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I am trying to understand the relationship between two measures of light: lux and candela. It is probably easy for those with longer experience, but here it goes:

I have learnt (here for instance) that lux equals candela per square meter. But this feels too "simplified" - doesn't the distance from the light source to the area have any impact? A thought experiment:

enter image description here

Imagine that we have a hollow cylinder (yellow in the figure above) with a length of L (say 1 meter) and a inner diameter of D, say 0.1 meter.

In the end of the cylinder there is a lux-meter, measuring the amount of light reaching the end of the cylinder (as measured in lux).

If we would place the tube in front of a light source with a light intensity of 1 candela, what would the lux-meter show?

If we set the inner diameter D to 0.1 meter, the end of the cylinder will have an area of ~0.0078 square meters. But is the lux then simply 1 candela divided with this area? Doesn't the distance from the light source to the area have any impact?

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If the source has luminous intensity of one candela in each direction, a type of normalized "candle" ("candela" is a counterpart of one watt per steradian, but with the color/frequency-dependent weights corresponding to the human vision), then it emits luminous flux one lumen in one steradian. One lumen is one candela times one steradian, a unit solid angle which measures the directions from the source. So the luminous flux of a source that has one candela in all directions is $4\pi\sim 12.57$ lumens.

The luminous flux (in lumens) has consequences – it illuminates areas. The larger areas (of the lux meter, for example), the more luminous flux we get. The luminous flux per unit area of the "lux meter" or another absorber is known as the illuminance. The unit of illuminance is one lux which is one lumen per squared meter (i.e. candela times steradian and per squared meter).

If the cylinder were transparent, i.e. if it were not there, the illuminance in luxes would be the luminous intensity times the solid angle over the area. But the area is $\pi D^2/4$ and the solid angle is apprroximately, for $D\ll L$, equal to $\pi D^2/(4L^2)$, so the illuminance in luxes is $1/L^2$ times the luminous intensity (e.g. one candela). Yes, $\pi D^2/4$ canceled.

The same is true if the interior walls of the cylinder are perfectly absorbing; the light going in different directions than to the lux meter won't affect the reading in either case. If the inner walls of the cylinder were perfect mirrors, the illuminance wouldn't decrease with $L$ – it could be calculated as $1/L_{\rm min}^2$ where $L_{\rm min}$ is the short distance between the center of the source and the left side of the cylinder – I was assuming $L_{\rm min}\ll L$.

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  • $\begingroup$ Great explanation! $\endgroup$ – user2078515 Jan 20 '14 at 9:49
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Candela is a unit of "luminous intensity" or simply "intensity", in the photometric system of units that apply to visible "light". Strictly it applies to point sources (which don't actually exist) but it is accurate for finite sources, when measured at a distance, more than ten times the source diameter, so it is lumens per steradian, and is invariant with distance from the source (beyond that 10x minimum; error less than 1/2%).

Lux on the other hand, is the SI unit of "illuminance" or "illumination" (loosely), and is simply lumens per square meter. It matters not what angle the flux arrives from, on the remote surface; it is simply flux (lumens) per unit area.

For a small (near point) source at distance d along the normal to a surface, at some off axis angle (A), if the source is Lambertian (diffuse or cos(A) off axis intensity) the illumination (lux) at the off- axis remote point is given by :

E = I(0).cos^4(A) / d^2. lux

For an isotropic small source, it is I(0).cos^3(A) / d^2 lux

If the source is Lambertian, then I(A) = I(0).cos (A) candela

and the total source flux is pi.I(0) lumens.

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protected by Qmechanic Jan 9 '17 at 23:56

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