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Thin Films

  1. Oct 27, 2005 #1
    This is not a textbook question. It is strictly some concepts which I am confused about.


    Suppose we have thin film situation.

    |||||||||||||||||||AIR|||||||||||||||||||
    -----------------------------------------
    |||||||||||||||||||FILM|||||||||||||||||
    -----------------------------------------
    |||||||||||||||||||AIR|||||||||||||||||||

    The ray enters the air from the top. Part of it will be reflected from the surface of the film (it has a higher index of refraction). The ray will reflect 180 degrees and thus be 180 degrees out of phase, assuming that the incident ray was normal to surface.

    Part of it will also pass through the air/film barrier. According to my professor, the ray will reflect of the film/air barrier, and then be 0 degrees out of phase. This is beyond me! How come the ray reflects of the boundary between film and air? Should it not pass through it without reflection? Also, why would the phase be 0?


    In another situation:
    |||||||||||||||||||AIR|||||||||||||||||||
    -----------------------------------------
    |||||||||||||||||||FILM|||||||||||||||||
    -----------------------------------------
    ||||||||||DIAMOND (highest n)|||||||||||

    The second reflected ray is said to be 180 degrees out of phase. Once again, why?!
     
  2. jcsd
  3. Oct 27, 2005 #2
    anyone?


    /0
     
  4. Oct 27, 2005 #3

    Chi Meson

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    This has a problem: the reflected ray will reflect according to the law of reflection and take off an an angle equal to the incoming ray. The 180 degree out-of-phase means that the refected wave is simply inverted. Look at a regular sine wave; if you flip the wave upside-down, notice that the crest has become a trough, and the trough becomes a crest. A full wavelength is chopped up into 360 "degrees" just like a circle; so the "distance" from a crest to a trough, is half a wavelength, or 180 degrees. Thus, the wave has shifted its phase by 180 degrees.

    Exactly why light waves invert when reflecting from a fast-to-slow interface but not when reflecting from a slow-to-fast interface is not an easy explanation. You can check out the mechanical wave analogy, but the real answer lies in quantum electrodynamics ("QED").

    All waves will partially reflect and partially transmit (refract) when they hit an interface of two media. The exception is for monochromatic light sources in special situations, such as what you are learning now.

    The "mechanical wave" explanation says that in thin film interference, the two rays that reflect (one internally and one externally) come back together out of phase and mutually destruct, causing no wave of that frequency to emerge. But the weird thing is even if you fire one photon at a time, not one single photon will reflect in that situation even if there is nothing to interfere with. Bizarre! but true. QED.
     
  5. Oct 28, 2005 #4

    ehild

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    When the light incides at the boundary of a two different materials part of it is reflected and the other part enters into the new medium, (it is called transmitted light) but will travell in different direction (it is refracted).

    The laws for light reflection and refraction at a boundary between different media can be derived from Maxwell's equations of electrodynamics, and these laws state that both the magnitude and phase of the reflected and transmitted waves are connected to the refractive indices of those media.

    In introductory Physics, it is enough to know that the phase change is 180 degree if the light ray incides from a medium of lower refractive index onto the surface of a medium of higher refractive index, and it is zero in the opposite case. Sometimes people call the refractive index "optical density" and say that the light reflects from an optically denser medium "out of phase" and in phase from an optically less dense one. The phenomenon is similar you experience with a wave travelling along a string: The wave turns over from fixed end and comes back in-phase from a loose end.
    Air has the lowest refractive index in your problems, and diamond has the highest. The refractive index of the film is between the other two. So the phase change of the reflected wave is 180 degrees when the light goes from air to film. The other part of the light which has not ben reflected goes through the film and reaches the boundary of the film with air. Here again, it is partly reflected and partly transmitted, and the reflected wave is in-phase now, as it is directed from a high-optical density medium to a low one.
    On the same way, the reflected wave is out-of phase when the light enters to diamond from the film.

    ehild
     
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