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Destructive interference - What happens to the energy?

  1. Jun 7, 2014 #1
    Hi Everyone

    This is a question which, in various forms has been asked in various places on the Internet but I have never seen a satisfactory answer. I have therefore decided to try and ask it in a unique way which will start off conventionally to begin with.

    If I have two side-by-side lasers of identical wavelength and amplitude completely overlapping at a far-away target with a half-wavelength phase difference, thus causing the target to be dark, where does the energy go?

    The following are intensity diagrams which show Lasers A and B, initially in phase and not overlapping and then being moved so that they overlap.

    These are ASCII diagrams using the Courier (fixed-width) font which hopefully will display correctly on your computer. Please ignore the underscores. I am using them because spaces at the beginning of a line are ignored. The height of the blocks represents intensity.

    In phase:

    1. No overlap:

    AAA BBB
    AAA BBB

    2. 1/3 overlap:

    __B
    __B
    AAABB
    AAABB

    3. 2/3 overlap:

    _BB
    _BB
    AAAB
    AAAB

    4. Total overlap:

    BBB
    BBB
    AAA
    AAA

    Now 1/2 a wavelength out of phase:

    1. No overlap:

    AAA BBB
    AAA BBB

    2. 1/3 overlap:

    AA BB
    AA BB
    AA BB

    3. 2/3 overlap:

    A B
    A B
    A B
    A B
    A B
    A B

    4. Total overlap:

    ?

    Can anyone complete the sequence? Is it the same as when the lasers are in phase? Is 1 to 3 correct?

    Thank you very much.

     
  2. jcsd
  3. Jun 7, 2014 #2

    mfb

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    Where is the point in a different, harder-to-read font for the text?
    You can use [code]-Tags for fixed width diagrams, by the way:
    Code (Text):
    AAA
      B
    This is impossible. There are always regions with a different phase difference where you will see light. There is no possible setup where you could avoid that.
     
  4. Jun 7, 2014 #3
    Hi mfb

    Thank you for your reply. Sorry, I did not know there was a fixed-width code.

    I know that in the real world, it is not possible for it to be completely dark but it seems obvious to me that with the two beams out of phase by 1/2 a wavelength, there will be much more destructive interference than there would be if the two beams were in phase with each other. What happens to the energy which can be seen when the beams are in phase but which becomes invisible when the beams are out of phase?
     
  5. Jun 7, 2014 #4

    Bill_K

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    Better yet, suppose the two beams are IN phase. Then they will interfere constructively, and you will have more energy than you started with!

    No, unfortunately, you can't have one without the other. Where the beams overlap there will be regions where they interfere constructively, other regions where they interfere destructively, and these occur in equal proportions. The two effects average out. Interference can only rearrange the energy from one place to another.
     
  6. Jun 7, 2014 #5
    Hi Bill K

    Thank you for your answer.

    I'm not sure I agree with you. 1 + 1 = 2. If the two beams are one unit of intensity each. How will constructive interference produce any more than two units of intensity?

    How can the ratio of constructive and destructive interference always be 50/50 in the interference pattern on the target when the phase difference between the two beams is changing?

    If the target is in a region of destructive interference, where does the constructive interference come from?
     
  7. Jun 7, 2014 #6

    Bill_K

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    The energy is the square of the intensity.
     
  8. Jun 7, 2014 #7
    I don't understand. Surely if you are seeing twice as much light, you are seeing twice as much energy. If I am wrong then what is energy? It is obviously not what I thought it is.
     
  9. Jun 7, 2014 #8

    Bill_K

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    The amplitude of an electromagnetic wave is E, equivalently B. The energy density for electromagnetism is quadratic, ½(E2 + B2). Double the amplitude and the energy density quadruples. When two waves interfere, you add their amplitudes, E1 + E2, remembering that these are complex quantities. Depending on the relative phase, the amplitude of the result can be anywhere from |E1 + E2| to |E1 - E2|, and the energy density is the square of that. If the waves have equal amplitudes, the result can be anywhere from zero (destructive interference) to four times as much (constructive interference).

    But you can't consider just one point in the interference pattern and draw any sweeping conclusion. The entire pattern must be taken into account, and it is impossible to overlap two plane waves without having regions of constructive interference as well as destructive ones, in equal amounts.
     
  10. Jun 7, 2014 #9
    What I meant to say in the point you are referring to is as follows:

    "If the two beams are 1/2 a wavelength out of phase with each other, where does the constructive interference come from?"

    Aren't the parallel rays of lasers or at least very good quality ones meant to be exactly in phase with each other? And thereby making it possible for two separate beams to be 1/2 a wavelength out of phase with each other?
     
