The intensity must be I(θ) = I0 / cos³θ, and the flux reaching the table is Φ = π R² I0 / h². With the book numbers (h = R = 1.0 m, I0 = 100 cd) that is 314 lm. For your sliders the flux is lm, and at the table edge the source sends cd, which is times I0.
Problem statement
A point source is suspended at a height h = 1.0 m over the centre of a round table of radius R = 1.0 m. The luminous intensity I of the source depends on direction so that the illuminance at all points of the table is the same. Find the function I(θ), where θ is the angle between the radiation direction and the vertical, as well as the luminous flux reaching the table if I(0) = I0 = 100 cd.
We assume the source is a point at height h above the centre. This means the distance to a table point at distance x from the centre is r = √(x² + h²).
We assume the table is horizontal and the only light comes from the source. This means the ray to any point meets the table at the same angle θ to the vertical, as the figure shows.
We assume the intensity depends only on θ, not on the direction around the vertical. This means the lamp has the same shape of light in every vertical plane.
Symbols: I(θ) is the luminous intensity in cd. I0 is the intensity straight down. h is the height. R is the table radius. x is the distance of a point from the centre. E is illuminance in lux. Φ is luminous flux in lumen.
The big idea in plain words
A point straight below the lamp is close, and the light hits the table square on. A point near the edge is farther away, and the light arrives at a slant. Both effects make the edge darker. To fix this, the lamp must send more light toward the edge. The more the direction tilts, the more extra light it needs. The calculation shows the extra factor is exactly 1 / cos³θ. Because every point gets the same E, the flux is just E times the area of the table.
The solution in each frame
Lab frame
The source and the table 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
Take a point P on the table at distance x from the centre. The distance from the source is r = √(x² + h²), and the ray makes angle θ with the vertical, so cos θ = h / r. The table is horizontal, so the ray also makes θ with the normal at P. The illuminance at P is
Put r = h / cos θ:
Straight below the source, θ = 0 and I = I0, so E = I0 / h². We need the same E at every point, so
At the edge tan θ = R / h. With R = h = 1 m, θ = 45° and I = 100 / (0.7071)³ = 283 cd, which is 2.83 times I0.
Every bit of light that reaches the table passes through the table, and E is the same everywhere. So the flux is the illuminance times the area:
For your sliders E = lx, and the edge is at θ = °, so Φ = lm.
The PDF gives I(θ) = I0 / cos³θ and Φ = π R² I0 / h² = 314 lm, and the page agrees. A separate check adds up I dΩ over 4000 thin rings of the cone, with dΩ = 2π sin θ dθ, and gets lm for your sliders, the same as the formula lm.
Five methods
1. Force method
This does not suit the problem. It is about light on a surface, not about forces.
2. Energy method
The flux that leaves the source in the cone toward the table all lands on the table. Uniform E over the area πR² gives Φ = E π R². This is the quickest way to the flux.
3. Graph method
The first graph shows I(θ) against θ. It starts at I0 and climbs steeply. The second graph shows E along the table: flat for this lamp, and falling for a lamp that sends I0 in every direction. The third graph shows Φ(x) = E π x² growing as x².
4. Calculus method
Integrate the intensity over the cone: Φ = 2π I0 ∫ sin θ / cos³θ dθ from 0 to θmax. The antiderivative is 1 / (2cos²θ), so Φ = π I0 tan²θmax = π I0 R² / h². This matches the energy method.
5. Check method
Units: cd per m² is lx, and lx times m² is lm. Limits: for a very small table, θmax is near 0, I is close to I0 and Φ is close to E π R². For a very wide table, θmax tends to 90° and I grows without limit, so no real lamp can light an endless flat table evenly.
Fastest: the energy method, Φ = E × area. Most reliable: the 4000-ring sum, because it does not need any algebra.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Intensity I0 | Yes | I(θ) at every angle, E, the flux and the graphs, in proportion |
| Height h | Yes | E = I0 / h², the edge angle, the edge intensity, the flux and the picture |
| Table radius R | Yes | The edge angle, the edge intensity, the flux and the picture |
| Target flux (given-answer mode) | Sets the answer | Solves I0 |
| Probe speed | No | Only how fast the probe point P moves out to the edge |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Intensity straight down | I0 | cd |
| Height | h | m |
| Table radius | R | m |
| Angle at the edge | θmax | ° |
| Intensity at the edge | cd | |
| Illuminance everywhere | E | lx |
| Flux on the table | Φ | lm |
Exam traps
- Using E = I / r² and forgetting the cos θ for the tilt of the table. That gives cos² and not cos³.
- Using the distance h instead of r in the inverse-square law.
- Writing I(θ) = I0 cos³θ. The lamp must send more light to the side, so the cube is in the denominator.
- Finding the flux as I0 times the solid angle. I is not constant, so that is wrong. Use E × area, or integrate.
- Thinking the table edge sets the flux through the angle only. The flux depends on R and h through R² / h².