(a) Φ = π L ΔS sin²θ. (b) M = π L. For your sliders Φ = lm, which is percent of the full-hemisphere flux, and M = lm per m².
Problem statement
A certain luminous surface obeys Lambert's law. Its luminance is equal to L. Find: (a) the luminous flux emitted by an element ΔS of this surface into a cone whose axis is normal to the given element and whose aperture angle is equal to θ; (b) the luminosity of such a source.
We assume Lambert's law means the luminance L is the same in every direction. This means a small element has intensity I = L ΔS cos α at angle α from its normal.
We assume θ is the half angle of the cone, as drawn in the figure. This means the edge of the cone is at θ from the axis, and the answer is π L ΔS sin²θ.
We assume ΔS is small. This means the element acts as a point that emits with intensity I(α).
Symbols: L is luminance in cd per m². ΔS is the area of the element. Φ is the flux into the cone, in lumens. M is luminosity, the flux per area into the whole hemisphere. α is the angle from the normal, and dΩ is a small solid angle.
The big idea in plain words
A Lambert surface looks equally bright from every direction. But a tilted view sees a smaller piece of the surface, so the light sent out sideways is weaker. The light sent in direction α goes as cos α. To find the flux into a cone, add up thin rings around the axis. Each ring has a solid angle and an intensity. The sum is simple because cos α sin α integrates to sin²α divided by 2.
The solution in each frame
Lab frame
The element is at rest. The full solution is in the next section.
Other frames
No second frame is needed. The element and the cone do not move.
The solution step by step
The element ΔS seen from the direction α has the projected area ΔS cos α. By the definition of luminance, its luminous intensity in that direction is I(α) = L ΔS cos α.
The directions between α and α + dα form a ring with solid angle dΩ = 2π sin α dα. The flux in the ring is I dΩ.
Add the rings from the axis (α = 0) to the edge of the cone (α = θ). Since the integral of cos α sin α is sin²α / 2:
For your sliders Φ = π × × cm² × sin²(°) = lm.
Part (b). The luminosity is the flux per area into the whole hemisphere. Put θ = 90°, where sin²θ = 1, and divide by ΔS:
For your sliders M = lm per m². It does not depend on ΔS or on θ.
The PDF gives Φ = π L ΔS sin²θ and M = π L, and the page agrees with both. A separate check adds 4000 thin rings of the cone and gets lm, against lm from the formula. The PDF solution writes the first step as "L ΔS cos α dΩ", which is the same ring flux used here.
Five methods
1. Force method
This does not suit the problem. It is about light, not forces.
2. Energy method
This is the main method. The flux leaving the element is the sum of the flux through every ring. With θ = 90° all the flux leaves, which gives M.
3. Graph method
The first graph shows what adds up: the curve 2π L ΔS cos α sin α. Its area up to θ is Φ. The right graph shows Φ as a share of the full flux, which is sin²θ. It is not a straight line.
4. Calculus method
Integrate cos α sin α with the substitution u = sin α. This gives sin²α / 2, as in the steps above.
5. Check method
Units: cd times sr is lm. Limits: for small θ, sin²θ is about θ², which matches the solid angle of a narrow cone, π θ², times the on-axis intensity L ΔS. For θ = 45° the flux is exactly half the full flux, since sin²45° = 1/2.
Fastest: the integral of cos α sin α. Most reliable: the 4000-ring sum, because it does not need the integral.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Luminance L | Yes | Φ, M and the intensity lobe, in proportion |
| Element area ΔS | Φ only | Φ and the lobe. M stays the same, because M is flux per area. |
| Half angle θ | Φ only | The cone, the cap, Φ and the share of the full flux. M does not depend on θ. |
| Target flux (given-answer mode) | Sets the answer | Solves θ, which needs the target to be at most π L ΔS |
| Probe speed | No | Only how fast the probe sweeps the cone |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Luminance | L | cd/m² |
| Element area | ΔS | cm² |
| Half angle | θ | ° |
| Intensity on the axis | L ΔS | cd |
| Full-hemisphere flux | π L ΔS | lm |
| Flux in the cone | Φ | lm |
| Luminosity | M | lm/m² |
Exam traps
- Using dΩ = π θ² for a wide cone. That holds only for small angles. The exact solid angle is 2π (1 − cos θ).
- Forgetting the cos α in the intensity. Without it the flux would be I0 × 2π (1 − cos θ).
- Using sin θ instead of sin²θ. The answer grows as the square of sin θ.
- Treating θ as the full aperture angle. With the full angle θ the result is sin²(θ/2).
- Saying M = 2π L, which is the result for a different case. For Lambert M = π L.
- Giving M a ΔS. M is flux per area, so ΔS has cancelled.