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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

    [tex]\frac{4L}{\lambda}[/tex]

    Next, I let

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

    and

    [tex]\frac{4L}{\lambda_b}=1.5[/tex]

    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?

    Thanks
     
    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

    [tex]8L-\lambda_a-1.5\lambda_b=0[/tex]

    Solving for L gives me

    [tex]L=\frac{\lambda_a+1.5\lambda_b}{8}[/tex]

    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)

    and

    [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

    [tex]\frac{4*75nm}{100nm}=3.0[/tex]

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

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

    [tex]\frac{4*75nm}{120nm}=2.5[/tex]

    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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