exposure - What is the difference between luminance and illuminance?

The more I read about this, the more confused I become. Illuminance, as I understand it, measured in lux, is the perceived brightness of light on a surface. It's what you'd measure with an incident light sensor. Luminance, then, is how much reflected light you'd perceive looking at that surface. This is what the light meter in a camera measures. Am I correct so far?

In both cases, the word "perceived" is important, because the scales are weighted using a luminosity function to map wavelengths to the human eye's particular sensitivities. However, in the case of illuminance, there's only actual perception if your eye happens to be the surface in question.

I can basically grasp this, but then I come across charts saying things like "family living room: 50 lux". Wait, hold on! Does that really mean that the typical lights in a house are that bright, or is it just confused and wrong, or am I confused and wrong?

If you're not taking pictures of light sources directly, why would an incident light meter reading ever be useful in photography? Reflected light recorded on film or by a sensor is usually what makes a photograph. So, if I have an incident light meter, how does that reading relate meaningfully to my camera settings?

Since incident light meters are sold and used, this implies that there must be some useful conversion. But this is where my brain explodes. Google tells me that 1 lumen equals 1 candela, so therefore 1 lumen/m² (i.e. 1 lux) must equal 1 candela/m2 (i.e. 1 nit). But clearly something is missing from this. There's something called a "steradian". Cones are involved. I'd never heard of this before; how does it fit in? I can see how it might be determined when calculating the usefulness of LED home lighting, but for a photo, I'm at a loss.

Some part of my (exploded, from the last paragraph) brain is trying to relate this to the difference between flash metering with TTL and with an incident meter. But how can the incident meter work without knowing the reflectance properties of the objects in the scene? Is this what the "C" in the incident light meter lux→EV standard is all about? Is it just an average-this-probably-will-work value, or is there more to it than that? And if it is just an average, what knowledge is necessary for compensating for off-averages scenes? (As with the K constant and 18% gray with reflective metering, where the photographer simply judges if the scene should be rendered brighter or darker than the average given by the meter.)

So yeah, so much confusion. In short:

• What's the difference?


• Can one meaningfully convert between the two?



• When and how are illuminance / incident light measurements useful for photography?

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Update: I appreciate Stan's answer, which covers the third point of when and how pretty well. And I think I've basically got the first point figured out, as described above. But I'd appreciate some answers covering the issue of conversion as well, both in the mathematical abstract and as the practical for photography. And I wouldn't mind more why and how, either.

Answer

Stan's answer is excellent in explaining metering from a practical standpoint. You also seem to be asking what the specifics are, particularly around steradians and conversion from Lux to EV. Using the Wikipedia articles you have linked, and a few child links from there, I believe I can explain a few things, and leave the rest up to extrapolation.

First off, steradians. Strange term, and odd concept, however once you understand what it really is, things start to make more sense. To take a step back, lets talk about radians first. A radian is an angular measure, which simply put means the following:

One radian is an arc that is equal in length to the radius of a circle.

A radian is measured in the two-dimensional plane. A steradian is similar to a radian, only measured in three dimensions. The definition of a steradian is as follows:

One steradian is a circular patch on a sphere's surface who's area is equal to the square of the radius of the sphere.

A steradian is an odd "projection" of a 2D subtended angle into three dimensional space, or what is called a solid angle. The intersection of the 2D angle with the sphere's surface intersects a circular patch (which is itself bisected by the radian arc.) Another term for it is squared radian. The solid angle that represents one steradian is computed as:

θ = A/r^2

Which, intriguingly, is simply r²/r², or has the units m² * m^-2, which makes a steradian, like the radian, a unitless specification describing a fixed area on the surface of a sphere relative to the sphere's radius.

To complete the definition of a steradian in relation to a sphere:

The solid angle of an entire sphere is equivalent to 4π sr.

One might look at that another way:

The surface area of a sphere has the unit 4π sr.

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Now that the definition of a steradian is out of the way, we can come to a clearer understanding of the relationship of a lumen to a candela. According to the linked wikipedia article:

1 lm = 1 cd sr

Or, one lumen is equal to one candela steradian. A candela steradian is the luminous power of light emitted from a steradian, which as we now know from the discussion above, is the area of a circular patch on a sphere equal to the square of the radius of that sphere.

If we bring a light source into the discussion, to make things more real, that would translate into the following. Assuming we have a light bulb with a 1.5" radius, that meters at 1 lumen, could be described as emitting 1 cd from any area on the surface of that bulb measuring 1.5"² (2.25" total area).

The full light bulb is actually emitting a total of 1 cd 4π sr, or a total of 12.57 lm from all angles. The light meter will not measure 12.57 lm though, because it is only measuring from one angle to the bulb, not all angles to the bulb. If we assume our light meter is effectively sensitive to roughly one steradian, then it will measure 1 lumen.

Further Questions?

Q: One may ask, why equate 1 lumen with 1 candela steradian, rather than just 1 candela?

A: The answer is geometry. Describing a candela is useful in telling us an amount of light, but not necessarily its concentration or the shape and size of the emission. The purpose of bringing steradians into the mix is that it involves a specific geometric shape and area to the light source emitting a lumen of light.

It becomes more important when you have a high density light source. For example, a low power laser pointer (milliwatts) could equate out to 250,000 w/sr. Now when you consider density capabilities of say an eye at 120,000 w/sr it become more than trivial - you see? Oh wait, you won't see!

According to wikipedia, one lux is the measurement of lumens per square meter. In unit terms, since one lm is equal to one cd sr, then:

1 lux = 1 cd sr/m^2

If I understand correctly, that could be read as 1 lux is equivalent to the amount of light received on a surface illuminated by a light source with a radius of one meter that emits 1 cd sr of luminous power towards the surface measured.

The conversion of lux to EV is a fairly simple one that involves a constant C. I can't say specifically how C is derived, however if we assume that the "common" value of 250 is accurate, a simple formula for converting from lux to EV would be:

EV = log₂(E*S)/C

Where S is the sensor ISO, and E is the illuminance in lux. If we assume a scene is illuminated with 1 lux, and our ISO is 100, then (long hand, translated to base ten logarithm for computation on a common calculator):

EV = log₂ (1 * 100) / 250
EV = log₂ 100/250

EV = log₂ 0.4
EV = log₁₀ 0.4 / log₁₀ 2
EV = -0.398 / 0.301
EV = -1.322

A pretty low exposure value, however that would be expected for the measly illumination provided by 1 lux. To go the other way, and figure out how much illumination you need to support a specific EV, we can rearrange the conversion between EV and E (long hand):

EV = log₂ (ES) / C
2^EV = 2^{log₂ (ES) / C}
2^EV = E * S/C

2^EV * C/S = E * S/C * C/S
2^EV * C/S = E
E = 2^EV * 250/100

That brings us to a nice simple formula to compute lux from EV (when ISO is 100):

E = 2^EV * 2.5

If our target EV is 1, then we compute lux as so:

E = 2¹ * 2.5
E = 2 * 2.5
E = 5

For an exposure of one EV, we need exactly 5 lux of illumination, or 5 cd sr/m^2, or 5 lm/m^2.

Further Questions?

Q: One may ask, why measure lux, which are lm/m^2, rather than simply measure lumens?

A: The answer would be units, or more specifically, units of area that a human can readily recognize. A candela steradian is useful in telling us amount and geometry, however a steradian is unitless itself. It purely defines geometry, but does not specify any specific area. A steradian is a steradian, regardless of the actual radius of the sphere. A candela steradian per meter squared, however, brings in enough unit specificity that we can more clearly understand exactly how much light 1 lux actually is (which, quite frankly, is not much at all.)