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Does the inverse square law begin to take effect the moment light leaves its source? For example, does light's intensity decrease, i.e. does the area in which the photons might land increase, at a few millimeters from the source?

I happened to come across an article about emergency lights and photometry from a few decades ago that appears to answer in the negative:

"The minimum test distance in photometry of these sources is called the 'minimum inverse-square distance.' The illumination from the light source, measured at distances greater than this minimum, obeys the inverse-square law which is a necessary criterion for the determination of luminous intensity. [...] The minimum inverse-square distance is determined by the type and size of the light source, lens, reflector, etc., and must be considered individually for each unit. If this distance is more than 100 meters (approximately 328 feet), a ranger larger than 100 meters must be used."

Source: Howett, et al. 1978. "Emergency vehicle warning lights: state of the art." USDC. NBS Special Publication 480-16.

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    $\begingroup$ No minimum distance is required, in the ideal case of a point light source in a vacuum. The area a distance $r$ from a point source is $4\pi r^2$, so the photon density at some distance $r^2$ away from a point source is $N/(4\pi r^2)$, obeying the inverse square law. I'm not sure how fog or other considerations might changes this in the real world. $\endgroup$ – danielsmw May 2 '16 at 19:12
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    $\begingroup$ @garyp From context we clearly need to address non-point sources, but from the first sentence of the question I wasn't convinced that OP realized that the "minimal inverse-square distance" is zero for point sources. It doesn't answer the question (if I thought it did, I would have submitted it as answer!), but I thought it was worth pointing out in case there was any confusion. $\endgroup$ – danielsmw May 2 '16 at 19:26
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    $\begingroup$ @DavidReishi, I definitely meant from a point source. I'm not sure that "the point of a source" is even well-defined except for in the case of a point source. That's the whole thrust of this question, right? When can you treat a non-point-source as a point source, i.e. subject it to the inverse square analysis? $\endgroup$ – danielsmw May 2 '16 at 21:53
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    $\begingroup$ Re, "... then isn't the inverse square law really just about the angle of diffraction and...?" No. The inverse square law is pure geometry. It's about the density of lines passing through a point in a three-dimensional Euclidean space. It's a useful model of light radiating from a source if the source is small enough/far enough away. It's a useful model for the gravitational field of an astronomical body if the body is compact enough/far enough away, etc., etc. $\endgroup$ – Solomon Slow May 3 '16 at 18:39
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    $\begingroup$ I'm saying that the intensity of light radiating from a sufficiently small source (small relative to the distance from the measuring device) can be adequately modeled by the inverse square law. I'm saying that it's true regardless of whether (and, completely independent of) whether the source is a diffraction-limited aperture or whether it is a red-giant star. I'm saying that there is no connection between "inverse square law" and "diffraction". $\endgroup$ – Solomon Slow May 3 '16 at 20:43
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As many have said, the inverse square law applies to point-sources. These are idealized light sources which are sufficiently small compared to the rest of the geometry that their size is of no importance. If a light source is larger, it is typically modeled as a collection of idealized light sources, potentially using integration. The exact definition of "sufficiently small" varies with application. The definition of a "point source" for astronomy is quite different from the definition of "point source" for a LCD projector.

There is actually a limit to this process. The inverse square law is only valid in its normal form if you are working on scales where light can be modeled purely as a wave. As you get very small, on the microscopic scales, those assumptions break down. You instead have to think about the statistical expectation of photons, which follows the statistical analogue of the inverse square law. Even smaller, and you start to enter the world of quantum mechanics, where you have to account for the actual waveforms of the objects under study.

Ignoring these corner cases, nearly all cases you find will have "sufficiently small" defined by macroscopic factors, like the sizes and locations of lenses. Its rare to find oneself in the world where the microscopic factors matter.

