Lasers, Flashlights, and Eyes - Part 2

More experiments:

The lasers and flashlights don’t have a standard distance, like the sun, so I’m just going to make increments on the distance.

General formula:

if pupil area is < beam spread area (still don’t know the terminology):

real power = \(\frac {pupil area}{π(\frac{distance(divergence rate)}{2})^2} * power\)

The fraction part of this formula is baiscally a ratio of how much power actually reaches your eyes.

else:

real power = power

The pupil area is \(π mm^2\)

0 meters:

The pupil diameter is greater than the beam spread in both cases(the beam spread is 0), so the laser will deliver a power of 1 mW and the flashlight will deliver 25 W. The flashlight delivers a lot more power.

1 millimeter:

The flashlight’s beam area will be \(π mm^2\), which is the same as \(π mm^2\). Since the laser diverges less than the flashlight, it will be less than \(π mm^2\).

They will still deliver the same power as before.

1 centimeter:

The flashlight’s beam area will now be bigger than the pupil area. The laser radius will be at \(0.006 mm\), so the area is \(π \times 0.00036 mm^2\), which is not smaller than the pupil area.

The laser will still deliver 1 mW.

The flashlight will be:

\[\frac{π mm^2} {π(2/1 \times \frac {10 mm}{2})^2} \times 25 W\] \[= \frac{mm^2}{100 mm^2} \times 25 W\] \[= 1/4 W\]

Which makes sense if you think about it, because the distance grew by a factor of 10. Therefore, the radius of the cone of light grew by a factor of 10, which means that the area grew by a factor of 100. That means that the power should be \(\frac {25}{100} W = \frac{1}{4} W\)

1 meter:

The pupil area (still assuming 2 mm diameter) will be smaller than both, since the beam spread for the flashlight is \(π m^2\) (not mm!) and the beam spread for the laser is π 0.0036 m^2, which is bigger than π/ 1,000,000 m^2 (the big denominator is from converting from mm^2 to m^2). From now on, only the first formula is used.

The flashlight will be 25/1000000 watts = 1/40000 watts. You can use the area scaling observation in the 1 centimeter section to find this.

The laser will be

\[\frac{(π/1000000) m^2}{π 0.0036 m^2} * 1 mW\] \[= \frac{1 mW}{(10000000 * 0.0036)}\] \[= 1/3600 mW\]

But how does a bright light actually harm your eyes?

When staring at the sun or something like a welding torch, ultraviolet light can go into retina and burn the flesh.

Does that mean I can stare at the sun with ultraviolet glasses on?

Still no. If the light is energetic enough, it can react with some chemicals to produce free radicals. Free radicals are basically atoms without a paired electron. They can damage cells by giving them “oxidative stress”, which I am guessing is the result of the radicals ripping off electrons from the cell wall. This can damage your eyes.