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This is a question I’ve been thinking about for a while. If I’m standing in the middle of a straight road, during night, I can see a car coming towards me because of its lights even if it is kilometers away. Notwithstanding, the driver can not see me because the car will brighten the road only few hundred meters further. What physics properties of the light causes this phenomenon?

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    $\begingroup$ will you say the same when a mirror is placed at your position ? $\endgroup$ – Ankit Sep 6 at 16:39
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    $\begingroup$ Closely related, and addressing the fact that reflected intensity goes like $1/r^4$ unless you have focusing optics that create some beam waist. $\endgroup$ – rob Sep 6 at 21:49
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    $\begingroup$ @rob: Yes. To continue the driving example, consider "cat's eyes" and corner reflectors in signs &c that change the inverse 4th power to approximately inverse square. $\endgroup$ – jamesqf Sep 7 at 2:44
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    $\begingroup$ Why do you not ascribe this effect to your eyes and the driver's eyes? You see bright pinpoints of light in a vast darkness, so your eyes are scotopic. The drivers eyes are adjusted to the brightness of the reflection from the first several meters, so are photopic. These give very different lower bounds on the brightness of visible objects. $\endgroup$ – Eric Towers Sep 7 at 4:57
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    $\begingroup$ you may find this video of interest: youtube.com/watch?v=Bi_Tp1H9CDs . Like others have said, if your shirt was a retroreflector it would be quite easy to see you. $\endgroup$ – eps Sep 7 at 19:49
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To clarify some things, The answer provided by @Vid is partially correct in that the information holds true only if you can assume an isotropic point source.

I design headlamp optics for a living and a low beam headlamp beam pattern is not anything close to an isotropic point source. It is more of an elliptical shaped beam that is very narrow in the vertical direction (~10° tall) and in the horizontal direction it is wider (~±30°) , but still narrow compared to an isotropic point source assumption.

The key photometric quanitity to consider when analyzing the transport of light flux from a headlamp is the luminance. The luminance is defined as the luminous flux per unit solid angle and unit source area.

$$\frac{F}{(\Omega)(m^2)}$$

The luminance of a source will remain constant from the source to the observer absent any intervening media (like fog) so that the perceived brightness of direct observation of the oncoming headlamp at night is very high. It should be noted here that brightness is not a photometric term, but is a psychophysical term that relates to a sensation - A headlamp will be perceived as not bright during the day, but very bright at night. Thus the luminance is the physical quantity we want to analyze as it relates directly to brightness.

When the light from the headlamp hits a person there are 2 factor that dramatically cut the luminance of the reflected light from the person. The first is that the total flux being reflected by typical clothing material is probably around 20% or less (I don't have exact data for that, but dyed textiles typically fall near this number). So immediately, you have cut the luminance of the reflected light down by ~1/5th compared to the direct oncoming light. The second factor is that textiles will have a diffuse reflection so the solid angle in the denominator of the luminance will be increased. If it is a perfectly diffuse reflection, the solid angle will be a hemisphere or 2pi steradians. Compared to the original solid angle which is approximately .005 steradians, you have now reduced the luminance of the returning signal by over 3 orders of magnitude.

So when you combine the reduced reflectance with the diffuse reflection of the clothing, the returning luminance is very low and likely not detectable under normal observation conditions.

If on the other hand you had a specular flat mirror at the same distance as the human, assuming parallel ground and the mirror normal to the ground, the driver would see almost no light reflected back to the driver. This is beacause the light for a low beam headlamp is nearly all aimed at the horizon or lower (there is a small amount of light above the horizon for seeing overhead signs), so the mirror reflection will maintain the luminance, but will not direct it back to the drivers eye. If you turn on the high beam in this case though, you will most likely see a very bright reflection back from the mirror because the light for a high beam is concentrated at angles near zero degrees to provide maximum seeing distance for a driver. The mirror then reflects that light back to the driver at almost the same luminance (a small reduction for reflectivity of the mirror) so it will be seen as being as bright as the human observer standing by the same mirror looking back at the car.

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  • $\begingroup$ Re, "...not anything close to an isotropic point source." Suppose we put an isotropic point source behind a mask with an elliptical-shaped hole in, such that it casts a beam that is ~10° tall and ~30° wide. I don't doubt that a designer of car headlights such as your self would find that to be a horribly inefficient way to light up the road, but would it change the answer to the OPs question? $\endgroup$ – Solomon Slow Sep 15 at 13:25
  • $\begingroup$ P.S., And what if we take away the mask? What if we ignore the rays that go off into the darkness in all directions, and we only talk about rays that actually hit the distant pedestrian? Would that change the answer to the OPs question? $\endgroup$ – Solomon Slow Sep 15 at 14:38
  • $\begingroup$ @SolomonSlow - to address your question, Vid's explanation describes how the flux density is changing from a point source. By adding optics to the system like a reflector to concentrate the light, we are changing the flux density in the direction of the observer, so that the apparent brightness of the original lamp is significantly higher for an observer. So in your scenario, the apparent brightness of the isotropic points source will be substantially lower (likely not seen at large distances) whereas the lamp with optics will be perceived as significantly bright as the OP observed. $\endgroup$ – lhb Sep 16 at 14:18
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    $\begingroup$ You're missing my point. I am trying to divorce engineering considerations (e.g., the efficiency of the headlamps) from the theoretical aspects that pertain to the OP's question. You can increase flux density from a bare bulb by adding optics? I can increase it by declaring my hypothetical point source to be brighter. But the numeric value of that flux density doesn't even matter: We want to find the ratio of intensity that the pedestrian sees from the headlights to what the driver sees reflected off the pedestrian's clothes, and we want to explain how that ratio scales with distance. $\endgroup$ – Solomon Slow Sep 16 at 14:44
  • $\begingroup$ P.S., I never was disagreeing with anything in your answer (which, by the way, I up-voted.) I was only saying that any talk about what the lenses and mirrors do is superfluous. Any light source—even a giant star with a radius of a hundreds of millions of kilometers—looks like a point source if you stand far enough away from it. If you simplify a problem by pretending that a car headlight is a point source, that simplification won't substantially change your answer if the headlight is 100m down the road. $\endgroup$ – Solomon Slow Sep 16 at 14:59
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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.

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    $\begingroup$ And also because the light is reflected diffusely (scattered in all directions), and not directly back to the driver. If you were a flat mirror at the correct angle, you'd appear much brighter. That's why staring at a person standing in sunlight is not blinding, but staring at the sun in a mirror is. $\endgroup$ – BlueRaja - Danny Pflughoeft Sep 7 at 3:51
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    $\begingroup$ @BlueRaja-DannyPflughoeft or, instead of being a mirror "at the correct angle", you could have a back-reflecting surface, similar to the high visibility vests. $\endgroup$ – Ruslan Sep 7 at 10:31
  • $\begingroup$ Also, headlights emit light in a pattern where most of it is aimed downward toward the road extending to a few hundred feet ahead. Much less light shines higher up and farther down the road. It's enough to be seen from miles away, but not enough to illuminate anything for the driver. This is by design so that an oncoming driver will see the car's headlights but won't be blinded by them. $\endgroup$ – Anthony X Sep 15 at 22:06
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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.

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    $\begingroup$ technology connections just did a nice short video on how retroreflectors work : youtube.com/watch?v=Bi_Tp1H9CDs $\endgroup$ – eps Sep 7 at 19:52
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    $\begingroup$ This is good, but you're still missing the effect of our eyes dynamically resizing to let different amounts of light in depending on the ambient brightness of the environment. If you added that as a #4, then this would be the most complete answer. $\endgroup$ – spacetyper Sep 8 at 0:17
  • $\begingroup$ @spacetyper That's a good point, thanks! $\endgroup$ – user1079505 Sep 8 at 0:21
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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 $1/r^2$.

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    $\begingroup$ Shouldn't it be $j \propto \frac{P}{4\pi r^4}$? Your equation is dimensionally incorrect. $\endgroup$ – Akshat Sharma Sep 6 at 22:53
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    $\begingroup$ I don't understand why it's $r^4$ with a mirror. A mirror at distance $r$ should behave the same as if there were a second driver $2r$ away, so the intensity should drop with $(2r)^2 = 4r^2$. With diffuse scattering, though, it's like a second light-source, so $r^4$ makes sense... $\endgroup$ – BlueRaja - Danny Pflughoeft Sep 7 at 3:56
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    $\begingroup$ This is similar to the radar range equation, which also has an $1/r^4$ dependency on the distance. $\endgroup$ – Bas Swinckels Sep 7 at 5:50
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    $\begingroup$ @BlueRaja-DannyPflughoeft In the far-field limit, the mirror is effectively a new aperture in the optical system. If you have a flat mirror that's a square of side $a$, you can ignore these effects if you're observing the reflected light from a distance like $a$, as when you're a meter from the big mirror in your bathroom. But if you're detecting the reflection from $1000a$, even light which left the mirror as a well-collimated beam will have diverged due to diffraction. $\endgroup$ – rob Sep 7 at 7:37
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    $\begingroup$ The whole point is that a car is not a mirror. That's why you get $1/r^4$ instead of $1/(2r)^2$. $\endgroup$ – TonyK Sep 7 at 18:31
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You are seeing light that comes directly from the car headlights. In contrast, the driver is seeing light that comes from the car headlights and bounces off of you. First of all, you are pretty dull (physically, not personality-wise) and don't reflect a lot of light*. Second, the light has to travel twice as far to reach the driver's eyes**.

