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Phase shift upon reflection

  1. Mar 2, 2006 #1
    Let's say I have light at normal incidence. Under what circumstances is there a phase shift? Under what circumstances is there no phase shift? My best guess is that there is normally a phase shift of 180 degrees. The exception is when n_incident > n_reflected, but I don't really know.

    To elaborate more, let's say I have monochormatic light normaly incident on a film. Why do I get a maximum when the film thickness is [tex]d = \frac{\lambda_0}{4n}[/tex] and a minimum when the film thickness is [tex]d = \frac{2\lambda_0}{n}[/tex].
     
    Last edited: Mar 2, 2006
  2. jcsd
  3. Mar 5, 2006 #2

    Doc Al

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    Staff: Mentor

    Right. When light goes from one medium ([itex]n_1[/itex]) to another ([itex]n_2[/itex]), the reflected light at that interface undergoes a phase change as follows:
    if [itex]n_1 < n_2[/itex]: 180 degree phase change
    if [itex]n_1 > n_2[/itex]: no phase change​

    Looks like you are talking about a situation, like a soap film in air, where [itex]n_1 < n_2 > n_1[/itex]. There are two reflections: the first has phase change; the second does not. So if the optical path length through the film is 1/2 [itex]\lambda[/itex] (your first example), then the total phase difference between the reflections is zero and you get maximum constructive interference. Similarly, if the optical path length is an integral number of wavelengths (as in your second example), the net phase difference is 180 degrees: maximum destructive interference.
     
  4. May 2, 2010 #3
    I have a question that is somewhat related. Regarding the phase shift upon reflection, how do you show that it is 180 degrees when n1 < n2 and 0 degrees when n1 > n2 using Fresnel's equations? Something along the lines of an informal proof.
     
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