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Why the first fringe is the biggest and brightest one?

  1. Sep 19, 2011 #1
    In the young experiment, why the first fringe is the biggest ang brightest one?

    I would desagree in both things, because a fringe would be bright if its amplitude were maximum, and ALL the constructive points have maximum amplitude, so wouldn't it be same bright?

    For the distance between one fringe and other, I've plot ed a graphic and for a VERY VERY small lambda, we get fringes aproximately of the same size (eventhough, the first fringe is ALWAYS smaller than the others ).

    So why?
  2. jcsd
  3. Sep 19, 2011 #2
    I'm new here and this is my first post, but I think I have an idea of why the middle fringe is the brightest...

    In the experiment the interference pattern is viewed in a plane screen. If you consider that the waves are emitted with spherical simmetry, then the fringes futher away from the center, are formed with further travelling distance, wich means that, even when they are form in maximium aplitude points, this aplitud is lower than that of the center.

    I hope this helps.
  4. Sep 19, 2011 #3
    But why a wave that has traveled more than other has a lower amplitude?

    Wasn't the "y" supposed to bethe same after a same period?
  5. Sep 19, 2011 #4
    think of it this way... you have a finite ammount of energy in a wave... le further the wave travels, the bigger the surface in wich this energy spreads, this means that energy per unit surface is lower at higuer distances, and so, the fringe is less bright.

    Sorry is my english is not perfect, but it is my second language.
  6. Sep 19, 2011 #5
    But in a wave, the energy is supposed to travel while the "wave" travel too. So why would the energy "spread"?
  7. Sep 19, 2011 #6
    BEcause is the same mount of energy but distributted in a larger area...

    IS basically the same thing that happens with the sun's radiation... the closer you are to the sun (the source of the wave) the hoter it is... as you get further away from the sun the energy expelled by it spreads along a larger surface...

    The distance between the source of the wave and the point you are considering is the radius of the sphere in which the energy is distributed
  8. Sep 20, 2011 #7
    You may be thinking about a plane wave. This is a good approximation for short distances of propagation. A real wave is more like a spherical wave (or even more complicated).
  9. Sep 20, 2011 #8

    No, it is more complex than that. Even if the imaging surface was a sphere rather than a plane, you would still get the middle fringe to be the brightest. It is because of the nature of interference. You can think of the typical interference pattern as two effects on top of each other: the interference of two waves which creates a cosine function distribution of intensity, and the spreading of a single wave which leads to an overall Gaussian shape with a peak at the center and dieing off to zero at infinity (actually more complex than this). If you managed to create two truly infinite plane waves that were in phase and have them interfere on an imaging surface, all the fringes would be the same brightness. So the answer is: single wave diffraction. You can think of it like a water wave pushing through a gap in a seawall. The wave will curve around the edges of the gap and spread out in all directions after coming through, but will be strongest in the forward direction because this was the original direction.

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