Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Path length of a light ray

  1. Aug 7, 2010 #1
    1. The problem statement, all variables and given/known data

    A glass optical fibre of length L = 3.2 m is in a medium of glycerine with a refractive index n0 = 1.47 . The fibre has a core of refractive index, n1 = 1.58 and diameter, d = 100μm surrounded by a thin cladding of refractive index, n2 = 1.53. The end of the fibre is cut square (see diagram).

    [PLAIN]http://img69.imageshack.us/img69/4976/imagenz.gif [Broken]

    For the aforementioned fibre, what is the total path length of a ray within the fibre that enters the fibre at the acceptance angle?

    3. The attempt at a solution

    Here are the stuff that I have already calculated:

    * The minimum angle, θ2, for total internal reflection at the core/cladding boundary: 75.5 degrees.

    * The angle, θ1, to the axis of the fibre that corresponds to the minimum angle, θ2: 14.5 degrees.

    * The acceptance angle θ0: 15.6 degrees.

    So, now to find the path length of a ray within the fibre that enters the fibre at the acceptance angle I tried using trigonometry:

    Since d=100.0 μm = 100 x 106 m. We have d/2 = 50 x 106.

    [tex]sin (14.5) = \frac{50 \times 10^6}{x}[/tex]

    x= a very huge number!!

    But the answer must be 3.305 m! What did I do wrong? :confused:
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 7, 2010 #2

    ehild

    User Avatar
    Homework Helper

    1 μm is not 106 m.

    ehild
     
  4. Aug 8, 2010 #3
    Oh, I meant 10-6. But why is it that I end up with 1.9 x 10-4? This is not the right answer...
     
  5. Aug 8, 2010 #4

    ehild

    User Avatar
    Homework Helper

    There are a lot of reflections in that 3.2 m long fibre. The light goes through the whole length of the fibre, but trawels along a zigzag path which is longer than the fibre length.


    ehild
     
    Last edited: Aug 8, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook