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How can you see light that is traveling away from the observer?

  1. Jan 30, 2004 #1
    You are on a salt flat, away from the city lights, no moon and it is cloudy. It is pitch black, you cant see a thing. It is a salt flat so there is nothing for miles. You shine a torch so the beam hits a hundred meters in front of you onto the ground. If the light is travelling away from you at 300 kps and there is nothing to bounce off,back to the observer, then how can you see the light? Do the same with a laser, 500 metres, how can you see the point where the laser hits the ground, when it is travelling away from your eyes, and in a focused beam?
     
  2. jcsd
  3. Jan 30, 2004 #2

    chroot

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    If there truly was nothing to bounce the light back, you would not be able to see the light.

    - Warren
     
  4. Jan 30, 2004 #3
    as experiments have proved that theory wrong would you like to try again?
     
  5. Jan 30, 2004 #4
    ...and if you replaced the 'salt flats' with a really good mirror, then you wouldn't see the light. But when it hits salt, or sand, or something bumpy, then some of the light bounces in every direction. That's why you can see it after the bounce.

    If there's enough water vapor, smoke, dust, or whatever in the air, you'll be able to see the beam even before it bounces off the ground.

    P
     
  6. Jan 30, 2004 #5

    chroot

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    Uh.. excuse me? Which experiments?

    - Warren
     
  7. Jan 30, 2004 #6

    Integral

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    When I do this sort of experiment, like every night I am driving, I see light scattered off of the road in front of me. The ground is something whether it be salt flats or asphalt.
     
  8. Jan 30, 2004 #7
    when refering to the "beam" I am not refering to the light contained within the beam travelling from observer to object. As stated, the point where it hits the ground. Mirror does the same, so your guess is wrong. And no tiny bits of light do not bounce off sand, if that were true then if you did the same thing inside a room the amount of light generated by the reflection would be blinding, yet there is no difference between out in the open or direct contact.
     
    Last edited: Jan 30, 2004
  9. Jan 30, 2004 #8
    the one with the torch the laser and the mirror on a dark clouded night, with no moon, what other experiment would you expect? The question wasnt can you see the point of impact, but why can you see it.
     
    Last edited: Jan 30, 2004
  10. Jan 30, 2004 #9

    chroot

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    You can see the point where the beam hits the ground because the ground scatters some of the light back to your eyes.

    - Warren
     
  11. Jan 30, 2004 #10
    Maybe I got the wrong conclusion from all this: but basically you couldn't see light if there weren't any particles for it to hit? (when its not shining directly at you).

    PS: I agree with the above answers.
     
  12. Jan 31, 2004 #11
    Yes you can as stated previously, in an experiment. You see the light that hits the ground be it torch or laser. Being at an angle whereby reflection is in the opposite direction.
     
  13. Jan 31, 2004 #12
    as already stated, in order for that to be true, it would mean that the value of the light energy of the one or two photons that magically come back to you would need to bee the same as the total value of all the other light that hits the surface, since you still see the point of impact with the same degree as if you were shining it directly at a wall.

    What you are basicaly saying is that if a torch was shon at a surface at an angle of 179 degrees, that somehow a small amount of light is always no matter what, returned in your direction, yet despite the miniscul amount it still allows you to see the light at the same level as if all the light was bouncing back. If this were true, then if you were to turn your back and look at your hand, then the "reflected" light you claim exists, would illuminate your hand, but it does not, therefore you are wrong.
     
  14. Jan 31, 2004 #13

    chroot

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    Sorry, simon. You're wrong. If you shine your torch or laser into an infinite vacuum (the blackness of space is a pretty good approximation), there will be no particles to scatter the light back to your eyes, and you will not see the beam.

    As has been said, in the real word, air molecules, dust, water vapor, and other particles scatter light. Of course, when your beam hits the ground, more light is scattered, in all directions.

    - Warren
     
  15. Jan 31, 2004 #14

    Umm....thats how human sight works, light bounces off of stuff. Well actually thats the simplified version. More specifically the light excites the molecules of said sand (and some of the photons are absorbed and others bounce off, giving the sight of the red dot) that then release a certain wavelength of light that is the color of the sand.

    However it is an inefficient process (as all antural processes are) meaning that the amount of light reflected is less than the amount recieved. Some of the recieved light is transformed into heat for example, rather than ebing reflected as visible light.
     
  16. Jan 31, 2004 #15

    Janus

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    Yes, the light does bounce off the sand. And it also does so off the walls of a room. If you were to place an object and a light in a room with white walls the object will be better lit than if the same light and object were place in open space. This is because the light reflecting off of the wall helps to illuminate the object. (This effect is call radiosity.)

    The light is not blindingly bright because even a white wall does not reflect all of the light, but only a portion. The rest is is absorbed and converted to heat.

    This heat is radiated off to both the inside and outside of the room. (Which is why the room doesn't just keep getting warmer and warmer. It will heat up just to the point where the heat being lost to outside the room equals the heat gained from the light.
     
  17. Jan 31, 2004 #16
    You cant seem to get this through your head, I am not taling about the BEAM of light that can be seen when it hits dust or smoke particles in the air. I am talking about the point where the beam of light hits the ground,that illuminates the surface area, that point and that point only, so vacuum is irrelevant as the same effect would occur on the dark side of the moon if you shon a torch onto the surface of the moon away from you.
     
  18. Jan 31, 2004 #17
    179 degree angle, so your assecrtion wont work.
     
  19. Jan 31, 2004 #18

    chroot

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    The angle is irrelevant. Some of the light is scattered directly back to the light source. Some of it hits your eyes.

    - Warren
     
  20. Jan 31, 2004 #19

    Integral

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    I think you are correct, we do not get your point.

    Are you telling me that I cannot see my head lights hit the road? Just what are you saying?
     
  21. Jan 31, 2004 #20

    Janus

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    The reflected light does illumnate your hand. But since this reflected light is only a small percentage of the light striking the spot of the ground, this illumination is very slight and in most cases will be below the eyes ability to see.(If the Light is very bright to start with, and you are very close to the spot on the ground, it will be visible. Shine a flashlight on a wall in a dark room, and hold your hand out of the beam but close to the wall and you will see it lit up by the light reflected off the wall.
     
  22. Jan 31, 2004 #21
    Umm....well actually if you had cared to read what it was refering to, it was refering the previous statement, therefore was not saying that light is not reflected off sand but only at that angle. And umm that is not how you see.
     
  23. Jan 31, 2004 #22
    No it does not, do the experiment yourself with a laser to make it easier instead of just guessing and making stuff up.
     
    Last edited: Jan 31, 2004
  24. Jan 31, 2004 #23
    No I am trying to point out that according to the current theory of light, the phenominom I have pointed out would be impossible, therefore that theory cant be correct, it has a flaw that has not yet been accounted for.
    What percentage of light from a laser could you possible expect to return to you, by hitting the odd angle of sand? .00000000000001% if you stretch your imagination to the limits? Yet no matter how you move the laser, or gat an assistant so that it is at a 90 degree angle to you and 1 degree angle from the ground, you will be able to see that point where it hits the ground, with the same degree of brilliance as if you shon it straight down.
    Also if you are able to see the point of impact of the torch light because of reflected light, then it is saying that there is sufficient light being reflected to illuminate that area sufficiently for you to see it , therefore the amount of light being "bounced back" would be sufficient to illuminate your hand to the same degree, but it doesnt.
     
    Last edited: Jan 31, 2004
  25. Jan 31, 2004 #24

    chroot

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    LOL.. you practically pushed the button for me.

    - Warren
     
  26. Jan 31, 2004 #25
    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 travelling 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
     
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