# light meter measurement: incident light and the inverse square law

Discussion in 'Exposure Discussion' started by gongman5000, Feb 8, 2011.

1. ### gongman5000Member

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As I understand it the inverse square law states that the intensity of light will diminish in value at a rate equal to the inverse of the square of the distance. My question is this: how can taking an incident meter reading of a subject be accurate without taking into account the distance of the camera from the subject?

Is the inverse square law only applicable to sources of illumination but not reflected light? This seems counter-intuitive but is the only way i can think to reconcile this apparent discrepancy in my head.

2. ### Paul SorensenMember

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Great question, I am trying to get my head around this myself. I do think that the inverse square rule applies to what are effectively point sources. When light goes through a lens, it becomes basically a point source at the middle of the lens (I am sure that there is a correct word, which I do not know). When metering in a normal scene, you are measuring light that is coming from any number of very diffuse sources. I am sure that the inverse square rule applies, but that it is not measurable because there is not a single point to measure from.

3. ### jeffreygSubscriber

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As i understand it an incident reading reads the light falling on the subject rather than the light reflected by the subject. You should be as close as possible to the subject and point the incident meter toward the light source. It does work but may not be best for every situation.

http://www.jeffreyglasser.com/

4. ### John KoehrerSubscriber

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Thinking about it, your light source(outside) is the sun. If you take a measurement and go 5, 50 feet or 50 miles the distance hasn't really changed.

And you do not have to be near the subject, just in the same light. So, if you're at the rim of the grand canyon & take a reading, the exposure will be the same on both sides.
Assuming no clouds or shadows or solar eclipse or errant asteroid or something else like falling into a black hole just south of Dallas.( I know it's not nearby!)

When you're using an incident reading you're not reading reflected light, that's what a spot meter or reflected light meter does. Point the meter at the subject.
Incident reads the light falling on the subject.

The inverse square law is pertinent to (generally) a point light source. Flash, street light, spotlight table lamp. etc. Whether you use either kind of meter.

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5. ### wiltwSubscriber

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The inverse square law applies to true 'point sources' of light. If you look up illustrations, the light which fans out from a true point source expands outward in all directions, which is why the number of photons falling onto an object decreases by the square of the distance (inverse square).

When the distance to the source is close enough, inverse square does not apply...a softbox used within about 3x largest dimension of the softbox behaves closer to inverse linear. Similarly if you were 2.8 million kilometers from the 1.4 million kilometer diameter sun, the sun is a huge 'softbox'.

Reflected light behaves a bit differently since the object is not a true 'point source', but conceptually is a large collection of points on a surface. So its behavior is not 'inverse square' either.

6. ### Q.G.Inactive

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The further the subject is away from the camera, the smaller it's image.
A smaller image contains less light than a larger image.
See?

(Throw on a longer lens, and the image gets bigger. But also darker.
Put a faster lens on, and the image will get brighter, but only because the lens has a larger image collecting surface, so captures a larger solid angle cone eminating from the subject.)

Reflected light does behave differently only in as far as the reflection could be (very) directional. A concave mirror, for instance, does not spread light, but does the opposite. So stand in its focus, and you'll find out why it's called focus.
But generally things don't reflect like that, and the inverse square law isn't a bad description of how also reflected light behaves.

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7. ### michaelbscMember

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No, point the incident meter at the camera lens, not the light source.

You need to measure the light falling on the subject from the direction of the camera.

8. ### Sirius GlassSubscriber

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As noted before, the inverse square law applied to point sources, not collimated sources nor extended sources. Those follow other laws. I do not have an optics book handy so that I can concisely state these laws, but if the rays of light are collimated [parallel, example sun light] or from extended sources [also large defuse sources, again sun light with atmospheric scattering] light does not fall of as the square of the distance from the source.

Steve

9. ### Q.G.Inactive

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Light also 'works' in accordance with the inverse square law when coming from extended light sources. It can't do anything else.

An extended light source is nothing other than a collection of point light sources. The distance from the object to each of those points in the light source will be different, so it gets complicated because of that: the total amount of light an object receives is the sum of what it receives from every point in that extended light source. But no matter from what point it comes, light will follow the inverse square law.

Unless, of course, the distance to the extended light source is large enough to make the differences in distance be so small that they do not matter.
Then you can ignore the complexity and do simple math again. And still that inverse square thing applies.
A good example of such an extended light source is the sun. Even from here, one astronomical unit away, it has an appreciable size. But that matters so little that it can be completely ignored.

