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Simple Newton's Rings problem

  1. Apr 24, 2008 #1
    1. The problem statement, all variables and given/known data

    If the apparatus for showing Newton's rings is illuminated with light at 6250 A (angstroms), what thickness of film underlies each of the first three light rings?

    2. Relevant equations

    2D (the thickness of the film) equals a multiple of 1/2 of a wavelength
    1 angstrom = 1 x 10^-10 wavelength


    3. The attempt at a solution

    I can't see how the angstroms connect to the fact that the thickness must be a multiple of 1/2.
     
    Last edited: Apr 24, 2008
  2. jcsd
  3. Apr 24, 2008 #2

    Doc Al

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    Staff: Mentor

    Generally Newton's Rings are formed when light is reflected off of a hemispherical lens resting on a sheet of glass. So I don't know what you mean by "film"--there is a space between the lens and the glass sheet.

    In any case, the phase difference (for constructive interference) must be an odd multiple of 1/2 wavelength, not just 1/2.
     
  4. Apr 24, 2008 #3
    I apologize for not clarifying wavelength, but I intended to. As for the film, I believe it means some form of glass, but its not really relevant to the solution of my problem.
     
  5. Apr 24, 2008 #4

    Doc Al

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    Do you still have a question?
     
  6. Apr 24, 2008 #5
    Yes, I dont see a way to connect the 6250 angstroms and the fact that each wavelength must be a multiple of a half. There must be some equation I'm missing, but I just can't figure out the problem.
     
  7. Apr 24, 2008 #6

    Doc Al

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    That's not quite right.

    6250 Angstroms is the wavelength; 2D must be an odd multiple of 1/2 wavelength.

    An Angstrom is just a unit of length. 1 Angstrom [itex]= 10^{-10}[/itex] meters.
     
  8. Apr 24, 2008 #7
    OK, but how can I use that information (the angstroms) to figure out the thickness for the first three light rings?
     
  9. Apr 24, 2008 #8

    Doc Al

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    By setting 2D equal to the first three odd integer multiples of 1/2 wavelength. (Solve for D.)
     
  10. Apr 24, 2008 #9
    Aha, thank you, I appreciate your help.
     
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