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  • $\begingroup$ In your first paragraph, you mention how if a light source is larger, it is typically modeled as a collection of idealized light sources. I found something similar a few minutes ago. "Real sources are extended sources which can be considered as a large collection of identical and uniformly distributed point sources." Does this not imply that the inverse square law, while in terms of actual sources of light applies strictly to point sources or their approximation, is in fact applicable also to light itself in general, at least in theory? $\endgroup$ – David Reishi May 3 '16 at 0:58
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    $\begingroup$ @DavidReishi You seem like you might be more comfortable with understanding why the inverse square law comes into being. It comes into being because the energy radiates outwards in a spherical pattern (or at least some subtended portion of a sphere, typically measured in steradians). The inverse square law comes because the amount of energy available at any given distance from the source is the same (because the light source is constantly emitting the same amount of light). The surface area of that sphere is proportional to the square of the radius. $\endgroup$ – Cort Ammon May 3 '16 at 1:46
  • $\begingroup$ So any energy which is well modeled as "radiating from a point" will obey the inverse square law. $\endgroup$ – Cort Ammon May 3 '16 at 1:47
  • $\begingroup$ I can't help being led by what you say to the thought that the inverse square is describing the progression of the angle of diffraction. Isn't "energy radiating outward" the same thing as light diffracting from a point-like aperture? $\endgroup$ – David Reishi May 3 '16 at 3:23
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Cort Ammon May 3 '16 at 4:45
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The inverse square law applies to point sources. For extended sources becomes accurate at distances that are large compared to the size of the source. At large distances the source looks like a point. What "large" means depend on the application. In the case of light fixtures, the Illuminating Engineering Society and other organizations have made judgments about what is large and what is not based on the use case. Is it room lighting? Is it illumination of products in a grocery store? Etc. There are published advice and tables to guide the lighting designer.

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  • $\begingroup$ But doesn't light from a point source diffract? Is the inverse square law merely about the continuation through space of the angle of diffraction? Or are we talking about a point source in a different sense? Also, if you can see the following image, link, can one demonstrate the inverse square law with this set-up, and, if so, is there a point source involved? $\endgroup$ – David Reishi May 2 '16 at 20:27
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    $\begingroup$ For distances much greater than the size of the lens, that will act mostly like a point source. For distances on order the size of the lens, it will deviate from the behavior of a point source. $\endgroup$ – BowlOfRed May 2 '16 at 20:37
  • $\begingroup$ @BowlOfRed, doesn't that imply that the inverse square law doesn't really require a point source in the literal sense? Light from a non-point source far enough a way acts like it's from a point source, but that doesn't make the source of light a point-source. $\endgroup$ – David Reishi May 2 '16 at 20:46
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    $\begingroup$ @DavidReishi Yes, exactly. But when it acts like one, we will usually describe it that way. If you're too close, it's not a point source. If you're far enough away, it is. $\endgroup$ – BowlOfRed May 2 '16 at 21:04
  • $\begingroup$ @BowlOfRed, well can you please take a glance at this: link. Where's there a point source involved here? Or are these guys not doing what they think they're doing? $\endgroup$ – David Reishi May 2 '16 at 23:36
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The inverse square law applies to point sources. A real emergency light is not a point source, and therefore the law appears to not apply at close distances, because any real point is at a varying distance from different parts of the emergency light.

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  • $\begingroup$ Would it be correct to say that the inverse square law applies to the light from any single point on the emergency light? $\endgroup$ – David Reishi May 2 '16 at 20:31
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    $\begingroup$ @DavidReishi In a certain technical sense, that's true (within the single-particle picture). But obviously it fails on, say, the dark side of the emergency light, due to interactions with other degrees of freedom (such as absorption of light by the emergency light device itself). $\endgroup$ – danielsmw May 2 '16 at 21:57
  • $\begingroup$ @danielsmw, only in a certain technical sense? Are you sure? In a way, a lot hinges on the question. Either the inverse square law is a phenomenon associated only with point sources of light, or it's a phenomenon associated, albeit perhaps obscurely, with all light. $\endgroup$ – David Reishi May 2 '16 at 23:27
  • $\begingroup$ And what about this? link Where's the point source here? Or do these guys have no idea what they're doing? $\endgroup$ – David Reishi May 2 '16 at 23:30
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The inverse square law says that the intensity of incident light falls off in proportion to the inverse of the square of the distance from the light source.

