It's not magical, it's logarithmic.
During the daytime, we are illuminated by 'blinding' amounts of sunlight: thousands of watts per square meter. It really is blinding when you step out into the sun from a dark room, but your eyes adjust until it doesn't seem that bright anymore.
At night, when there are thousands of kilometers of Earth and iron between us and the sun, none of it hits our eyes directly. But some of it bounces off of the moon. It's traveling away from us, and it bounces off of the moon and scatters in every direction. Some of it bounces in our direction, and so we see the moon by reflected light.
The amounts really are miniscule, simon. Roughly speaking, a sunny day is one hundred times as bright as indoor lighting, and a hundred thousand times as bright as moonlight. If this doesn't seem right, it's because our eyes are not linear detectors. We'd never survive if they were, so we have the ability to adjust our aperture size, and use a redundant system of cone and rod cells of various sensitivities and distributions.
You asked what percent of the laser I think gets reflected back to my eye. I can calculate that amount for you. What most of us are saying is that the reflection at the air-salt interface should be treated as a diffuse, as opposed to a specular, reflection. That means that the light energy will be divided equally among all directions, all 'one hundred and eighty degrees', as you put it. But 180 makes a half circle, what we really mean is the half of a sphere that makes up every direction above the ground. This is measured not in degrees but in steradians.
So one hundred percent (roughly) of the light is spread out over 2 pi steradians. (4 pi steradians = one whole sphere; also, white salt will reflect close to one hundred percent of incident red light.)
How many steradians is your eye at a distance of (say) 200 m? Divide its cross-sectional area by the square of the distance from the source.
I found several websources listing the aperture area for the human as 1000 square millimeters, but this is 10 square centimeters, and far too large. Perhaps this is the surface area of the entire eye?
I'll use one square centimeter, or 0.00001 square meters. Then the solid angle occupied by a 200 m distant eye is
0.00001/(200)^2 = 2.5 * 10^-10
Divide this by the total solid angle, 2 pi, and you have the fraction of light from the diffuse reflection that actually makes it into the eye: about 4 * 10^-11, or one part in 25 billion.
I understand simon's point completely: how can one part in 25 billion be enough to see? But remember that the whole amount of laser light shining into your eye is much brighter than the sun, and the diffuse reflection can be dimmer than the moon and you could still see it. In fact, many people can perceive brightness levels that correspond to only tens of photons per second.
How many photons would make it into your eye from the laser in the salt flats?
The power of the laser is probably half a watt, or half a joule per second. It is made up of red photons, with a wavelength of 650 nm, and thus a frequency of (c/650*10^-9) = 4.6 * 10^14 Hertz. This means that each photon has an energy of
E = h f = (6.626 * 10 ^ -34 Js) * 4.6 * 10^14 Hertz = 3 * 10^-19 Joules.
Thus, to make 0.5 Joules, it takes 1.64 * 10^18 photons per second of red light. But only one in 25 billion of these bounces off of the salt and comes exactly back to your eye: this means that your eye is being struck by 'only' 65 million photons per second. As I've mentioned above, this is easily above the minimum levels that the human eye can perceive.
Your doubts were well-founded, but the eye is much more robust than you give it credit for. It can handle the intense sunlight, but can get by on a few million or even a few hundred photons per second. You are correct in saying that if you keep subjecting light to diffuse reflection, you won't be able to perceive it any more--maybe you'd see your hand by the light of the reflected torch, maybe not. (I've seen my hand, and even read street signs, by the light of the sun diffusely reflected off of the moon and then the street sign.) But this doesn't mean the light isn't there, it just means it's finally been spread to thin to perceive.
P
