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Frank
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[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.

[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.

Updated calculations and corrected a few errors
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Frank
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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)20mm2=40mm one side, 160mm21600mm2 area of the light rectangle at the window.) Assuming uniform energy distribution, we get 30mW/160mm21600mm2 = 181.75mW875mW/cm2

  • 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*0750.39=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=1.39=77*(45 degree in radians)=1.3mW7785mrad->1.335cm. This is the dia1meter of the spot on your retina. The area is a=pid^2/4=1. If this was a straight line, you would end up blind since the lens would collimate that straight line, however in this case,4cm2 so the light comes with 45 degree angle, therefore eye will only be focusing on a very small portion of this energy properlyretinal irradiance becomes E=7. I am trying to figure out is this the reason why Kinect denoted a Class 1 certification3mW/1. Even if I am off by 50%4cm2=5.2mW/cm2. This still seems exceptionally high to me, thereeven if it is still tons of energy coming out of Kinect laserdiverging 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)10002=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.

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)20mm2=40mm one side, 160mm2 area of the light rectangle at the window.) Assuming uniform energy distribution, we get 30mW/160mm2 = 18.75mW/cm2

  • 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.

  • 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)10002=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.

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)20mm2=40mm one side, 1600mm2 area of the light rectangle at the window.) Assuming uniform energy distribution, we get 30mW/1600mm2 = 1.875mW/cm2

  • 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 18.750.39=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)=1.7785mrad->1.335cm. This is the dia1meter of the spot on your retina. The area is a=pid^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)10002=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.

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Frank
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