Problem statement

An illuminant shaped as a plane horizontal disc of radius R = 25 cm is suspended over a table at a height h = 75 cm. The illuminance of the table below the centre of the illuminant is E0 = 70 lx. Assuming the source obeys Lambert's law, find its luminosity.

We assume the disc is a Lambert source. This means every part of it has the same luminance L in every direction, and its luminosity is M = π L.

We assume the table area dA below the centre is tiny and parallel to the disc. This means a ray at θ to the vertical is at θ to the normal of both the disc and dA.

We assume only the disc lights the table. This means no light comes from the walls or the ceiling.

Symbols: L is the luminance of the disc. M is its luminosity, the flux leaving each square metre. R is the disc radius. h is the height above the table. x is the radius of a ring on the disc. θ is the angle between a ray and the vertical. α is the angle to the edge of the disc, with tan α = R / h.

The big idea in plain words

Every piece of the disc sends light along a straight ray to dA. Pieces at the same distance x from the centre form a ring. All pieces of one ring are at the same distance from dA and make the same angle θ with the vertical, so a ring has one single dE. Add the rings from the centre, x = 0, out to the edge, x = R. This gives E0 = M sin²α. The table is lit by the disc up to the angle α, so E0 is smaller than M. Knowing E0 and the angle α, we find M. The page cuts the source into 10 rings of equal angle and gives each ring its own colour. Each ring adds its own share of E, and the coloured stack in the meter is the total.

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

Stand at dA. A piece of the disc of area dS is at distance d = h / cos θ along a ray at angle θ to the vertical. Its face is tilted by θ, so the area seen from dA is dS cos θ, and the solid angle it covers is

A source of luminance L, seen at the angle θ to the normal of dA, sends dA the flux dΦ = L dA cos θ dΩ. Divide by dA:

The ring at angle θ has radius x = h tan θ and width dx = h sec²θ dθ. Its area is dS = 2π x dx = 2π h² tan θ sec²θ dθ, and d² = h² sec²θ. So

The disc ends at x = R, so θ runs from 0 to α, with tan α = R / h. Use sin θ cos θ = (1/2) d(sin²θ):

For a Lambert source M = π L, so E0 = M sin²α. From the right-angled triangle, sin²α = R² / (R² + h²). Solve for M:

Put in E0 = 70 lx, h / R = 75 / 25 = 3:

(1 lx is 1 lm/m², so the units agree.) For your sliders sin²α = 0.1, M = 700 lm/m² and L = M / π = 222.8 cd/m².

Five methods

1. Force method

This does not suit the problem. It is about light on a surface, not about forces.

2. Energy method

Add the flux that reaches dA from all the rings and divide by dA. This is what the rings do, and it gives E0 = M sin²α.

3. Graph method

The first graph plots dE/dθ = M sin 2θ, and the area under it up to α is E0. The second graph shows the running total M sin²θ reaching E0 at α. The third graph shows the luminosity needed as the disc is raised: M grows as h² for a fixed E0.

4. Calculus method

Integrate 2π L sin θ cos θ from 0 to α. Put u = sin θ, so du = cos θ dθ, and the integral becomes 2π L ∫ u du from 0 to sin α, which is π L sin²α.

5. Check method

Units: cd/m² × sr is lm/m², which is lx. Limits: if R is very large compared with h, α goes to 90°, sin²α goes to 1, and M goes to E0. That matches the infinite plane in 5.10. If R is small, α is small and M is much bigger than E0, as it should be, since the table sees only a small part of the sky.

Fastest: the formula E0 = M sin²α, with sin²α from the triangle. Most reliable: the 4000-ring sum, because it needs no algebra.

What each slider changes

Control	Changes the answer?	What it changes
Disc radius R	Yes	The angle α, E0 / M = R² / (R² + h²), M and the picture
Height h	Yes	The angle α, E0 / M, M and the picture
Table illuminance E0	Yes	M in proportion, and every curve
Luminosity M (given-answer mode)	Sets the answer	Solves E0
Probe speed	No	Only how fast the probe ring sweeps

Parameters

Quantity	Symbol	Value used
Disc radius	R	25 cm
Height above the table	h	75 cm
Table illuminance	E0	70 lx
Edge angle	α	18.43°
Fraction E0 / M	sin²α	0.1
Luminosity	M	700 lm/m²
Luminance	L = M / π	222.8 cd/m²

Exam traps

• Writing E0 = M. The table sees only the angle up to α, so E0 = M sin²α.
• Using sin α = R / h. The correct value is sin α = R / √(R² + h²), because the edge ray is the hypotenuse.
• Using L instead of M for the answer. The question asks for the luminosity M = π L.
• Using the point-source rule E = I / h². The disc is wide, so the rule does not apply here.
• Mixing cm and m. The ratio h / R is the same in either unit.
• Forgetting that lx and lm/m² are the same unit.