(a) The mean illuminance is E0 / 2. (b) It is (I/R²) (1 − √(1 − (R/l)²)) / (1 − R/l), which is 50 lx for I = 36 cd, l = 100 cm and R = 60 cm. For your sliders part (a) gives 50 lx and part (b) gives 50 lx.
Problem statement
Find the mean illuminance of the irradiated part of an opaque sphere when it receives (a) a parallel luminous flux that gives an illuminance E0 at the point of normal incidence, and (b) light from a point isotropic source at a distance l = 100 cm from the centre of the sphere. The radius of the sphere is R = 60 cm and the luminous intensity of the source is I = 36 cd.
We assume the sphere is opaque and only its side facing the source is lit. This means the lit part is a cap, and the far side gets no light.
We assume the point source radiates equally in all directions. This means the flux in a cone is I times the solid angle of the cone.
We assume no light is reflected on to the sphere from anywhere else. This means only the direct flux counts.
Symbols: E is illuminance in lux, which is flux per area. Φ is luminous flux in lumens. I is luminous intensity in candela, which is lumens per steradian. Ω is solid angle. α is the half angle of the cone from the source that just touches the sphere.
The big idea in plain words
The mean illuminance is the total flux that lands on the sphere divided by the area it lands on. Find the flux that lands, find the lit area, and divide. The flux that lands is all the flux inside the cone that just touches the sphere. The lit area is the cap the cone touches. The light is strongest at the point facing the source and falls to zero at the edge of the cap, so the mean is less than the value at the centre of the cap.
The solution in each frame
Lab frame
The source and the sphere are at rest in the lab. The full solution is in the next section.
Other frames
No second frame is needed. Nothing moves in the problem.
The solution step by step
Part (a). A parallel beam that hits the sphere at normal incidence on its axis gives E0 there. The flux that lands on the sphere is the flux that passes through the shadow of the sphere, which is a circle of area π R².
For your slider E0 = 100 lx, so the mean is 50 lx.
Part (b). Draw the two tangents from the source to the sphere, as in the figure. The half angle α of this cone has sin α = R / l. The solid angle of a cone of half angle α is Ω = 2π (1 − cos α), with cos α = √(l² − R²) / l.
All of this flux lands on a cap of the sphere. The tangent points are at an angle 90° − α from the axis, as seen from the centre. A cap of half angle θ has area 2π R² (1 − cos θ). Here cos(90° − α) = sin α, so
Divide the flux by the area:
Put in I = 36 cd, R = 0.6 m and R/l = 0.6. Then √(1 − 0.36) = 0.8, so the top is 0.2 and the bottom is 0.4:
For your sliders the mean is 50 lx. At the point of normal incidence the light is 225 lx, so the mean is 0.222 of that.
A separate check adds the illuminance over 2000 thin rings of the cap, using E = I cos θ / d² at each ring. It gives 50 lx, against 50 lx from the formula.
Five methods
1. Force method
This does not suit the problem. It is about how light spreads, not about forces.
2. Energy method
This is the main method. The luminous flux in the cone is all received by the cap, so flux in equals flux landed. The mean is that flux divided by the cap area.
3. Graph method
The first graph shows the illuminance along the cap. It is highest at the middle and falls to zero at the edge. The amber dashed line is the mean, which lies between the peak and zero.
4. Calculus method
At a point at angle φ from the axis, E = I cos θ / d² with d² = l² + R² − 2 l R cos φ and cos θ = (l cos φ − R) / d. Integrating E over the cap, with dA = 2π R² sin φ dφ, gives the same total flux as I Ω.
5. Check method
Units: cd divided by m² is lx. Limits: when l is much larger than R, the beam is nearly parallel and the mean tends to E0 / 2, which matches part (a). When the source nearly touches the sphere, the lit cap shrinks and the light there becomes very strong.
Fastest: flux divided by area. Most reliable: the ring sum, because it does not rely on the cap formula.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Intensity I | Yes | Φ, the mean E, the local E at the probe, the graphs. Locked in given-answer mode. |
| Distance l | Yes | R/l, the cone, the lit cap, Ω, the mean E and the picture |
| Radius R | Yes | R/l, the cap, its area, the mean E and the picture |
| E0 for part (a) | Part (a) only | The mean of a parallel beam, E0/2 |
| Target mean E (given-answer mode) | Sets the answer | Solves the intensity I |
| Scan speed | No | Only how fast the probe walks over the cap |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Intensity | I | 36 cd |
| Distance to the centre | l | 100 cm |
| Radius | R | 60 cm |
| Ratio | R/l | 0.6 |
| Half angle of the lit cap | 90° − α | 53.13° |
| Flux on the sphere | Φ | 45.24 lm |
| Lit area | 0.9048 m² |
Exam traps
- Dividing by the whole sphere area 4π R² instead of the lit cap.
- Using the full hemisphere for the lit area in part (b). The cap is smaller than a hemisphere, because the tangents touch it before the half-way circle.
- Using the distance l from the source to the surface instead of to the centre in cos α.
- Mixing cm and m. I/R² needs R in metres to give lux.
- Quoting the illuminance at the centre of the cap as the mean. The mean is less, because the edge of the cap is dark.
- Mixing the cone half angle α with the cap half angle 90° − α.