  11. Jun 7, 2014 #10
    Sure, but how are you going to arrange these lasers so that their beams overlap? You cannot put the two lasers in exactly the same place. If you use some tricky half-silvered mirror setup then you just end up cancelling out one of the beams. And then of course in the real world the laser wavefronts are never perfectly parallel either, and certainly not infinite as is an idealised plane wave.
     
  12. Jun 8, 2014 #11

    mfb

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    It is impossible to make perfect, planar laser waves. In the best case, you are limited by diffraction with the width of the laser beam. You can model that as double slit (where each slit is a laser) with an additional phase shift. And this won't give or lose energy. You can show this mathematically.
     
  13. Jun 9, 2014 #12
    I have calculated that if two 650nm laser beams which are each 1mm in diameter are arranged side-by-side 1mm apart and they converge to a single spot on a target 62 meters away then the difference between the phase differences on the left hand side of the spot and the right hand side of the spot is about a tenth of a wavelength. This theoretically means that if the left hand side of the spot is in total destructive interference then the rest of the spot will be almost in total destructive interference as well. What, if anything, am I getting wrong here?
     
  14. Jun 9, 2014 #13

    stevendaryl

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    The issue of total destructive interference is sort of like the issue of perpetual motion machines. On the one hand, you can argue from first principles (conservation of energy) that it's impossible. On the other hand, for specific scenarios, it can be difficult to pinpoint exactly where the flaw is.

    The scenario that had many people puzzled for a long time was the use of beam-splitters and mirrors to get total destructive interference, as shown in this diagram
    HoxRTn5.png

    (The discussion is here: http://physics.stackexchange.com/questions/88422/fully-destructive-interference)

    It would seem that by adjusting the lengths between mirrors, we could get perfect cancellation for the parts of the beams shown in dotted lines (if we assume an ideal case of a beam splitter that passes 50% and reflects 50%).

    I don't remember the definitive resolution, but in discussions about the problem, someone pointed out something that I didn't realize (I'm assuming that this person knew what he was talking about): The phase shift resulting from reflection off of a partially silvered pane of glass is different, depending on whether the light reflects off the outer surface or the inner surface. For the mirror in the upper right corner of the setup, one beam must reflect off the inner surface, and the other beam must reflect off the outer surface. So they won't have the same phase shifts. That is enough to spoil the possibility of perfect destructive interference.

    I'm going to do a little Google research to see if I can find a definitive statement about the phase shift from reflection off inner and outer surfaces.
     
  15. Jun 9, 2014 #14

    stevendaryl

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    There's a statement from a previous Physics Forums discussion:
    https://www.physicsforums.com/showthread.php?t=535525
     
  16. Jun 9, 2014 #15
    Hi Stevendaryl

    Thank you very much for your reply. Do you have a specific answer to the scenario I have described in Post #12. I know that it is not a perfect scenario but with a phase-difference variation across the spot of just one tenth of a wavelength then how can there be any constructive interference in the spot when part of the spot is in total destructive interference?

    Isn't the type of interference wholly dependent on the phase-difference? If not, what else is it dependent on?
     
  17. Jun 9, 2014 #16

    mfb

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    Laser spots of 1mm initial diameter will have a size of at least a few centimeters, at a distance of 62 meters. This is just given by diffraction. You can't just look at one interference effect and neglect the other one.
     
  18. Jun 9, 2014 #17
    Hi mfb

    Thank you very much for your reply.

    To be honest, I don't know all that much about diffraction.

    I am getting the impression from the replies that I have received so far that whenever you think you have managed to create ideal circumstances which will create a ratio of constructive and destructive interference other than 50/50, nature has a way of bringing it back to exactly 50/50. Am I correct?

    Can anyone recommend a very easy to understand book or online text which is an idiots guide to interference patterns and diffraction? My main aim is to try and get my head around wave-particle duality.

    Edit: The book/online text I am looking for needs to explain not just what is happening but it needs to enable the reader to fully understand exactly why it is happening.
     
    Last edited: Jun 9, 2014
  19. Jun 10, 2014 #18
    Hi Everyone

    It has been a bit more than 24 hours since this thread has received a reply. Does anyone have an answer to the question in my previous post (#17) about nature bringing the ratio back to 50:50? (I think that I made a mistake with the ratio format in my previous post. It should be 50:50, not 50/50.)

    Thank you very much.
     
  20. Jun 11, 2014 #19

    mfb

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    Somehow I missed that post.

    Right*. And this is pure mathematics - you cannot beat mathematics.

    *there is something between "100% destructive" and "100% constructive", but that does not change the main idea.
     
  21. Jun 11, 2014 #20
    If I may, perhaps, co-opt the conversation for an alternate hypothetical situation. Let's imagine we have a ONE (spatial) dimensional system. It has homogeneity in time so it has conservation of energy. I then have two wave sources pointing at each other with identical frequencies but phases and distances such that I have destructive interference.

    Where then did the energy go in this one-dimensional case?
     
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