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enter image description here

What is the dependence of my led's output with respect to distance.

I am guessing that since the LED inside the glass pane it is somewhat like a wire. Thus, it might follow inverse proportion with distance.

If I'm not to think of it as a wire then the best choice would be to think of it as a point source which would then make the proportionality inverse of distance square.

Is any of the two right? If not, then please give some details of the right proportionality.

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closed as unclear what you're asking by duskwuff, Bence Kaulics, PeterJ, dim, Daniel Grillo Jun 27 '16 at 12:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ What do you mean by "inside the glass pane"? It sounds like you're asking about a very specific situation here, but you haven't explained what it is. \$\endgroup\$ – duskwuff Jun 27 '16 at 7:02
  • \$\begingroup\$ I think you're really asking about the light that the LED produces, not the manner of its production, aren't you? \$\endgroup\$ – Roger Rowland Jun 27 '16 at 7:38
  • \$\begingroup\$ @RogerRowland Yes, I want to know the dependence of the light that the LED produces \$\endgroup\$ – Ekdeep Singh Lubana Jun 27 '16 at 7:42
  • \$\begingroup\$ If you're worried about what's happening inside the plastic, it sounds like you are thinking about sawing the plastic, and trying to couple it to an optical fiber? All point sources produce light with an inverse square intensity vs. distance variation. A long line of point sources will be reciprocal with distance, up to the distance where the line ceases to to be 'infinitely long', and then will transition down to inverse square as the angle subtended by the line gets to be < 1 radian (ish). A plane will be uniform, until the angle it subtends ... you get the idea. \$\endgroup\$ – Neil_UK Jun 27 '16 at 7:56
  • \$\begingroup\$ Well, maybe not strictly a point source, because the LED will have some viewing angle that depends on the optics of the encapsulation. Have a read of this web page for some background info. \$\endgroup\$ – Roger Rowland Jun 27 '16 at 7:56
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"Dependence" is not a thing, but...

There is no physical concept of "dependence of energy", but I believe you are just trying to describe the relationship between energy output and energy received at a given distance.

Since energy is a cumulative term, we usually refer to its accumulation rate (flux) instead. By analogy, think of energy as the amount of water in a bucket. We could have a bucket with a small opening, that admits only a small flow, we leave under the faucet for a long time -- or -- we could have a wide-open pail that will accumulate the same amount of water in it in a very short time.

To make a meaningful water measurement, we need to standardize, not only time (instantaneous energy is called power), but the size (area) of the opening we are using to make our measurement. In radiative energy, we call this measurement Intensity.

Intensity = Power / Area

It's that simple. This has nothing to do with LED's, glass, DIY, or anything other than basic optics.

In three dimensions, for a source emitting energy uniformly in all directions the intensity drops off as $$\frac{1}{R^2}$$ where r is the distance from the source.

Your assumption about an inverse-square law is not valid here.

It's called a Lens

See that part above about uniformly in all directions? That plastic lens on your LED makes that assumption incorrect. Lenses focus the radiated energy into a smaller area, thereby increasing the initial intensity and decreasing the rate at which that intensity is lost to distance.

In RADAR and other EM antenna design, the expression dBi is used to define the gain of an antenna system relative to an isotropic radiator at radio frequencies.

So...

Intensity = Power / Area

In your question, the "area" is actually the area of a sphere of radius R that surrounds the source of power output P:

$$I = \frac{P}{4\pi R^2}$$

That's the surface area of a sphere in the denominator.

However, you are not emitting into the entire sphere, just a narrow cone that the lens constrains. Let's define a term L, which represents the reduction factor in illumination caused by the Lens.

$$L = \frac{A_{lens}}{A_{sphere}}$$

We already know A_sphere (above). To find A_lens consider:

enter image description here

A_lens is the surface area of the sphere above the orange latitude.

To find "h" in the figure, we need to know the lens' radiating angle (this is typically 30degrees half for the type of LED in your picture, but exact value will be available in the part's datasheet).

Using trigonometry we can find...

$$R-h=R*cos(\theta)$$

$$h = R - R*cos(\theta)$$

and therefore the lens area...

$$A_{lens} = 2*\pi*R^2*(1 - cos(\theta))$$

Ta-da! (e.g. the answer)

Putting it all together, we see that power does not decrease at the rate of the inverse-square law, but substantially better -- by a factor of A_lens:

$$I = \frac{P}{A*L}$$ $$I = \frac{P}{\frac{A_{sphere} * A_{lens}}{A_{sphere}}} = \frac{P}{A_{lens}}$$

...and thus:

$$I = \frac{P}{2*\pi*R^2*(1 - cos(\theta))}$$

where theta is the half-angle of radiation of the lens. Most datasheets will report the full-angle, so be sure to divide by 2 before plugging into this formula.

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  • \$\begingroup\$ I'm sorry but isn't there a mistake on the I formula at the end? If you have P/(Asphere/(Alens/Asphere)) it's equal to (PAlens)/(Asphere²) and not P/Alens, I think it's because I=P/(AL) and not I=P/(A/L) \$\endgroup\$ – damien Jun 27 '16 at 9:47
  • \$\begingroup\$ @damien -- Yes, you're right. My apologies. The final answer is still correct. \$\endgroup\$ – DrFriedParts Jun 27 '16 at 21:04

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