Luminance of a lambertian sphere (formula)

[henksp]

I need formula's to calculate the luminance from a sphere.

When a flat Lambertian surface is illuminated then:
luminance [cd/m2] = illuminance [lux] / pi (3.1415..).
The light falling on the surface radiates back into 2 * pi, and integrating the cosine of the lambertian surface results in pi. So the end result is pi / (2 * pi) = 1/pi.

But now I put an isotropic light source of 1 candela inside a homogeneous translucent sphere. What is then the luminance at the outside of the sphere (assuming 100% transmission of the translucent material)?

Now a infinite piece of surface on the sphere radiates into more than 2 *pi depending on the curvature of the sphere. But how to calculate this?

Secondly: does the cosine law for the Lambertian surface apply also for a curved surface? I think it's also dependent of the curvature.

Can someone help me out ?

Many thanks,
Henk

 

[henksp]

Problem solved ! Re: Luminance of a lambertian sphere (formula)

I solved the problem by integrating the surface of the sphere and by projecting the ray from the surface towards a plane at some distance from the sphere. The intensity of the ray at the surface of the sphere obeys the cosine-law (Lambert) and at the plane also the cosine to calculate the perpendicular flux.

The result is that at large distances the flux is pi times the luminance of the sphere. So putting a bulb of c candela in a translucent sphere with radius R gives c/pi/R^2 cd/m2 (the translucent sphere has 100% transmission). A plane at distance D (D >> R) from the surface of the sphere receives pi/D^2 * c/pi/R^2 = c/(D^2*R^2) lux .

Integrating this for a sphere at distance D gives the total lumen of 4 * pi * D^2 * (c/D^2*R^2) = 4 * pi * c / R^2 lumen. So no light is lost.