# Why can we see light further than it shines?

Mainly because the driver sees the much brighter road right in front of the car, which is reflecting a greater portion of the light from the headlights. The light reflected from kilometers away is much less intense. The drivers eyes are not sensitized to the less intense light from farther away when they are mainly seeing the more intensely lit road in front of them. There is also the fact that the light you see from the headlights is much more intense than the light that the driver sees reflected off of you. This is due to doubled distance of the reflected light traveling with intensity lowering with the inverse square law https://en.wikipedia.org/wiki/Inverse-square_law#Light_and_other_electromagnetic_radiation , and because part of the light that reaches you is absorbed and part of it is not reflected directly back to the driver.

Let me try to combine the existing answers into a full picture.

There are four effects which make you perceive light reflected from a distant object much weaker than a direct light at the same distance:

1. Only a small fraction of light is reflected from most objects.
2. The light is reflected isotropically, so only small fraction of light is reflected in the direction of the car.
3. The total distance traveled by light is twice the distance between the car and the object.
4. Pupils adjust the vision sensitivity to the brightest object in your view: that will be the objects close to your car. This makes farther objects harder to see, even if they are somewhat lit. If you in turn look at a distant car, the brightest objects will be its lights and you will see them easily.

(@Vid provides calculations explaining points 2. and 3.)

This is why retroreflectors (like street signs, or reflective vestments) are so much more visible, because they address points 1 and 2: they reflect the light well, and they reflect it at the direction of its source. At the same time presence of a street sign, or incoming car lights in your view may make other objects even more difficult to see.

Well, we know that the density of light (flux density) decreases with $$r^2$$. We can assume that the car light is an isotropic source with power $$P$$. So we can see, that the flux $$j$$ at given distance, where there is a mirror is: $$j_0=\frac{P}{4\pi r^2}$$ So, for understanding of the problem we can assume, that the light isotropically reflected from the mirror back to the driver. The flux density of reflected light, when it reaches the driver will be: $$j=j_0 \frac{1}{r^2}=\frac{P}{4\pi r^4}$$

The density of reflected flux depends on size, shape and albedo of mirror. But the main phenomena, why driver can't see the mirror, while the mirror can "see" the driver is, that the first flux decreases with $$1/r^4$$ and the other with $$1/r^2$$.