The wall is brightest at the angle θ = 60° from the vertical, at depth R / √3 below the ceiling, where E = 3√3 ρ E S / (16 π R²). With the given values that is 0.21 lx. For your sliders the maximum is 0.2067 lx at d = 1.155 m.
Problem statement
A vertical shaft of light from a projector forms a light spot S = 100 cm² in area on the ceiling of a round room of radius R = 2.0 m. The illuminance of the spot is E = 1000 lx. The reflection coefficient of the ceiling is ρ = 0.80. Find the maximum illuminance of the wall produced by the light reflected from the ceiling. The reflection is assumed to obey Lambert's law.
We assume the spot is at the centre of the ceiling and is small compared with R. This means it acts as a point source at the ceiling.
We assume the ceiling reflects by Lambert's law. This means the lit spot has the same luminance L in every direction, so its intensity goes as cos θ.
We assume the room is tall enough for the best point to be on the wall. This means only the light reflected once from the ceiling matters.
Symbols: S is the spot area. E is the illuminance of the spot. ρ is the fraction of light the ceiling reflects. R is the room radius. L is the luminance of the spot. θ is the angle from the vertical at the spot. d is the depth of a wall point below the ceiling.
The big idea in plain words
The lit spot acts like a small lamp pointing down. Straight down it shines most, but the light does not reach the wall there. Toward the ceiling level it reaches the wall, but it grazes the wall and a spot on the lamp seen sideways looks thin. So two things fight. As you go down the wall, the light arrives more squarely and travels a shorter path at first, but the lamp itself shines less in that direction. In between there is a best point. The calculation puts it at θ = 60°, which is R / √3 below the ceiling.
The solution in each frame
Lab frame
The projector, the room and the wall 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
The light that hits the spot is E S lumens. The ceiling reflects the fraction ρ, so the reflected flux is ρ E S. A Lambert source has luminosity M = π L, which is the flux leaving each square metre. So M S = ρ E S, and
The intensity in the direction at angle θ from the vertical is the luminance times the area the spot shows from there:
Take a wall point P at depth d. The horizontal distance to it is R, so tan θ = R / d and r = R / sin θ. The wall is vertical, so its normal is horizontal. The ray makes 90° − θ with that normal, so the cosine of the angle of incidence is cos(90° − θ) = sin θ. The illuminance at P is
For the maximum, set the derivative of cos θ sin³θ to zero:
At 60°, cos θ = 1/2 and sin³θ = 3√3 / 8, so cos θ sin³θ = 3√3 / 16. The point is at d = R cot θ = R / √3 below the ceiling. Then
Put in ρ E S = 0.8 × 1000 × 0.01 = 8 lm and R² = 4 m²:
For your sliders the reflected flux is ρ E S = 8 lm, L = 254.6 cd/m², and E_w,max = 0.2067 lx at d = 1.155 m.
A separate check scans 4000 angles between 0° and 90° and finds the maximum 0.2067 lx, the same as the formula 0.2067 lx.
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 ceiling reflects ρ E S lumens, and Lambert's law spreads it as I = (ρ E S / π) cos θ. The wall illuminance is the flux per area on the wall. This gives the same E_w.
3. Graph method
The third graph shows the two factors. The factor cos θ falls from 1 to 0, and sin³θ rises from 0 to 1. Their product peaks at 60°. The first two graphs show the same peak against depth and against angle.
4. Calculus method
Differentiate cos θ sin³θ and set it to zero. This gives tan²θ = 3. The second derivative is negative there, so it is a maximum.
5. Check method
Units: lm per m² is lx. Limits: at θ = 0 the ray goes straight down and never reaches the wall, so E_w = 0. At θ = 90° the ray runs along the ceiling and sin³θ cos θ = 0. So a peak must lie between.
Fastest: the calculus method. Most reliable: the 4000-angle scan, because it does not need any algebra.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Spot area S | Yes | The reflected flux, the wall E and E_max, in proportion |
| Spot illuminance E | Yes | The reflected flux, L and the wall E, in proportion |
| Reflection coefficient ρ | Yes | The reflected flux, L and the wall E, in proportion |
| Room radius R | Yes | The wall E as 1 / R², the best depth and the picture |
| Target wall maximum (given-answer mode) | Sets the answer | Solves E of the spot |
| Probe speed | No | Only how fast the probe point moves down the wall |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Spot area | S | 100 cm² |
| Spot illuminance | E | 1000 lx |
| Reflection coefficient | ρ | 0.8 |
| Room radius | R | 2 m |
| Spot luminance | L | 254.6 cd/m² |
| Best depth | R / √3 | 1.155 m |
| Wall maximum | E_w,max | 0.2067 lx |
Exam traps
- Using E S as the reflected flux. The ceiling reflects only ρ E S.
- Writing M = L instead of M = π L for a Lambert surface.
- Forgetting that the wall is vertical. The angle of incidence is 90° − θ, so the factor is sin θ and not cos θ.
- Using the distance d instead of r = R / sin θ in the inverse-square law.
- Mixing cm² and m². S = 100 cm² is 0.01 m².
- Measuring the best depth from the floor. It is R / √3 below the ceiling.