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Optics: Simple question

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Please take a look at page 4 in this PDF, namely problem 35.54: http://faculty.physics.tamu.edu/hu/221-fl08/YF-ch35-exmpls-new.pdf

    My questions is: Why is it that the equation [itex]2t = (m+\frac{1}{2})\pi[/itex] is changed to [itex]2tn = (m+\frac{1}{2})\pi[/itex]?

    I know it is because we are in air now, but the light still travels a distance of 2t, and not 2tn? Can you explain this to me?


    Thank you very much in advance.

    Best regards,
    Niles.
     
  2. jcsd
  3. Nov 12, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The issue is not the distance, but how many wavelengths it corresponds to. The distance in the glass is 2t, so you'd express that as some multiple of wavelengths in the glass. If instead you want your equation in terms of wavelengths in air, then you need to multiply by n. (Since the wavelength decreases by a factor of n when entering the glass.)
     
  4. Nov 12, 2009 #3
    I am not entirely sure I understand your post. Am I correct when I say that what you wrote is essentially

    [tex]
    2t = (m+\frac{1}{2})\lambda_{glass}
    [/tex]

    AND


    [tex]
    2tn = (m+\frac{1}{2})\lambda_{air}
    [/tex]

    and this is because we have [itex]n=\lambda_{air}/\lambda_{glass}[/itex] (approximately)?
     
    Last edited: Nov 12, 2009
  5. Nov 12, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Exactly.
     
  6. Nov 12, 2009 #5
    Thanks. I had a hectic day today, and it means alot that you responded so swiftly.
     
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