The important word here is "the distance" — the inverse square law implicitly assumes that all parts of the light source are at the same distance from the measurement point, or at least approximately so. For real-world light sources, which are not infinitesimally small points, this approximation must necessarily fail when you get close enough to the source — you can pick a measurement point arbitrarily close to some part of the source, but you can't get it arbitrarily close to all parts of the source at the same time.


So, how close is too close, then? For that, we can come up with all kinds of rules of thumb (like, say, "no closer than $x$ times the maximum diameter of the source", for some value $x$), but if you want a precise, quantitative answer, we're going to have to do some math.

For simplicity, let's consider the (in some sense worst) case where the extended light source consists of two identical, very small pointlike light sources spaced a distance $2d$ apart. (We assume that the diameter of the individual point sources is very small compared to the distance $d$, so that it can be safely neglected.) We'll take the midpoint between the two point sources (i.e. at distance $d$ from each of them) as the nominal center of the extended light source, and place our measuring device at a distance $r > d$ from it.

Let's first look at the case where the two point sources and the measuring device are all collinear (or just very slightly staggered, so that the two point sources won't eclipse each other). Then one of the point sources will actually be at the distance $r_1 = r - d$ and the other one at the distance $r_2 = r + d$ from the measurement point. Thus (since each individual point source is assumed to be negligibly small, and so follows the inverse square law very closely) the combined intensity of the light from the two sources will be proportional to:

$$\frac12 \left( \frac1{r_1^2} + \frac1{r_2^2} \right) = \frac12 \left( \frac1{(r - d)^2} + \frac1{(r + d)^2} \right) = \frac1{r^2 - d^2} \\\approx \frac1{r^2} \left(1 + \left(\tfrac{d}{r}\right)^2 + \left(\tfrac{d}{r}\right)^4 + \dots \right) $$

where the dots denote higher-order terms ($O(\frac{d^6}{r^6})$ and higher).

We can also look at the opposite case, where the line between the point sources is perpendicular to the line from its midpoint to the measurement point, so that by Pythagoras' law, $r_1 = r_2 = \sqrt{r^2 + d^2}$. Then the actual light intensity at distance $r$ from the midpoint is:

$$\frac12 \left( \frac1{r_1^2} + \frac1{r_2^2} \right) = \frac1{r^2 + d^2} \approx \frac1{r^2} \left(1 - \left(\tfrac{d}{r}\right)^2 + \left(\tfrac{d}{r}\right)^4 - \dots \right). $$

In both cases, the relative error in the $\frac1{r^2}$ approximation is approximately proportional to the square of $\frac dr$ (and the absolute error is thus inversely proportional to the fourth power of $r$), although the sign of the leading error term is different.

Other configurations of point sources (with the same maximum diameter $2d$) will generally fall somewhere in between these two extreme cases. Thus, when the distance $r$ to the light source is, say, 10 times the half-diameter $d$ of the source, we can pretty confidently say that the relative error in the light intensity calculated using the simple inverse square approximation, compared to the true intensity obtained by integrating over the full extended light source, is at most $\left(\frac{1}{10}\right)^2 = \frac{1}{100} = 1\%$.

