I am looking at Kinect Laser and couple of things that are not clear, so I thought I bring up here. First the known facts: - Kinect has a laser projector that works at 840nm - Output power of the laser is 60mW - The laser projector has a diffuser. I didn't break one and look inside but my guess is laser diode->collimator->diffuser is the path. - Kinect is a Class 1 safety device.

My primary interest is understanding Kinect's laser safety. For interested parties, the following links discuss some of these stuff and I will not repeat them here. - A very informative link - An independent Kinect Laser output measurement

My approach is as follows: - Assume 60mW output power is correct - Diffuser efficiency is 50% and therefore 50% of the energy is lost - Diffuser output projects a rectangle image with angles of 45 degrees on both axis (vertical and horizontal)[Probably the angle is different but good enough for us to make some calculations) - Diffuser output is 20mm away from the Projector window (this is the glass in front of the laser diode, light output leaves this window and projects the image)

Now, given 45 degree angle, the size of the rectangle at the output of the window is still 30mW spread over the size of the rectangle. (the size of the rectangle is tan(45)*20mm*2=40mm one side, 1600mm2 area of the light rectangle at the window.) Assuming uniform energy distribution, we get 30mW/1600mm2 = 1.875mW/cm2

This looks high to me. I am trying to figure out why it is given Class 1.

Now the questions:

  • This light is not collimated, in fact is diverging, if you think of your eye, the eye will focus only a very small portion of this light to the retina. (model for eye is just a lens with focal length of 17mm) Assuming pupil size is 7mm, the area of the pupil is 0.39cm2. So the energy enters the eye would be 18.75*0.39=7.3mW. If this was a straight line, you would end up blind since the lens would collimate that straight line, however in this case, the light comes with 45 degree angle, therefore eye will only be focusing on a very small portion of this energy properly. I am trying to figure out is this the reason why Kinect denoted a Class 1 certification. Even if I am off by 50%, there is still tons of energy coming out of Kinect laser. I did more checking on this, using the attached pdf, one can conclude the beam from Kinect will have an area of d=ftheta, d=1.7(45 degree in radians)=1.7*785mrad->1.335cm. This is the dia1meter of the spot on your retina. The area is a=pi*d^2/4=1.4cm2 so the retinal irradiance becomes E=7.3mW/1.4cm2=5.2mW/cm2. This still seems exceptionally high to me, even if it is diverging beam.

  • My second suspicion is that, Kinect is indeed dangerous however no sane person would stick his eye to the projector and at normal working distance (say 1meter) the energy entering to the eye would be very small (tan(45)*1000*2=2000mm, area is 4m2. 30mW/4m2 =7.5mW/m2=7.5 10e-4 mW/cm2 which is safe for eye. However if this is how Kinect got the Class 1 certification, it is pretty scary, since I am pretty sure some idiot out there will stick his eye to see what is inside.

Any insights would be appreciated..

[Update] I have done some measurements today.. Here they are:

Using a PD (Osram BPW34FA) and a 10K resistor. I manually touch to the surface of the projector's window with the PD and 10K resistor. I measure voltage across resistor using a scope. The result is approximately 500mV. (I took out the 4mW ambient measurement, it was 504 but when Kinect isn't there it is 4mW, so the delta due to Kinect is 500mW)

Now, 500mV translates to 500uA current on this diode. (500mV/10K=500uA). At 7mm2 (assuming pD is uniform energy absorption, which is a reasonable assumption) per mm2, I get 71uA. The diode's efficiency is 0.65A/W, I simply use this to calculate incoming power per mm2, it translates into 109uW/mm2.

I also measured the angles of the diverging pattern. They are 50 degrees vertically and 54 degrees horizontally. (Use a sony cam in nightshot mode and a ruler. The patter is not perfect rectangle but rectangle is a reasonable assumption)

So, assuming the output of the diffuser is 12 mm away from the window, you get an approximate area of 12*tan(50)= 14.4mm and similarly 17.7mm. So the total area of the beam at the window is 254mm2. Since we established per mm2, this thing emits 109uW, total energy output becomes 27.6mW, which is inline with 60mW laser output with 50% optical efficiency. I didn't break the Kinect to measure the depth of the diffuser output but I guess it could easily be 1mm, in this case the output power translates to 18mW at the window. Anyway it is high and I am still not understanding why it is Class 1. Hope you guys can help.