Only when light is directional enough will the inverse square law not apply. Lasers, for instance, do spread, but not in such a way that the inverse square law would be applicable.

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10. ### Ed SukachMember

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As the distance in question increases, the area decreases proportionately (I THOUGHT cube of distance relative to the inverse cube of the area - but, I don't know - possibly my memory cells are getting rusty). The amount of light given off for each unit of area remains the same. An example: If a measurement limited to a one degree circular area of a gray card from one cm indicates an EV of 10, increasing the distance to the card to 100 meters and limiting the measured area to the same area as before will also indicate 10EV.

Reflected metering will indicate an AVERAGE light output of the entire scene, limited only by the acceptance angle of the meter, which may be one degree, five degrees, ten ... or ?? in "Spot" metering.

Incident metering measures the light falling on the subject - the distance to the CAMERA has no effect. Properly, the meter should be directed, generally, at the camera; it is also useful to point the meter at individual light sources to determine their effect on the entire scene (balance).

11. ### Q.G.Inactive

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We are indeed talking areas, not volumes, so it is the square, not cube.

The question is: why?
Light has to travel from the subject to the camera to be captured. And along the way, why would it not behave in accordance to the inverse square law?

It does.

12. ### GalahMember

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The inverse law is irrelevant in incidental light readings (after all, if the light comes from the sun -92 million miles- does it matter which side of the Grand Canyon your'e standing on in relation to your subject?)

However, what does matter is that both your subject and the meter are in the same light (not subject in the shade, you in the sun etc).

Point the dome on the meter towards the camera (away from subject) along a line parallel to that between the camera and subject.

Can't be any simpler!

13. ### phaedrusMember

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People, it just works, leave it alone. Don't go all quantum-mechanic on me

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15. ### Morry KatzMember

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It's amazing that something as simple as incident light measurement is so confusing to so many people. There are similar threads currently on Rangefinderforum and l-camera-forum. Galah's previous post is 100% accurate. I've been doing it that way for about 50 years and it works just fine.

16. ### Q.G.Inactive

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But you don't show that you understand, and certainly do not answer the question.

Incident light metering meters the light incident on the subject.
So far so good. It indeed does not matter, if the light falling on the subject comes from very far away, where exactly you hold the meter, an inch (or any other unit that's relatively small, compared to the distance to the source of light) closer to the subject or closer to the source.

But after falling on the subject, that light has to travel from the subject to the camera to be captured on film.
The subject acts as the source of the light you will allow to fall on the film in your camera.

The question is why, while traveling that distance, it would not spread out and behave according to the inverse square law.
And if it does, why the reading would not change, depending on how far your camera is from the subject. Why do those mountains lit by a distant sun, appear equally bright when you move a considerable distance away?

And the answer to that is that it certainly does behave according to the inverse square law.
With increasing distance, less light reaches the light gathering area of your lens.
The same 'mechanism' (distance), the same geometry however is also responsible for the apparent size of the subject: it diminishes at the same pace as the amount of energy available to create the image of the subject gets less.
The thing that varies with distance is the size of the solid angle, the size of the area that both determines how much light is captured and how big the thing will appear.
The result is that the same exposure will create an equally bright image, no matter how close or far away you are.

Perhaps it's easier to see if you think of light as paint: You do get less of it if you put a bucket of a given diameter further away from a paint spray. But that lesser amount will suffice to cover a lesser sized area in an equally thick layer of paint.

17. ### 2F/2FMember

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The location of the camera is only relevant in incident metering as it relates to composition. There is no technical reason that the location of the camera should even be a factor in how you take an incident light reading, except as it relates to what parts of the subject are shown in the frame. (E.g. one generally would measure the light falling on a part of the subject that is visible in the frame.) Think of using an incident meter only in terms of where the light is in relation to the subject, not where the camera is in relation to the subject. They are great tools because they take the camera's location and the composition out of the equation. No need to compromise their usefulness by reintroducing the factors that you have deliberately nixed by choosing an incident meter in the first place.

Do that if you want to average the light that is falling on both sides of the subject (i.e., from the area covered by the dome). But when using light that is uneven, why would you want to average it as a matter of course? Sometimes you do, yes. But not always; not even often, I would say, for my own pix. Do this in deliberate ratio lighting, e.g. a side-lit portrait, and you overexpose. The more contrasty the lighting ratio, the more overexposed you will be.

