The illuminance at the centre O of the plane is E = π L, and it does not depend on the radius R of the dome. For L = 1000 cd/m² that is 3142 lx. For your sliders E = 3141.6 lx.
Problem statement
A luminous dome shaped as a hemisphere rests on a horizontal plane. Its luminance L is uniform and does not depend on direction. Find the illuminance at the centre O of the plane.
We assume the dome is a Lambert source. This means every part of it has the same luminance L in every direction.
We assume the lit area dA at O is tiny compared with R. This means every ray to it is a radius of the dome.
We assume nothing else lights the plane. This means only the light from the dome counts.
Symbols: L is the luminance of the dome. R is its radius. dS is the area of a small piece of the dome. dA is the small area at O. θ is the angle between the radius to a dome piece and the vertical at O. Ω is a solid angle.
The big idea in plain words
Every piece of the dome sends light along a radius to O. Pieces near the top arrive straight down on the plane, so they count fully. Pieces near the edge arrive almost flat along the plane, so they count very little. The weight is cos θ. There are also more pieces near the edge, since a ring at larger θ is longer. The two effects combine to a neat result: the dome gives E = π L, whatever its size. 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. A bigger dome has more area but each piece is farther away, and the two cancel.
The solution in each frame
Lab frame
The dome 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 and look at the dome. A piece of area dS covers a solid angle dΩ = dS / R² of the hemisphere. A source of luminance L, seen from dA at the angle θ to its normal, 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 sin θ and width R dθ, with area 2π R² sin θ dθ. Its solid angle at O is
R has cancelled. Add the rings from the top, θ = 0, to the edge, θ = π/2. Use sin θ cos θ = (1/2) d(sin²θ):
With L = 1000 cd/m²:
For your sliders E = π L sin²θmax = 3141.6 lx, and for the full dome π L = 3141.6 lx.
A separate check adds 4000 rings of the dome 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, so the dome lights O as strongly as it would shine on a plane right at its surface.
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 dome and divide by dA. This is what the rings do, and it gives π L.
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: R 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 dome cap near the top (small θmax) gives E = π L sin²θmax, which goes to zero as it should. Using the Lambert rule M = π L gives the same number, since the dome's inner surface is a Lambert emitter.
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 |
| Dome radius R | No | Only the ring size, the labels and the picture. R cancels in E. |
| Dome angle θmax | Yes | How much of the dome shines, 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 |
| Dome radius | R | 1 m |
| Dome angle | θmax | 90° |
| Solid angle of the dome | Ω | 6.283 sr |
| Illuminance at O | E | 3141.6 lx |
| Full-dome 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.
- Writing E = L Ω with Ω = 2π. That ignores the slant of the rays on the plane.
- Thinking a bigger dome gives more light at O. R cancels.
- Using the diameter instead of the radius for the distance to O. It does not matter here, since R cancels.
- 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 dome is only a hemisphere.