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Why does a laser pointer contain a laser diode, and not just an LED?

A laser pointer contains a laser diode, which essentially shines coherent light over a large angle, and a collimating lens, to take that light and focus it into a beam.

If the collimating lens works for coherent photons coming from a laser diode, why wouldn't it work for non-coherent photons coming from an LED? As I understand it, photons don't interact with each other, and so shouldn't care if other photons nearby are in phase with them or not.

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    $\begingroup$ While you can collimate an LED, the fact remains that the because the LED has a much (much) larger angular spread than a laser, you just can't collimate it nearly as well. So, perhaps you can get an LED down to a few degrees, but the laser will be in tenths of a degree. That means the laser can go across a (big) room and still make a small dot on the screen. $\endgroup$
    – Jon Custer
    Apr 27, 2016 at 13:59
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    $\begingroup$ This is a case where comprehension is much harder to obtain using a particle picture of light than in the wave picture. $\endgroup$ Apr 27, 2016 at 13:59
  • $\begingroup$ @dmckee: Maybe we could go with the NRA version: an LED is like a shotgun, a laser is like a sniper's rifle... :-) $\endgroup$
    – CuriousOne
    Apr 27, 2016 at 19:02
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    $\begingroup$ Everyone is answering the general question in the title but ignoring the real question, which is "If the collimating lens works for coherent photons coming from a laser diode, why wouldn't it work for non-coherent photons coming from an LED?" I think an answer to this question needs to explain some wave optics. Also, OP should change the title to be more specific, as recommended in our FAQ on writing titles. $\endgroup$
    – DanielSank
    Apr 27, 2016 at 19:56
  • $\begingroup$ @DanielSank - I've clarified the title. $\endgroup$ Apr 27, 2016 at 21:18

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Semiconductor light emitters are made of such materials, which have quite large index of refraction. This makes it hard for light to exit the emitter — due to Fresnel equations and low index of refraction of air.

In a laser the light mostly goes back and forth between two mirrors, and reflections only help the lasing. So the light either exits from a tiny area on the side of a laser, or reflects back and continues the lasing cycle.

In a LED, on the other hand, light is incoherent and is emitted in all directions. Only some of the photons generated go at low angle to the normal of the crystal's surface to be efficiently transmitted outside. Most others are reflected back and are likely to eventually be absorbed back, leading to nothing but generation of heat.

To circumvent this, LED crystals are generally packed into a lens, which acts as a buffer between index of refraction of the crystal and air. But this also makes the complete device a much larger light source, so collimation of the light it produces is considerably harder.

The lasers, on the other hand, are used directly as bare crystals behind the collimating lens, so they look much like point light sources, which can easily be collimated.

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  • $\begingroup$ I would also guess that besides overheating, the LED cavity with a small hole would behave like a black body source, making you not have a single frequency as is often desired from lasers. $\endgroup$ Jun 22 at 21:51
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    $\begingroup$ @CiroSantilliПутлерКапут六四事 not sure what you mean. A non-laser LED doesn't even have a cavity (at least functionally). And if the crystal overheats so much as to emit blackbody light with frequencies close to the spectrum of its electrolumiscence (hundreds of terahertz), it'll quickly degrade and cease to be a diode. $\endgroup$
    – Ruslan
    Jun 22 at 22:34
  • $\begingroup$ By cavity I meant trying to add an (insulated) box with a small long hole in an attempt to collimate the light. But as you say, the LED would stop working, I wasn't sure about that. I guess just a blackbody oven with no LED would be a better thing to try and compare for educational purposes. $\endgroup$ Jun 22 at 22:48
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The light from a typical laser emerges in an extremely thin beam with very little divergence. Another way of saying this is that the beam is highly "collimated". An ordinary laboratory helium-neon laser can be swept around the room and the red spot on the back wall seems about the same size at that on a nearby wall.

The high degree of collimation arises from the fact that the cavity of the laser has very nearly parallel front and back mirrors which constrain the final laser beam to a path which is perpendicular to those mirrors. The back mirror is made almost perfectly reflecting while the front mirror is about 99% reflecting, letting out about 1% of the beam. This 1% is the output beam which you see. But the light has passed back and forth between the mirrors many times in order to gain intensity by the stimulated emission of more photons at the same wavelength. If the light is the slightest bit off axis, it will be lost from the beam.

Italics mine.

So laser light is not only coherent but also highly collimated. Have a look at the lunar ranger experiment

The ongoing Lunar Laser Ranging Experiment measures the distance between Earth and the Moon using laser ranging. Lasers on Earth are aimed at retroreflectors planted on the Moon during the Apollo program (11, 14, and 15) and the Lunakhod 2 mission. The time for the reflected light to return is measured.

 .....

At the Moon's surface, the beam is about 6.5 kilometers wide and scientists liken the task of aiming the beam to using a rifle to hit a moving dime 3 kilometers away. The reflected light is too weak to be seen with the human eye: out of 10^17 photons aimed at the reflector, only one will be received back on Earth every few seconds, even under good conditions. They can be identified as originating from the laser because the laser is highly monochromatic.

The highly collimated laser beams have many more applications than just pointers.

Edit after comment:

A LED typically has milliwatt power, which is dispersed in 4pi with 1/r^2 fall off in intensity. A same power pointer laser has all the watts concentrated in the beam by initial production of photons.

One can get coherent light from an incoherent source by passing it through a slit, this would give the tiny 1/r^2 fraction of power to be collimated using the laser geometry. The laser in the collimation process does not lose intensity, but gains , because of the lasing action.

the light has passed back and forth between the mirrors many times in order to gain intensity by the stimulated emission of more photons

The LED has a one off contribution to the beamm, and the end point would have orders of magnitude less intensity at the dot than the concentrated intensity laser action delivers.

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    $\begingroup$ I'm not sure this answers the real question, which is why the collimating lense works better with coherent light. $\endgroup$
    – DanielSank
    Apr 27, 2016 at 19:53
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    $\begingroup$ Thanks anna, this is an interesting answer, but I don't see how it answers the question. Is the same thing true of a laser diode? $\endgroup$ Apr 27, 2016 at 20:38
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    $\begingroup$ You don't seem to get what I mean: due to the small aperture, the laser in a laser pointer also has highly divergent beam. It's not collimated in any way, that's why the construction does include a collimating lens. The difference from a LED is that for a laser the actual light source is small, close to being a point source, which makes collimating it trivial. $\endgroup$
    – Ruslan
    Apr 28, 2016 at 4:35
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    $\begingroup$ 1. There are LED spotlights which do have small angle of divergence (although their lenses are an order of magnitude larger). 2. The angle is not $4\pi$, it's about three-four times smaller — and — almost the same for a LED and an edge-emitting semiconductor laser. See also Why does a laser beam diverge?. $\endgroup$
    – Ruslan
    Apr 28, 2016 at 5:53
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    $\begingroup$ Given good enough (large enough etc.) collimating optical system, yes. The only problem is the size of the light source. Well, this is of course unless you want a diffraction-limited spot. $\endgroup$
    – Ruslan
    Apr 28, 2016 at 6:05

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