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Light problem

  1. Jul 25, 2005 #1
    Here is something though that has been haunting me recently. If lights from two different sources meet and the distance that the lights have travelled from each source differs a half of the wave length of the one photon the light waves will “destroy” each other and if they meet on a screen then it should be impossible to see any light spot at all. That is what we call interference. So much is clear. Here is the part that bothers me.

    If light waves are anything like water waves then they really can’t be destroyed in the sense that they stop to spread or permanently stop their movement, the rather “stop” to exist at that particular spot they meet the other wave and interact with it. But they continue their movement just like nothing ever happened after that spot.

    So if we look at it that way, the light waves still should reflect against the screen the hit even if they both get “neutralised” and destroyed by each other right? So why is it that we cant see them any more? And they no longer can provide the eye any picture of the surface they have been reflected from? Or do they get completely destroyed (don’t travel anymore) after they have meet each other? And if so, why is my water wave theory wrong?

    Thanks a lot everyone for your contribution
  2. jcsd
  3. Jul 25, 2005 #2
    I have got another question, if light can be described as a wave. Then how can the wave top and the wave bottom be described, I mean what sort of qualities do they have? Is it the wave top that creates light? What does the wave bottom do then?

    Sorry for my foolish question, but I don’t think it’s correct to see light waves exactly as mechanical waves. I would also like to know more about light waves.
  4. Jul 25, 2005 #3
    The wave is two waves really: a wave in the electric field and a perpendicular wave in the magnetic field, each one changing causing a change in the other. A peak would correspond to a maximum flux of electric/magnetic field, and a trough to zero. Maybe this is your answer (I don't know for sure): if a peak and a trough of, say, the electric field coincided, you would have a large change of electric field causing a large change in magnetic field (and vice versa), hence the continuing propogation of the wave. Something like that? Maybe?
  5. Jul 25, 2005 #4
    The light from two different sources will be extended in space. Even though
    they will cancel in one place as you correctly imagine, a little further away
    they will add up and be double. The energy will be preserved.

    The top and bottom are like the left and right swings of a pendulum.
    They are parts of a complete wave motion. You can't say that the top
    is visible and bottom is not, just like you can't say that the left swing
    of a pendulum "does more" than the right.
  6. Jul 25, 2005 #5
    Thanks for your replies, some of the things you hade said has been useful.
    It was stupid to question the roll of the peak through, and I has been very useful to what light waves are made of.

    As I understand it from your replies, my theory about how light waves should behave like water waves is correct. The two light waves that meet on the screen don’t completely "destroy" each other, rather neutralize each other wave abilities at that spot and then continue to be "reflected" on the screen, as any other wave (that has not been trough a interference) would and even reach the eye of the beholder.

    But something in the process of the interference make the two light waves loose their ability to transfer images to the eye, even if they reach the eyes, it is like no light at all has been reflected to the eye from the object at all. I want to understand this process. Please explain.
  7. Jul 26, 2005 #6
    I'm not sure I get you. Do you mean two waves destructively interfering when they his the same spot on a mirror, then someone sees the reflection? Given that they must have had different angles of incidence, otherwise they would be parallel, I imagine you'd see the reflection as if they'd never interfered at all. If you're talking about intereference patterns on a screen, the light you receive optically when you look at that screen is not the same as that which caused the interference pattern.

    Images are not transferred... they're made on your retina. If two waves pass through your lens and destructively interfere at your retina, then yes that's as if neither of those waves were received as far as I know. I imagine the effect is not noticeable.
  8. Jul 26, 2005 #7
    Hello Hombre

    What i mean was the two light waves that meet on a screen and cause an interference pattern. A dark spot on the screen, this two waves cancel each other on that particular spot (on the screen, not a mirror), but the two waves don’t completely destroy each other there.

    They will be reflected on the screen (Yes light can reflect on a screen that’s not a mirror) and maybe even reach the eye of a beholder. But something that happened during the interference they went trough makes the two light waves loose the ability to make up images. I want to know what that is.
  9. Jul 26, 2005 #8
    It seems to me that, when the two waves interfere at the screen, they sum up zero and this means that the atomic stuff of this location of the screen doesn't feel any action (any mechanical action). Therefore, the screen at that location cannot act as a secondary source and thus, no reflection can be seen.
  10. Jul 26, 2005 #9
    Oh, I get you. I was thinking by 'screen' you meant some kind of film onto which the interference pattern is exposed then later viewed. You mean something more like a piece of paper on which you can view the pattern directly? Sorry, I didn't understand.

    I don't think the reflection of the inteference pattern can really be described in the same way as the cause of the interference pattern. Essentially, when the light from the slits hits the screen and reflects, you are looking at separate sources of light, rather than a continuation of the interference pattern. Using the wavefront method, which is used in EM to describe diffraction, once the light is incident upon the screen, the reflected light is sourced from new wavefronts coming from the point of reflection. The 'dark' fringes are not observed as dark simply because light from the screen is still interfering as it comes towards you. It is dark because no new wave fronts are created at those points. Wavefronts from points are spherical, and we detect the source of light in the direction perpendicular to the tangent of the part of the wave front detected - i.e. always in a straight line back to the source (in this case, a light fringe). So if no new wavefronts are created at that dark fringe, no light will be detected as coming from it. This is not to say that light reflected from the screen is not interfering at all - it just isn't any different to the way light reflected off everything around us isn't interfering.
  11. Aug 10, 2005 #10
    I've wondered about this myself. As I understand it, the light at the dark spot does indeed 'stop existing', ie if opposing waves 'hit' the screen, there is no 'invisible' reflection, since the light has cancelled itself out and there is simply no light to reflect. *However*, the trick is that in order to create a dark spot in the first place requires that other parts of the light *must* become brighter somewhere else. If you remove the paper from the dark spot and let the light continue on as if nothing happened, you will find that other parts of the light grow dimmer to maintain the total energy. How do the other parts 'know' if the screen is there or not? I'm not sure if *anyone* truly understands it ... another bizarre example of quantum behavior.
    Last edited: Aug 10, 2005
  12. Aug 11, 2005 #11
    On second thought, the light doesn't exactly 'stop existing' in the dark spot ... a better way to describe it would be to say that the light contains a certain number of photons, and the 'waves' affect the probability of finding the photons in certain places, thus the light doesn't 'disappear', but instead 'decides' that it can't allow photons to appear in the dark spot.

    If you perform a 'two-slit' experiment with a very dim light source, (like 1 photon/second), you will still get an interference pattern over time. The 'wave' of light seems to 'explore' the entire path (and 'knows' that there are two slits). But of course you will only have one photon at a time hit the screen.
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