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I'm trying to use LED's as a point source of light for a project I have. a crucial idea is that the intensity of the light of the led falls off as 1 over the distance squared as we move away from it. i.e. it should obey the inverse square law. I ran some measurements on an LED and tried to fit the results on a curve but did not get that result,(the intensity was more proportional to $1/d^3$). my measurements could be non-accurate.

I haven't found any helpful sites that point out that LEDs actually follow the inverse square law, but I'm still skeptic that it does.

so my question is, can an led be used to some extent as a point source? how does an LED light vary with distance and with direction? if unfit, what other light emitting circuit components can be used instead if any?

I appreciate any kind of help.

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  • $\begingroup$ @MahendraGunawardena a laser is nothing like a point source. A laser is a collimated source of arbitrary diameter. A point source, as the OP suggests, radiates uniformly over a hemisphere or sphere. $\endgroup$ – Carl Witthoft Oct 11 '16 at 13:53
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All EM radiation from a point source falls off in intensity with the square of the distance. LEDs can't magically change this basic physics.

LEDs can be quite directional in their emission pattern. One possibility is that the orientation of the LED relative to the receiver wasn't exactly the same in all your measurements. Another is that you were fairly close to the LED so that the point source approximation was not valid. Yet another possibility is that scattering and reflection from surrounding objects was contributing a significant fraction of the received light. Or, the receiver could just be calibrated badly.

There are many possible explanations, but light intensity falling off with the cube of the distance from a point source is not one of them.

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As Olin said, LEDs, even without their lens cap, are quite directional. If you want a decent point source, the typical approach is to use an external lens to refocus the LED output as best you can onto a pinhole (20 microns if possible). This will cause some light loss but the output of the pinhole should be much closer to an ideal point source.

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As Olin mentioned, the emission pattern is quite anisotropic. It depends on the LED. Most have primary lenses which all have unique emission patterns. One example of such can be seen in the Cree X-Lamp XP-G Datasheet (p.10)

Emission pattern and LED On the left is the 'typical emission pattern' given by the manufacturer. On the right is the type of LED I'm assuming you are talking about when talking about a 'point light source'.

The best case scenario in my mind would be measuring it by yourself if you have access to a goniometer. The second best (and the easiest) would be searching the manufacturers datasheet for information, keeping in mind that reality is often quite different from manufacturer datasheet figures. If you have nothing else and must move forward a Gaussian distribution would be my best guess.

Note that this only applies to LEDs such as illustrated. I do not know how COB -type LEDs radiate.

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Technically, any EM radiation that has a reflecting surface can fall off with distance cubed. It is the same thing that happens with radio transmission and ground bounce.

Having said that, I would highly recommend checking your test equipment. Make sure it is linear.

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  • $\begingroup$ say what? Have you looked at the radar equation? $\endgroup$ – Carl Witthoft Oct 11 '16 at 18:46
  • $\begingroup$ Was referring to the picture here 186sci2h13vn1wj5802isw8h.wpengine.netdna-cdn.com/wp-content/…. the website I saw says it diminishes to the fourth power, but I would have to do the math. $\endgroup$ – PJazz Oct 11 '16 at 18:57
  • $\begingroup$ Nothing in that diagram is related to signal strength. The standard radar equation is certainly $ 1/R^4$ $\endgroup$ – Carl Witthoft Oct 11 '16 at 19:51
  • $\begingroup$ the radar equation has two distance-squared terms in it, one for the transmitted pulse and a second one for the return. that's why for radar, the return intensity falls off as 1/distance^4. for a light source illuminating a target, the intensity at the target falls off as 1/distance^2. $\endgroup$ – niels nielsen Oct 28 '17 at 3:03

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