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Homework Help: Phase shift between two waves

  1. Jan 26, 2008 #1
    I am working on the following problem.

    http://img244.imageshack.us/img244/9777/homeworkun4.jpg [Broken]

    3. The attempt at a solution

    I realize that the relative phase shift between ray 1 and ray 2 will be


    Next, I let




    I did this in hopes to make the phase shift in wavelengths for [tex]\lambda_a[/tex] an integer number and the phase shift for [tex]\lambda_b[/tex] an integer plus 0.5 (to put the waves exactly out of phase).

    I tried various paths from this point but cannot get a valid length for [tex]L[/tex] that puts [tex]\lambda_a[/tex] in phase and [tex]\lambda_b[/tex] out of phase. Could anyone offer a suggestion on how to proceed?

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 27, 2008 #2
    Does anyone have any suggestions?
  4. Jan 27, 2008 #3

    Doc Al

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    Staff: Mentor

    Hint: Since [itex]\lambda_b > \lambda_a[/itex],

    [tex]\frac{4L}{\lambda_b} < \frac{4L}{\lambda_a}[/tex].
  5. Jan 27, 2008 #4
    I have two equations with three unknowns so I should be able to solve for one variable in terms of the other two. I use these two equations

    [tex]4L - \lambda_a=0[/tex] (1)

    [tex]4L - 1.5\lambda_b=0[/tex] (2)

    Adding both equations gives me


    Solving for L gives me


    When testing this equation, I find that I don't get an integer number of wavelengths for the phase shift for [tex]\lambda_a[/tex] and an integer + 0.5 wavelengths phase shift for [tex]\lambda_b[/tex].

    I have also tried a few different paths but still get nowhere.
  6. Jan 27, 2008 #5

    Doc Al

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    Staff: Mentor

    Reread my hint and correct these equations:
  7. Jan 27, 2008 #6
    Since [tex]\lambda_b > \lambda_a[/tex] we should have

    [tex]\frac{4L}{\lambda_a}=1[/tex] (1)


    [tex]\frac{4L}{\lambda_b}=0.5[/tex] (2)
  8. Jan 27, 2008 #7
    When I subtract (2) from (1) I get the equation

    [tex]\frac{4L}{\lambda_a}-\frac{4L}{\lambda_b}=1-0.5 \implies [/tex]

    [tex]4L\bigg(\frac{1}{\lambda_a}-\frac{1}{\lambda_b}\bigg)=0.5 \implies [/tex]

    [tex]4L\bigg(\frac{\lambda_b - \lambda_a}{\lambda_a \lambda_b}\bigg)=\frac{1}{2} \implies [/tex]

    [tex]L=\frac{1}{8}\frac{\lambda_a \lambda_b}{\lambda_b-\lambda_a}[/tex]

    Say [tex]\lambda_a=100nm[/tex] and [tex]\lambda_b=120nm[/tex]. Using the above equations, we find L to be 75nm. The relative phase shift between the two rays for a 100nm wavelength is


    3 wavelengths phase shift puts the 100nm wavelength exactly in phase!

    For the 120nm wavelength, the relative phase shift between the rays is


    The relative phase shift between the two rays is 0.5 wavelengths and they are exactly out of phase. This appears to be a correct answer. I am hoping this is the least L that satisfies the problem.
    Last edited: Jan 27, 2008
  9. Jan 27, 2008 #8

    Doc Al

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    Staff: Mentor

    Looks good to me!

    The most general way to express the relationship would be:

    [tex]\frac{4L}{\lambda_a}-\frac{4L}{\lambda_b}= n + 0.5[/tex]

    The smallest value of L would be when n = 0.
  10. Jan 27, 2008 #9
    Thanks for the help Doc Al!
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