Problem statement

Determine the luminosity of a surface whose luminance depends on direction as L = L0 cos θ, where θ is the angle between the direction of the radiation and the normal to the surface.

We assume the surface emits only into the outward hemisphere, from θ = 0 to 90°. This means the integral runs over a half space.

We assume L0 is the luminance along the normal and the same at every point of the surface. This means M is the same for every small area.

We assume luminosity M means the luminous flux emitted per unit area of the surface, into all directions. This means M is in lm per m².

Symbols: L is luminance in cd per m², the flux per area per solid angle, measured per unit of projected area. M is luminosity. Ω is solid angle. θ is the angle from the normal.

The big idea in plain words

A surface looks bright or dim depending on the direction. To get the total light leaving one square metre, add up the light into every direction. Directions at the same angle θ from the normal form a ring. Add up the rings from straight ahead to the horizon. The ring at θ has the weight cos θ, because a tilted surface shows a smaller area to that direction.

The solution in each frame

Lab frame

The surface is at rest. The full solution is in the next section.

Other frames

No second frame is needed. The surface and the eye are not moving.

The solution step by step

By the definition of luminance, the flux that leaves an area dS into a small solid angle dΩ at angle θ is dΦ = L dS cos θ dΩ. Dividing by dS gives the luminosity as a sum over all directions:

The directions between θ and θ + dθ form a ring. Its solid angle is dΩ = 2π sin θ dθ, since the ring has circumference 2π sin θ and width dθ on a sphere of radius 1.

Put in L = L0 cos θ. Let u = cos θ, so du = − sin θ dθ, and u goes from 1 to 0:

So M = (2/3) π L0 = 2.094 L0. Your sliders use the general form L = L0 cos^n θ up to θmax:

For n = 0 and θmax = 90° this gives π L0, the value for a Lambert surface. The ratio M / (π L0) for your sliders is .

Five methods

1. Force method

This does not suit the problem. It is about light emission, not forces.

2. Energy method

This is the main method. The flux through each ring is added up, so M is the total energy flux per area, as the integral shows.

3. Graph method

The first graph shows what adds up: the curve 2π L0 cos^(n+1)θ sin θ. The area under it is M. The polar plot shows why the curve is not flat: the surface dims towards the edge.

4. Calculus method

The substitution u = cos θ turns the integral into a power of u, which integrates at once. This is the method used in the steps above.

5. Check method

Units: cd per m² times steradian gives lm per m². Limits: n = 0 gives π L0 for the Lambert case. A larger n gives a narrower beam and a smaller M. Because L never exceeds L0, M is always below π L0 for n ≥ 0 and θmax up to 90°.

Fastest: the substitution u = cos θ. Most reliable: the 4000-ring sum, because it does not need the integral.

What each slider changes

Parameters

Exam traps

• Forgetting the extra cos θ. The integrand is L cos θ, so for L = L0 cos θ it has cos²θ.
• Writing M = 4π L0 or π L0. Those belong to other cases. For L = L0 cos θ the answer is 2π L0 / 3.
• Integrating over the full sphere. Only the outward hemisphere emits.
• Using dΩ = sin θ dθ without the factor 2π.
• Mixing luminance L (cd/m²) with luminosity M (lm/m²).
• Reading "2/3 π L0" as 2/(3π) L0. It means (2/3) π L0.