Point it at the light for which you want to correctly expose - the "main light." You have gone to all this trouble to craft light, or choose a location and time of day, that will sculpt the subject the way you want it, why would you then average that light with the dark side, which you have intentionally made dark? You crafted, or chose to use, the light that way because you want the dark side to be dark. If you don't want it to be dark, then change the fill ratio. Don't average the exposures for the light side and the dark side as a matter of course; it does not make sense. For best results, one meters the main light source, unless in very flat, even light, in which case one could point the dome practically anywhere and get the same reading.

Overexposed negs are not a problem to print down, so it is easy to get by without realizing that this method is a problem. But try the "point at the camera every time" method with positive film, and you are screwing yourself in anything but even light.

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18. ### benjiboySubscriber

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Just what I was going to write, the explanation is simple the light source the Sun is a fixed distance away, and since you are measuring the light falling on the subject not reflected of the inverse square law is irrelevant.

19. ### DiapositivoSubscriber

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So we have a floodlamp in the midst of the night, in the Grand Canyon, there is no moon, and it is bloody dark. The floodlight illuminates a piano, that we placed near the floodlight. The camera is 300 m away, it has got a 1000 mm lens, and we suppose for the sake of the argument that it frames the piano quite exactly.

The floodlamp light falling on the piano follows the inverse square law. We measure the light coming from the lamp with an incident light meter, without dome, pointing at the light source. We calculate the exposure based on that reading.

Then we walk 300 m to our camera, and set that exposure on it. Will it work?

Or, even, we don't have a 5000 mm but an ordinary 50 mm and the piano looks quite tiny in our frame. Nonetheless, is the tiny spot that represents the piano going to be correctly exposed?

"Experience" seems to suggest that the piano is going to be well exposed.
My (shaky) understanding of physics seems to suggest to me that the light reflected by the piano should indeed obey to the inverse square law, and the reading I take with the incident light meter should produce underexposure. If we had placed the lamp at twice the distance, the light we would have measured would have been 2 stops less. Now we walk 300 m away from the piano reflecting our light in all directions, we should observe a fall of light reflected from the piano with distance.

Fabrizio

PS In this example, the piano fills the image. We observe the light coming under that smaller angle that is given by our tele lens. We are not gathering on film more light than is necessary to describe the piano. A reflected light meter should give indicate a less bright object.

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20. ### Q.G.Inactive

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You're still not getting the question.
And besides, incident metering works the same, whether the light source is millions of miles away, or two feet from your subject.
The question is why, if you measure at your subject (which you certainly should with a light two feet away from the subject), will the reading stil be correct if you set the cmaera up two miles from your subject.
The light seen by the camera has to travel those two miles after being reflected off the subject.

So before anyone else gives the same non-answer you gave, tell us why you can use the same setting you should use with the camera not two miles, but six feet from your subject.

It's easy enough to do that, since the answer has been given a couple of times now.
And it's definitely not "the light source the Sun is a fixed distance away, and since you are measuring the light falling on the subject not reflected of the inverse square law is irrelevant".

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21. ### Q.G.Inactive

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We do!
There is less light, reflected off the piano, reaching the camera than at 150 m, or 10 m.
But don't forget that the image of that piano will be proportionally smaller.
As mentioned before: both image size and light intensity follow the same geometry. So though there's less light, that light has to fill in a smaller spot on film. The intensity per area unit will be the same at 300 m as it would be at 150 m, or 10 m.
So the setting to use is the same too.

22. ### Steve SmithSubscriber

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Light at double the distance is reduced to a quarter of its previous level. However at twice the distance it also covers four times the area.

Not sure what that has to do with the actual question though!

Steve.

23. ### Stephen BenskinMember

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According to the Encyclopedia of Photography, "The luminance of an extended source or surface is invariant with viewing distance. Luminance is also invariant within a lossless optical system because changes in image size are balanced by inverse changes in solid angle. The luminance of a Lambertian diffuser is invariant with viewing angle because the reduced power reflected off-axis is balanced by a reduction in projected area."

24. ### Ed SukachMember

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25. ### DiapositivoSubscriber

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If at 300 m we use a tele lens and we fill the image with the piano, a lesser quantity of light fills the entire image, so we have a different exposure.

If I place on the piano a grey card and read it with a spot reflected meter, which only reads light reflected from the card (that is, supposing the card "fills" the reading angle of the spot lightmeter), my understand of physics tells me that my spot meter will give me a different exposure than the incident light meter used near the piano.

I am not convinced that my understanding of physics is right, though. Actually I very much doubt it.

If I take a picture of a lit monument at night, the exposure for the monument isn't the same even if I am far from the monument?

I'm really puzzled.

Fabrizio

26. ### Q.G.Inactive

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That's it.