Lasers, Flashlights, and Eyes - Part 1

This is about the strength of lasers, flashlights, and the sun and their effects to your eyes.

Lasers, flashlights, and the sun are comparable to white surfaces and mirrors. They both reflect most of the light coming at them, but looking at a bright light shining at a white surface is definitely better than looking at a bright light through a mirror. This is because the white surface scatters light while a mirror doesn’t. The flashlight is like the white surface (because it has a wide spread) and the laser is like a mirror (and the sun is a really, really, really bright flashlight).

Divergence:

I’m going to assume that a laser’s divergence rate is \(12cm/100m\) (I don’t know the terminology). A flashlight can vary greatly, but I’m going to assume that it’s 90 degrees, which is definitely an overestimation, or \(200m/100m\)? (still don’t know the terminology). The sun’s divergence rate doesn’t really make sense, since it emanates all 360˚ around.

Power:

Laser: \(1mW\) maybe? One classroom laser pointer I found on the internet was that power.

Flashlight: \(25 W\) maybe? A random thing on the internet said that.

Sun: \(3.86 * 10^26 W\), or \(386,000,000,000,000,000,000,000,000 W\), or \(386\) septillion watts (if the prefix labeling still works like that).

The sun produces \(43%\) visible light, so there is \(165.98 * 10^{24} watts\) of visible light energy.

Experiments:

Uh oh.

Sun viewing(without atmosphere):

The arc length of your pupil = \(2mm\) (really bright lights affect pupil size)

distance to sun = \(93\) million miles (approximately)

sun radius = \(432,690 mi.\)

sun surface area = \(6.07 \times 10^18 m^2\)

If we use similarity on the pupil diameter, we get

\(\frac{2mm \times 432,690 mi.}{93,000,000 mi.}\) \(\)

\(= 0.00930516129 mm\) as a diameter, or \(0.00465258064 mm\) as a radius.

Finding the area of the circle with that radius, we get \(0.0000680045 mm^2\), which is your pupil’s area, if it was projected onto the sun’s surface. That means that without atmosphere and that stuff, your eyes would receive

\(\frac{0.00465258064 mm^2 \times 165.98 \times 10^{24} watts}{6.07 \times 10^{21} mm^2}\) \(\)

\(= \frac{4.65258064 \times 165.98}{6.07}\) \(watts\)

\(= 127.221636677\) \(watts\)

of visible light reaching your eyes if you look at the sun without the atmosphere. That’s slightly more than 5 times the regular 25 watt flashlight, if it was right in your pupil and was perfectly efficient!

To be continued…

(sorry to all of the math people for using the x symbol instead of the dot or just parentheses to symbolize multiplication)