Electronic – Safety when working with CD burner laser

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.

A laser diode is usually a three terminal device: a common point, a supply pin for power to the laser diode itself, and a photodiode output for feedback. The device you have looks like it has either a built-in controller or is running in straight open-loop (uncontrolled) mode.

If you keep the current within the Safe Operating Area (SOA) for the laser diode, you won't hurt anything, but you end up with an uncontrolled laser. It's not as scary as it sounds, but you probably aren't lasing or, if you are, you have no idea what your actual laser power output level is, which can be dangerous to eyeballs.

Laser diodes are designed to be operated in closed-loop mode where the power supply senses the output level of the laser through that photodiode and adjusts its output current on the fly in order to keep the device laser diode operating in the region where it is actually a laser and also to keep the power output level where it is designed to be. The link you provided isn't a laser power supply in that sense; it's a simple open-loop power supply which is probably better at maintaining a specific voltage/current level, but not actually maintaining a specific laser output power level.