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  • $\begingroup$ I like your answer, and if I understand it correctly you're giving the mathematical basis for the difference in the result of the inverse square law for two points located on a non-point source...or, in other words, the relative error if we were to regard to the points as a point-source. And in doing so you bring back the original question which I appreciate, namely what happens closer and closer to the source. But as Cort Ammon and I discussed in chat, when dealing with light in the singular, i.e. a single light-wave or photon, it also "loses intensity" with distance...(cont) $\endgroup$ – David Reishi May 3 '16 at 17:09
  • $\begingroup$ (2 of 2) ...in the sense that the area in which its energy might be delivered at a point spreads out from the point at which it is aimed, and that this also obeys the inverse square law. So what about taking your answer further and considering what would occur with this single light-wave or photon closer and closer to the source. Would it begin losing intensity according to the inverse square law from the very moment and point from which it is emitted? In other words, e.g. aiming a laser at a point and analyzing it at a micrometer distance...would the energy land often slightly off aim? $\endgroup$ – David Reishi May 3 '16 at 17:17
  • $\begingroup$ (insertion) In the question above, "e.g. aiming a laser at a point and analyzing it at a micrometer distance...would the energy land often slightly off aim?", I forgot to add that we're talking about a laser firing off one photon at-a-time. $\endgroup$ – David Reishi May 3 '16 at 17:27
  • $\begingroup$ This is all classical ray optics, no photons (or even explicit waves) involved at all. I'm effectively taking the inverse square law for negligibly small sources as given (either as an empirical postulate, or as derived from a lower-level theory), and calculating how much a light source with a non-negligible spatial extend will deviate from it at close distances. (I'm also implicitly assuming that light intensity is an additive scalar quantity, which is not strictly true for coherent light sources due to wave interference.) $\endgroup$ – Ilmari Karonen May 3 '16 at 17:35
  • $\begingroup$ You could certainly calculate the time evolution of the quantum mechanical wave function for a single photon emanating from a point source, and find that its square (i.e. the expected probability of observing the photon at a particular location) does obey the inverse square law (after which you can forget about quantum mechanics, and just use the inverse square law you derived). But that's a lot of work to derive a basic classical result, not to mention that trying to visualize it quickly gets into tricky territory like wave-particle duality. $\endgroup$ – Ilmari Karonen May 3 '16 at 17:35
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The quote from the reference says it all: (I added caps) "The minimum test distance IN PHOTOMETRY of these sources is called the 'minimum inverse-square distance.'"

The minimum distance is therefore a photometry issue, in other words, a measurement problem.

The essence of the measurement problem is how far away you have to be before you can approximate the light source as a point source. That is the minimum distance.

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  • $\begingroup$ You are undoubtedly correct. When I ran into the problem that brought forth the question, I had stupidly forgot about the point-source aspect of the inverse square law. So the quote from the reference read different to me at the time. $\endgroup$ – David Reishi May 3 '16 at 18:27
  • $\begingroup$ It is indeed a complex read. With all this hindsight, it might have been better to state it as: "In photometry, the minimum practical distance for testing of these sources is called the 'minimum inverse square distance'." $\endgroup$ – neonzeon May 3 '16 at 18:47
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Cort and Ilmari have given good answers about the practical issue: the inverse square law is for point sources, and so a non-point source (like an emergency light) will only appear to have the same properties at some minimum distance that depends on the geometry of the real source.

However, it seems nobody has mentioned a different "minimum distance" that applies to even point sources (such as the electromagnetic field generated by a single electron). It turns out that in quantum electrodynamics (QED), the electromagnetic gauge coupling (which determines the strength of electromagnetic forces) is only approximately constant. At very high energies (corresponding to very small distance scales), the coupling strength increases, so that photons at these scales would appear to not obey the inverse square law, but would instead appear to lose their "brightness" even faster. This is of course not at all relevant on the scale of things like emergency lights, but rather on scales even smaller than the proton.

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  • $\begingroup$ That's very interesting. I'm surprised that it's the opposite effect to what I pondered, i.e. that within some very small distance from its source, a light-wave or photon (e.g. fired off one-at-a-time) remains 100% intense, or, in other words, 100% on-aim. $\endgroup$ – David Reishi May 3 '16 at 19:45
  • $\begingroup$ @chase , In regards to the photon could you elaborate on this a little more. Thanks $\endgroup$ – Bill Alsept May 10 '16 at 2:55

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