Problem statement

An illuminant shaped as a plane horizontal disc of area S = 100 cm² is suspended over the centre of a round table of radius R = 1.0 m. Its luminance does not depend on direction and is equal to L = 1.6 × 10⁴ cd/m². At what height over the table should the illuminant be suspended to provide the maximum illuminance at the circumference of the table? How great will that illuminance be? The illuminant is assumed to be a point source.

We assume the disc is small enough to act as a point source at height h. This means the distance from the disc to the rim is r = √(R² + h²).

We assume luminance does not depend on direction. This means the disc is a Lambert source with intensity I(θ) = L S cos θ at angle θ from the vertical.

We assume the table is horizontal and the light comes only from the disc. This means the light meets the rim at the same angle θ from the vertical.

Symbols: L is luminance in cd per m². S is the disc area. h is the height of the disc. R is the table radius. θ is the angle from the vertical, with tan θ = R/h. E is illuminance in lux.

The big idea in plain words

Two things change as you lift the disc. It gets farther from the rim, which makes the light weaker. But it also looks more tilted from the rim when it is low, and a tilted disc, and a tilted table, both reduce the light. Low is bad because of the tilt, and high is bad because of the distance. In between there is a best height. The calculation shows it is at h = R, where the ray makes 45° with the vertical.

The solution in each frame

Lab frame

The disc 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

The disc seen from the angle θ to its normal (the vertical) shows the area S cos θ. So its intensity in that direction is

The rim point P is at distance r = √(R² + h²) from the disc, and cos θ = h / r. The ray meets the table at the same angle θ to the vertical, as the figure shows. So the illuminance at P is

To find the best h, divide the top and the bottom by h²:

E is largest when the bracket R²/h + h is smallest. Its two parts have the product R², which is fixed, so their sum is smallest when they are equal: R²/h = h, so h = R. At that height the bracket is 2R, and

Put in L S = 1.6 × 10⁴ × 0.01 = 160 cd and R = 1 m:

For your sliders L S = 160 cd, so E_max = 40 lx at h = 1 m. At your height, θ = 45° and E = 40 lx, which is 1 of the best value.

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 disc spreads over the table. The illuminance at the rim is the flux per area there. This gives E = I cos θ / r², the same formula as above.

3. Graph method

The first graph shows E at the rim against h. It rises, peaks at h = R, and falls. The third graph shows that the same curve fits every table, if the height is measured in units of R.

4. Calculus method

Set the derivative of h² / (R² + h²)² to zero. This gives R² + h² = 2h², so h = R. The trick of dividing by h² avoids the derivative and uses only that a sum with a fixed product is smallest when the parts are equal.

5. Check method

Units: cd per m² times m² gives cd, and cd per m² gives lx. Limits: for h much smaller than R, E goes as h² and tends to zero. For h much larger than R, E goes as 1/h² and tends to zero. So a peak must lie between.

Fastest: the sum-with-fixed-product trick. Most reliable: the 6000-height search, because it does not need any algebra.

What each slider changes

Control	Changes the answer?	What it changes
Luminance L	Yes	I, E at the rim, E_max and the graphs, in proportion
Disc area S	Yes	I, E at the rim and E_max, in proportion. The best height stays R.
Table radius R	Yes	The rim point, the best height, E_max = L S / (4R²) and the picture
Height h	Yes (E at the rim)	θ, E at the rim, E / E_max, the picture and the graphs. The best height and E_max do not change.
Target E_max (given-answer mode)	Sets the answer	Solves L
Probe speed	No	Only how fast the probe moves out to the rim

Parameters

Quantity	Symbol	Value used
Luminance	L	16000 cd/m²
Disc area	S	100 cm²
Table radius	R	1 m
Height	h	1 m
Angle from the vertical	θ	45°
Intensity toward the rim	I	113.1 cd
E at the rim		40 lx

Exam traps

• Treating the disc as an isotropic point source. That would give I = constant and h = R/√2, which is a different problem.
• Forgetting the cos θ in the intensity of a Lambert disc.
• Forgetting the second cos θ for the tilt of the table, so E = I cos θ / r² and not I / r².
• Using the distance h instead of r in the inverse-square law.
• Mixing cm² and m². S = 100 cm² is 0.01 m².
• Maximising E at the centre of the table instead of at the rim.