The plane gives an illuminance E = π L on a small area parallel to it, and it does not depend on the distance h. For L = 1000 cd/m² that is 3142 lx. For your sliders E = 3141.6 lx.
Problem statement
A Lambert source has the form of an infinite plane. Its luminance is L. Find the illuminance of a small area element dA placed parallel to the source.
We assume the plane is a Lambert source. This means every part of it has the same luminance L in every direction.
We assume dA is tiny and parallel to the plane, at a distance h. This means a ray at θ to the normal of the plane is also at θ to the normal of dA.
We assume nothing else lights dA. This means only the light from the plane counts.
Symbols: L is the luminance of the plane. h is the distance from dA to the plane. dS is the area of a small piece of the plane. θ is the angle between a ray and the normal. r = h tan θ is the radius of a ring on the plane around the foot of the perpendicular. Ω is a solid angle.
The big idea in plain words
Every piece of the plane sends light along a straight ray to dA. Let F be the foot of the perpendicular from O to the plane. Pieces of the plane that are the same distance r from F form a ring around F. All pieces of one ring are at the same distance from O and make the same angle θ with the normal, so a ring has one single dE. Add the rings from F outwards. The rings add up to πL, and h drops out. The page cuts the source into 10 rings of equal angle and gives each ring its own colour. Each ring adds its own share of E, and the coloured stack in the meter is the total.
The solution in each frame
Lab frame
The plane and the plane are at rest. The full solution is in the next section.
Other frames
No second frame is needed. Nothing moves.
The solution step by step
Stand at O. A piece of the plane of area dS is at distance d = h / cos θ along a ray at angle θ to the normal. Its face is tilted by θ, so the area seen from O is dS cos θ, and the solid angle it covers is
A source of luminance L, seen at the angle θ to the normal of dA, sends dA the flux dΦ = L dA cos θ dΩ. Divide by dA:
All pieces at the same θ have the same dE. They form a ring of radius r = h tan θ and width dr = h sec²θ dθ, with area dS = 2π r dr = 2π h² tan θ sec²θ dθ. The distance squared is d² = h² sec²θ, so
h has cancelled. So dE = 2π L sin θ cos θ dθ. Add the rings from θ = 0 to θ = π/2. The plane is infinite, so the rings reach the horizon. Use sin θ cos θ = (1/2) d(sin²θ):
With L = 1000 cd/m²:
If the plane stops at the angle θmax, that is at radius h tan θmax, then E = π L sin²θmax. For your sliders E = 3141.6 lx, and for the infinite plane π L = 3141.6 lx.
A separate check adds 4000 rings of the plane, using the true distance and tilt of each ring, and finds 3141.6 lx, the same as the formula 3141.6 lx. The same number π L is the luminosity M of a Lambert surface: a small area facing an infinite Lambert plane is lit by exactly the flux that leaves each square metre of the plane.
Five methods
1. Force method
This does not suit the problem. It is about light on a surface, not about forces.
2. Energy method
Add the flux that reaches dA from all pieces of the plane and divide by dA. This is what the rings do, and it gives π L. It is also the luminosity M = π L of the plane, since all the flux that leaves a patch of plane reaches the facing area.
3. Graph method
The first graph plots dE/dθ = 2π L sin θ cos θ. The area under it is E. The second graph shows the running total, π L sin²θ, rising to π L at 90°. The third graph is flat: h does not change E.
4. Calculus method
Integrate 2π L sin θ cos θ from 0 to π/2. Put x = sin θ, so dx = cos θ dθ, and the integral becomes 2π L ∫ x dx from 0 to 1, which is π L.
5. Check method
Units: cd/m² × sr is lm/m², which is lx. Limits: a tiny plane patch right below O (small θmax) gives E = π L sin²θmax, which goes to zero as it should. As θmax goes to 90°, the rings reach out to infinity and E reaches π L.
Fastest: the calculus method with x = sin θ. Most reliable: the 4000-ring sum, because it needs no algebra.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Luminance L | Yes | E in proportion, and every curve |
| Height h | No | Only the ring radius h tan θ, the picture and the labels. h cancels in E. |
| Plane angle θmax | Yes | How far the plane extends (radius h tan θmax), E = π L sin²θmax |
| Target illuminance (given-answer mode) | Sets the answer | Solves L |
| Probe speed | No | Only how fast the probe ring sweeps |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Luminance | L | 1000 cd/m² |
| Height of O above the plane | h | 1 m |
| Plane angle | θmax | 90° |
| Solid angle of the lit plane | Ω | 6.283 sr |
| Illuminance at O | E | 3141.6 lx |
| Infinite-plane value | π L | 3141.6 lx |
Exam traps
- Forgetting the cos θ. Without it the sum gives 2π L, which is double the right answer. Also counting only one tilt factor in dΩ.
- Writing E = L Ω with Ω = 2π. That ignores the slant of the rays on the plane.
- Thinking a plane closer to dA gives more light. h cancels.
- Using h instead of h / cos θ for the distance to a piece. Each piece is farther than h.
- Mixing up luminance L (cd/m²) with luminosity M (lm/m²). For a Lambert surface M = π L.
- Integrating θ up to π instead of π/2. The plane is only on one side of dA.