I'm new to this forum. This is half a question, half a challenge. And it's more engineering then physics but I thought I might get good insights from a physics forum.

I would like to cure a UV activated resin using a collimated $390$ to $420~\rm nm$ light source. The area I need to cure is $3$" ($7.5~\rm nm$) by $3$". ( although the light beam can be bigger then this area and of any shape) so long as it's collimated and as efficient and low cost as possible. I want to get at least $3$ watts of light power over that area and it has to have light everywhere and ideally as evenly distributed as possible. Ouff!

I'll have to build some sort of DIY apparatus to collimate light coming from $390$ to $420~\rm nm$ LED's.

There are different types of LEDs on the market. Single point high power ones ($1$-$3~\rm W$) with viewing angles $40$ to $160$ degrees (the angle at which the light cone expands out) or some even higher power arrays of diodes : $5~\rm W$+

Or low power $5$mm diameter ones with witch I could build a large array of evenly spaced LEDs. ( but not necessarily) these have viewing angles of $25$ to $160$ degrees.

In all cases, the actual light emmiting part is generally square and very small, then it is encapsulated in another package.

I don't know much about optics and I have been pondering about this problem for a while now.

How would you go about doing this? From lenses to mirrors, to what type of led to use, there's pretty much an infinite different ways to approach this and I needed to get the idea out there, so others could share their insight.

Feel free to throw any ideas but try to remain as technically and scientifically correct as possible.

Also, although the problem makes sense to me maybe I failed to share it with you appropriately, feel free to ask questions or clarifications!

EDIT: I intend to pass the light through an LCD screen before hitting the resin. Which brings me to another question. At these wavelength, will diffraction come I to play if the LCD "holes " the light is passing through are 100 microns large?

Edit wavelength can be between $390$ and $420$ blow that will damage the LCD. Above will not cure the resin.

  • $\begingroup$ LED output is typically not isotropic, hence the specialized collimators mentioned in D_D's answer. $\endgroup$ – Carl Witthoft Feb 24 '14 at 19:42
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    $\begingroup$ It doesn't sound like you need collimated light. It seems like you need uniform irradiance over a 3" x 3" square. (Correct me if I'm misunderstanding.) Consider an array of LEDs of moderate beam divergence. Say, a 6" x 6" array, with your sample held about 4" away from it. $\endgroup$ – garyp Feb 24 '14 at 19:42
  • $\begingroup$ Sorry I forgot to specify the reason why it has to be collimated is that the light will pass through a LCD screen with pixels 100 microns large before hitting the resin. It can be not perfectly collimated though. If anything slightly convergent, as it will travel about 1000 microns of glass after exiting the 100 micron pixel hole. $\endgroup$ – Ethienne Feb 24 '14 at 19:52
  • $\begingroup$ I'd rather suggest you looking at lasers. LED with specification of 400 nm output will give you about half of its energy in visible range, not UV. With laser you can have all the power in tiny spectral line, and, as a bonus, you can easily control its angle of divergence by adjusting the lens. See something like this, although it's not 3W (you'd need 6 of them), and quite expensive. $\endgroup$ – Ruslan Feb 24 '14 at 20:37
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    $\begingroup$ To close voters: there is specific and important physics in this question, as its full answer gets down to the second law of thermodynamics (no passive decreasing of optical grasp). It could though be construed as a duplicate of physics.stackexchange.com/q/91507/26076 $\endgroup$ – Selene Routley Feb 27 '14 at 8:47

One cannot collimate light from an LED accurately without loosing a great deal of light and / or being happy with a very wide collimated beam, because the source is often quite a wide extended source (sometimes up to 1mm across). This may or may not be a helpful answer depending on exactly what you mean by collimated, i.e. how accurately you need to collimate, or, otherwise put, what the acceptable angular spread of your "collimated" output is and how much light you're willing to lose. You will need to do some calculations to find out.

