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?

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.

  • \$\begingroup\$ You're making it a whole lot more complicated than it needs to be. Steradians, for instance, are the next-best thing to immaterial; they merely explain (in exposure terms) why an object can have a given apparent brightness even though you aren't receiving the full blast of all of the light it gives off. And an incident light meter is just a simplified camera and grey card in one unit, with 18% transmissivity instead of 18% reflectivity. A given lux level reflected from a grey card will give the same exposure value as the same level falling on the dome of an incident meter. \$\endgroup\$
    – user2719
    Commented Mar 26, 2011 at 5:23
  • \$\begingroup\$ After researching those wikipedia articles you linked, I think it can be definitively stated that Google is wrong if it tells you 1 lumen is equal to 1 candela. On the contrary, 1 lumen is equal to 1 candela steradian, which is a more specific unit, and quite different than 1 candela. \$\endgroup\$
    – jrista
    Commented Mar 26, 2011 at 7:49

2 Answers 2


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.

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
2EV = 2log2 (E
S) / C
2EV = E * S/C
2EV * C/S = E * S/C * C/S
2EV * C/S = E
E = 2EV * 250/100

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

E = 2EV * 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.)

  • \$\begingroup\$ Wow, I really hope this explains things...it looks far more complicated now that I read it than it sounded while I was writing it... \$\endgroup\$
    – jrista
    Commented Mar 26, 2011 at 7:20
  • 2
    \$\begingroup\$ A little complicated at first, but careful reading brings sense to it all... Phew - - brings me memories of many years ago where I studied steradians during my Space Engineering course... \$\endgroup\$
    – ysap
    Commented Mar 26, 2011 at 16:23
  • \$\begingroup\$ Yeah, steradians are an odd concept, but they bring a useful element into units. Sometimes knowing the geometry of something is essential to fully understanding it. \$\endgroup\$
    – jrista
    Commented Mar 30, 2011 at 18:07

You've got the problem space pretty much pegged. Incident light metering measures what's falling on the subject independent of its characteristic reflectivity, etc., while reflected light metering measures what's being reflected from the subject, independent of the characteristics of the incident light. The recording medium in the camera, as you pointed out, records what's reflected from (or, if a light source is included in the image, what it transmitted by) the subject. No matter which type of metering is used, the object of the game is to record the light coming from the subject appropriately for the image.

It's easy to picture what's going on with a reflected light meter, whether that's the in-camera meter or a separate hand-held spot or average meter. Incident light metering is just a little more difficult to picture.

Most incident light metering is done with a dome-type invercone. Essentially, you have a hemispherical diffuser over the light-sensitive element that accepts light from 180 degrees (90 degrees off of the axis in all directions), and transmits 18 percent of that light to the light-sensitive element. (The element can usually be used as a reflected light meter as well, sometimes requiring accessories for the meter.) You can measure light sources separately, either by shading the dome from individual sources, or by selectively turning the sources on and off (as with studio lighting) to determine lighting ratios (the difference in illumination between areas that will be highlights and shadows on the subject).

While the method of working is different, what you are accomplishing is almost identical to placing a grey card at the subject position, angled to be tangent to the subject, and talking a spot meter reading of that grey card. It tells you nothing about the subject -- but that can be a good thing. If your metering depends on the subject, then you need to know what the subject's reflectivity is (or, in Zone System terminology, where to place the subject on the exposure scale). Without that bit of knowledge, you won't be able to tell the truth about the subject using only the meter reading -- you may be able to capture the whole range of tonalities perfectly, but that may mean, say, severely under- or over-exposing the one thing you were really trying to capture.

That's why there's such a divide between incident and reflected light metering. You'll see incident light metering used most often in the studio, but often in location portraiture or fashion/product work as well. There is usually something in the image that needs to be absolutely bang on, and that's best achieved by leaving it out of the equation altogether when metering, then letting its actual reflectance and color characteristics fall where they may on the sensor. The rest of the scene is atmospherics -- you can play with them to your heart's content, and as long as the subject is correctly lit and exposed, the picture will be what you want it to be.

Working in less controlled environments, especially when shooting things like landscapes, you generally want to get it all in (so to speak) without worrying too much about the "truth". In these cases, the "truth" is more subjective, and you are looking to collect the data that support your version of the truth. That may mean making that painfully bright highlight a citizen of Zone VII or forcing the inky blackness up into a barely dark lower midtone in order to preserve the detail that's important to you. But you have to know where things go in your version of the world. Or you can let the camera's scene-recognition algorithms make that determination for you.

In either case, though, the meter is just the starting point. It gets you into the ballpark, then your experience and vision will let you adjust the objective reality to meet your subjective requirements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.