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    \$\begingroup\$ Rushing out. More later maybe. Sounds risky to me "at a glance" (pun almost intended). I have some 160 mW white phosphor LEDs with good efficiency (120+ lumen/Watt). Nichia evaluated these at my request and concluded that they constituted a potential eye hazard at the blue end of the spectrum under current standards. They have a double half angle (aka cone angle) of 80 degrees. That evaluation was probably about at the LED surface (I'd have to check) but I suspect they'd be safer than your projector. \$\endgroup\$
    – Russell McMahon
    Commented Aug 22, 2011 at 2:40
  • \$\begingroup\$ A cure for the projector would be a safety frame that extended far enough to make the system safe, did not occlude the expanding beam, did not allow head entry and needed to be there to allow system to work. Could collapse to minimal size for carrying. \$\endgroup\$
    – Russell McMahon
    Commented Aug 22, 2011 at 2:40
  • \$\begingroup\$ @russel mcmahon, do you have experience with such calculations? I am looking almost a walkthrough. \$\endgroup\$
    – Frank
    Commented Aug 22, 2011 at 10:19
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    \$\begingroup\$ no experience of formal calculations for LASERs. But extensive doses of common sense and applied engineering and years of technical calculation experience. :-) - and lots of LED experience. I design these and similar and they too need to be safe. I'll try to look at your figures and see if they make sense. I may also have some references (only possibly). See my ID page for email. I didn't have time to check your figures in detail before but the approach seemed reasonable. Worst case I'd take the eye-lens area and not just pupil for irradiated area. \$\endgroup\$
    – Russell McMahon
    Commented Aug 22, 2011 at 10:33

3 Answers 3


I see a fundamental units problem in your math. You are using the symbol m which stands for milli (10^-3). The actual energies are µ (10^-6). You are off by a factor of 10^-3. Since most keyboards do not have a symbol for µ (micro), people often revert to an m (milli), and end up confusing the units. Hey, they both start with the same letter, right? :)

Additionally, there is the question of "wall plug efficiency". While a diode may 60mW, even military grade lasers only have a 10-15% wall plug efficiency. Thus that means an output of around 6mW. Then any optical element will reduce power by around 50% for each element. Assuming at most 2 optical elements (nu-naturally low number), that means the output can at max be 1.5mW.

In the answer you quoted, use this paragraph as a point of comparison:

For the sake of comparison, sunlight is one kilowatt per square meter and perhaps 5% of that is near infrared i.e. 700 to 1000 nanometers. Just going outside will expose you to much greater power densities of SWIR than the Kinect.

Also, remember that even though the generator is 60 mW (yes, I used the correct units), there is a series of diffusers, optics, and such so that at the very extreme of the exit aperture, the power density is <25 μW (again, note the symbol). The series of steps required to get at the 60 mW generator would indicate a willful intent to cause self harm, and be beyond simple mechanical failure.

Your initial assumption is incorrect.

My approach is as follows: - Assume 60mW output power is correct - Diffuser efficiency is 50% and therefore 50% of the energy is lost

The diffusers and optics reduce the power to <25 μW at the aperture. Run your math with that figure and you'll have an accurate representation.

  • \$\begingroup\$ than as I understand, the measurement is done at the worst case condition. (i.e. at the output of the window) and the optical output must be less than 25uW. (is this for the eye aperture of 7mm?) And final question, does the diffuser angle play any role in this? \$\endgroup\$
    – Frank
    Commented Aug 21, 2011 at 16:21
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    \$\begingroup\$ @Frank, Yes, the measurement is done at the exit lens (the side we can put out eye up to). And yes, the diffuser angle does play a role, but considering the actual energy levels involved, the effect is really irrelevant. \$\endgroup\$ Commented Aug 21, 2011 at 17:43
  • \$\begingroup\$ are you sure on the wall plug efficiency. We are talking about optical output directly, not just conversation. This seems inconsistent. \$\endgroup\$
    – Frank
    Commented Aug 22, 2011 at 14:34
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    \$\begingroup\$ @Frank, well, it's hard to tell. The input to the diode (or whatever element they are using) is probably well over 60mW, so that efficiency loss may be moot in the calculation. I don't have the schematics for a Kinect, so I really can't tell. :) \$\endgroup\$ Commented Aug 22, 2011 at 22:46

mW - milliWatt
uW - microWatt

LLQ is correct about the factor of 1000 error. I'm not so sure about a number of his other assertions which do not have references and which are not necessarily true.