Put simply: the main factors here are that if you design your collimation optics to collimate output from the centre of the chip (i.e. a light emitted from a point on the chip centre will become an on-axis plane wave at the collimator output), light from point sources at the sides of the chip will be mapped to slanted plane waves. The angular spread is then of the order of $w / f$, where $w$ is the LED chip's width and $f$ the focal length of the system. You can make this spread smaller by increasing $f$, but then the collimated beam becomes very wide and low intensity and the optics begin to get very big if you don't want to lose light. This may or may not be a problem.

The above is an imaging optics argument, but the idea is very general as we're stepping into the optical version of the second law of thermodynamics. In optical terms this is that the source's optical grasp (sometimes called optical extent or étendue) cannot be lowered by passive optical processing. Optical grasp is roughly the angular spread of a beam multiplied by the beam's width (why the simple word "optical spread" never took hold is beyond me and testament to how badly we English-speaking scientists treat our mother tongue). So you can see my imaging optics argument working here again: you can lower the angular spread in a beam at the expense of widening it. Of course you can simply throw away most of the beam if you want the beam narrow, but the system becomes highly inefficient. So you need to do some calculations to see what works for your application.

You can see the second law form more clearly with the following argument. If your LED chip were a blackbody radiator, and if you could make an arbitrarily narrow collimated beam from it, you could focus all the blackbody radiation in the collimated beam down onto a much smaller spot than the chip. The smaller spot's temperature would rise until steady state were reached, i.e. the power into the chip were equal to that radiated back. By the Stefan-Boltzmann law, if the image were smaller than the LED chip, the image would need to be at a higher temperature than the source to balance power flows, and this violates the Carnot/Clausius statement of second law of thermodynamics that heat cannot continuously, spontaneously be shifted from a lower temperature body to a higher temperature one.

Now the above doesn't rule out some fancy future active technology that can truly collimate an LED chip's output, needing work input of $k_B\,T\,\log 2$ joules for each bit of light state forgotten in accordance with the Landauer Principle form of the second law of thermodynamics. I say more about this in my answer here.

I've just recalled a question very like yours Is there any optical component that uniformizes the incoming light? and my answer to it is here.

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A couple things: this site is not specifically for engineering questions, so there's a small chance this may be closed by mods.

Second, many commercial high-power LED's are listed not by luminous flux, but by nominal diode power consumption; a 1 watt LED actually emits far less than 1 watt of radiant power, so you'll need to think about what radiant flux you actually need.

Third, what is the wavelength range that the resin is usable over? 400 nm is sort of at the border of violet and UV. I've seen some resins that cure at "royal blue" wavelengths; this has the advantage that commercial InGaN high-power emitters are very efficient at 450 nm, with energy-to-light efficiencies approaching 50%; for example, see this very cheap 500 mW emitter. If that's usable, you can buy a few of them for $10-20. The site also sells optics specifically designed for being attached to these emitters to collimate the light.

You can also get 410 nm ones, although they're a little less efficient (although power efficiency is probably not an issue unless your unit is battery-powered).

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  • $\begingroup$ Thanks for your response DumpsterDoofus. The resin is curable up to 420 nm. The reason why I'm staying in the visible range is that I will use and LCD as a photomask. The intent is to build a stereolythographic 3D printer. $\endgroup$ – Ethienne Feb 24 '14 at 19:46
  • $\begingroup$ Which brings me to another question. At these wavelength, will diffraction come I to play if the LCD "holes " the light is passing through are 100 microns large? $\endgroup$ – Ethienne Feb 24 '14 at 19:48
  • $\begingroup$ @Ethienne your 100-micron holes are nominally 200 waves wide, so basically no diffraction to worry about. $\endgroup$ – Carl Witthoft Feb 24 '14 at 20:49
  • $\begingroup$ Thanks Carl. Good thing to know its not something I'll have to worry about! $\endgroup$ – Ethienne Feb 24 '14 at 21:20
  • $\begingroup$ Your link to the 500 mW emitter is broken. $\endgroup$ – Ruslan Sep 14 '18 at 16:23

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