1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Phase difference and number of wavelengths

  1. Oct 9, 2007 #1
    1. The problem statement, all variables and given/known data
    A light wave is propagated from point A to point B in space. We introduce along the way a glass lens with parallel faces of index 1.5 and width L=1mm. The value of teh wavelength is 500μm in space. How many wavelengths are between A and B with and without the glass lens? What is the phase difference introduced as we intercalate the glass lens?


    2. Relevant equations
    f=v/lambda
    v=c/n

    3. The attempt at a solution

    With the lens, I just used f=v/lambda which gave me f = 6x10^14 Hz.
    Without the lens, I first used v = c/n and then used v in f=v/lambda to get 4x10^14 Hz.

    I hope I atleast got the first part right. I'm not sure how to go about finding the difference in phase. Any pointers?
     
  2. jcsd
  3. Oct 9, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    What's the distance between A and B? You are given the wavelength in vacuum (although I suspect that should be nm, not μm) so just divide into the distance to find the number of wavelengths.

    How does the wavelength change when the light passes through the glass?
     
  4. Oct 9, 2007 #3
    I don't have the distance between both points. Is there a way to calculate it or just use a variable?
     
  5. Oct 9, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I see no way to calculate it. Are you sure you are presenting the problem exactly as given?

    But you can certainly calculate the phase difference.
     
  6. Oct 9, 2007 #5
    yep. I mean, I did translate it from French but I'm usually pretty good at it. It doesn't mention any sort of distance or anything. Just says "from point A to B".


    In regards to the phase difference:

    with the lense I found: 500nm
    without: 750nm

    so the phase difference is 750/500 = 1.5?
     
  7. Oct 9, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    I don't quite understand what you've calculated. In vacuum, how many wavelengths are in the 1 mm width? And in glass?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Phase difference and number of wavelengths
  1. Phase Difference (Replies: 3)

  2. Phase difference. (Replies: 2)

  3. Phase difference (Replies: 5)

Loading...