He gave a similar answer on April 29th here on Stack Exchange skeptics but sadly did not reference it in his current reply.

His path from 60 mW to 25 uW is uncertain. He cites a number of optical steps with a 50% loss in each (and there is no reason that loss needs to be anything like 50% depending on the function performed.)

Frank's window size measure meant has a factor of 1o error (40mm x 40mm = 1600 mm^2, not 160 mm^2) but also there is a jump from mm^2 to cm^2. All easily done but needs checking.

Going back to Frank's original assumptions, IF the LASER is 60 mW optical output then I think it feasible that it may output around 30 mW.

Re the youtube video - the person doing the measurement does not state (afaik) what the aperture of the sensor used is. This may be available elsewhere. The aperture has a very substantial effect on the result. At small distances he was measuring in excess of 1 mW of LASER energy while not intercepting all of the beam.

I have an instrument which may be able to make a useful measurement of this - but I do not have a Kinect. If all else fails I could go and find one.

However apparently Microsoft have sold 8 million + of these and I saw a figures of 14M in the first year mentioned. Microsoft not being stupid, whatever else they may be, and employing more lawyers than most of us would expect to meet in out worst nightmares, are very unlikely to sell a product which they expect to run the risk of blinding anyone at all. Strapping a Kinect on your eye as a fashion accessory probably lies within the range of normal that a Microsoft lawyer would like to protect against.

Some consistent calculations and some properly described measurements still seem like a good idea.

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    \$\begingroup\$ I meet lots of lawyers in my worst nightmares. Just how many lawyers does Microsoft employ? \$\endgroup\$ Commented Aug 22, 2011 at 14:19
  • \$\begingroup\$ @Russell mcmahon, I agree with you that this must be safe. I am just unable to understand how. We are basically reverse engineering their calculations. \$\endgroup\$
    – Frank
    Commented Aug 22, 2011 at 14:31
  • \$\begingroup\$ It appears (from video comment) that the guy in the youtube video is using the The Newport 818-IR-L detector which claims a cylinder shaped active area of 0.071cm^2 and a diameter of 0.3cm. \$\endgroup\$
    – Mark
    Commented Aug 22, 2011 at 16:35
  • \$\begingroup\$ @russell mcmahon Russel I just updated the question with measurements. Appreciate if you can help. thx. Frank \$\endgroup\$
    – Frank
    Commented Aug 23, 2011 at 7:10

Larian LeQuella's answer is obviously wrong and can be dismissed out of hand. He's clearly simply assuming the questioner doesn't understand the difference between micro and milli prefixes because he read on the referenced page that stated Class 1 lasers are restricted to <25 μW, and apparently to him, that's that, case closed. Well, I don't see any prefix misuse in the questioners analysis, and frankly, the idea that you're looking at less than 25 MICROwatts illuminating an entire room in this video taken with cheap, non-image-intensifying camera is completely LAUGHABLE. This device is easily emitting tens of MILLIwatts of IR light. The questioner's concerns are justified and I found this page as a result of sharing them.

However, I believe we have the answer to why the Kinect is eye-safe already right here in front of us as a result of the questioner's treatment (which is the most thorough I can find anywhere on this subject so far). If the questioner's considerations of beam divergence and optical path losses are correct, and the beam intensity at the retina really is 5.2mW/cm2, that is actually a pretty safe, low level. Note that other sources cite the focused intensity of sunlight on the retina as 10 WATTS/cm^2 and a 1 mW collimated laser beam at ~1.7 KW/cm^2. We know that even though you definitely shouldn't do it, looking at the sun and looking at even a 5 mW laser pointer do not immediately damage the retina. So if we're right about divergence being the key to Class 1 classification here, a mere 5 mW/cm^2 is really quite safe...unless someone puts a lens of some kind in front of the output....

  • \$\begingroup\$ Thanks Blake.. I actually find out the real answer to this.. No time to go into details but here is the simple rule.. The eye can focus only at 70mm, therefore even if you stick your eye to the laser window, due to focus problem, your eye deflects most of the laser. At 70mm, eye aperture is 7mm diameter (0.39mm2). At that moment, the eye should not get more than 0.8mW energy to be qualified as class one. I didn't run the calculations but if you do, I bet you will find out that Kinect is just at the border of that value. \$\endgroup\$
    – Frank
    Commented Sep 5, 2011 at 4:02

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