Edit:

*I think it's actually more important that you reflect light diffusely. If you were a mirror, and you were oriented so that you reflect light directly back to the car, then most of the light that reached you would reach the car. In contrast, since you're probably not a mirror, a lot of light is reflected in other directions, and only a bit of the light returns to the car.

**As commenters have mentioned, the intensity of light does not significantly fade with distance if the beam is narrow (as with a laser beam). This answer assumes that the car headlight does not produce a narrow beam of light, in which case the intensity of the light fades as $\frac{1}{r^2}$.

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  • $\begingroup$ The distance has nothing to do with the visibility of light as long as we assume there are no atmospheric effects. There might be, but I would point that out as separate scenario. $\endgroup$ – B M Sep 6 at 17:59
  • $\begingroup$ For narrow wave, the light doesn't faint with distance. You have to evaluate the power/distance relation. $\endgroup$ – Vid Sep 6 at 18:08
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    $\begingroup$ @BM The intensity (or illuminance or irradiance) of light or other linear waves radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period). Per Wikipedia en.wikipedia.org/wiki/… $\endgroup$ – Adrian Howard Sep 6 at 21:01
  • $\begingroup$ A mirror would be brighter if it's perpendicular to the path to the car. On average, over all angles, it would be the same brightness. $\endgroup$ – Acccumulation Sep 8 at 0:58
  • $\begingroup$ You don't even need an adjusted mirror to become visible, reflector stripes on your clothes suffice. And they are an excellent example of how important the distinction between directed reflection and diffuse reflection actually is. $\endgroup$ – cmaster - reinstate monica Sep 8 at 8:54
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The headlights that you see appear to have a brightness at your position on the road that is large enough for you to detect them. For the driver to detect you, he has to make use of the light reflected off of you back in his direction. That amount of light is small because you are not very reflective (as @Ankit alludes to in his comment) - your clothing absorbs and scatters a lot of the light incident on it.

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  • $\begingroup$ Albedo of any "normal" object is still quite low... So it reflects quite a lot of light... We are not all coated in venta-black or any special ekstra black color. You can observe phenomena from question even with a mirror $\endgroup$ – Vid Sep 6 at 18:45
  • $\begingroup$ @Vid If something has a low albedo, that means it doesn't reflect much light. $\endgroup$ – Acccumulation Sep 8 at 1:00
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There are two main reasons, as alluded to throughout the other answers:

  1. Your eyes and the drivers eyes being open different amounts--due to having to calibrate with the amount of light in the environment in order to not over or underexpose the environment.
  2. The amount of light coming from you vs. coming from the car.

In the case of 1, look at your pupils when in a very dark room, vs very bright room. You'll notice that in the dark room, they're much larger. Why is that? Well, there are less photons coming into contact with your eyes when its darker, so in order to maximize the light coming into your eyes, they open up more. In your scenario, the person standing on the dark road will have larger pupils than the person in the car, because they're calibrated to the dark environment with little light.

In the case of 2, you're less bright than the car. Even if your pupils and the drivers pupils were collecting the same amount of light, there's much more light coming from the car (which is producing light), than you (which are only visible due to reflected light).

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I have never seen such a load of unnecessary verbage in answer to a question! We see the light from stars millions of light years away from us but only a complete dum-dum would expect an observer on a planet orbiting a distant star to see that star's light reflected off say an astronaut doing an EVA in Earth orbit. The answer is that there is simply not enough light reflecting back towards the source. It's as simple as that.

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Strictly speaking the following statement is wrong:

the car will brighten the road only few hundred meters further

The fact that you can see the light emitted by the car is evidence for the light illuminating all the way between you and the car. Two main reasons can be named why you see the car but the driver does not see you:

  1. Scattering of light
  2. The sensitivity of the human eye (and brain) for scattered light

The first effect leads to the car's light being scattered, the second to the driver not seeing this light.

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    $\begingroup$ Re, "sensitivity of the human eye...for scattered light." There's nothing special about scattered light. Your eye is no more sensitive to scattered light than it is to light coming directly from the headlights of an oncoming car. The reason those two situations are different is that the radiosity of light leaving the headlamps of a car is several orders of magnitude greater than the radiosity of light that is reflected off of distant objects that are illuminated by the headlights. $\endgroup$ – Solomon Slow Sep 6 at 17:04
  • $\begingroup$ "There's nothing special about scattered light". Yes there is: it has less radiant flux, since scattered. :-) But you put that into more precise terms and I agree with your explanation. Bottom line is: if you put a perfect mirror next to the observer and assume the car's light travels in a perfectly straight line, the driver and the observer would see the light the same way. $\endgroup$ – B M Sep 6 at 18:05
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    $\begingroup$ @BM Yeah true, but light sources are generally isotropic rather than narrow, with exception of lasers. $\endgroup$ – Vid Sep 6 at 18:23

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