming a Lambertian light distribution is default, if no beam pattern is provided. Anyway, I would really appreciate your explanation on why these two different geometries produce the same irradiance.
Thank you very much for your insight,