1. Limited time only! Sign up for a free 30min personal 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!

Optics Question -- small glass spheres to couple to fiber optic cables

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data
    hAvPSJM.png

    2. Relevant equations
    I know that yin = 0.7mm, y2 = 0mm, the index of refraction for the sphere is 1.8 and the indices of refraction for the air surrounding it is 1.

    3. The attempt at a solution
    Not sure how to even begin with the given information. I was thinking I could ray-transfer matrix analysis, but how would I set that up for this question?
     
  2. jcsd
  3. Sep 14, 2015 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    What equation comes to mind when one is dealing with refraction between two media?
     
  4. Sep 14, 2015 #3
    Snell's Law: n1sinθ1 = n2sin2
    But how would I find θ1?
     
  5. Sep 14, 2015 #4

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    For the first refraction, try to draw a line from the sphere center to the entrance point of the incoming ray. You may not need to calculate ##\theta_1##, from the geometry it may be possible to get ##\sin \theta_1## directly in terms of the known data.
     
  6. Sep 14, 2015 #5
    Okay I've determined that sinθ1 = 0.7. Using Snell's law, the refracted ray inside the sphere's
    θ2 = sin-1(0.7/1.8) = 22.8854°.

    But then how do I get the angle of the incident ray when it exits the sphere? I'm just not seeing the geometry.
     
  7. Sep 15, 2015 #6

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Now connect three points, the entrance point, exit point, and sphere center, what kind of triangle do you get?
     
  8. Sep 15, 2015 #7
    An isosceles triangle. So the angle of the incident ray angle when it exits the sphere is the same as the refracted angle when it enters the sphere, θ2 = θ3 = 22.8854°. Trying to find the height it leaves the sphere now, I cut the triangle in two and using the refracted angle and radius, determine that the distance the ray traveled in the sphere is 1.84257 mm. The bottom angle of the isosceles angle is 180-22.8854°-22.8854° = 134.229°

    If I draw a line straight from where the ray left the sphere I can determine the height. Using the geometry of the line, the left angle of this triangle should be 180° - θ1 - 134.229° = 1.34376° Then the height is 1mm*sin(1.34376°) = 0.023451 mm.

    Is that correct so far? The height seems too small to me.
     
  9. Sep 15, 2015 #8

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    I think you are correct.
     
  10. Sep 15, 2015 #9
    Okay, and then I can use snell's law again to determine, θ4, the refracted exit ray's angle. Then I can say 0.023451mm*tan(θ4) = f [mm]
     
  11. Sep 15, 2015 #10
    Oh that's not quite right. I need to take 180 - θ4 - upper angle of height triangle from before. Then I can say 0.023451mm*tan(θ) = f [mm], and I'm getting 0.0025 mm as the focal length.
     
  12. Sep 16, 2015 #11

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    I guess you put one zero more than it is needed behind the decimal.
     
  13. Sep 16, 2015 #12
    Yep my mistake, 0.025 mm. Thanks for your help; much appreciated.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Optics Question -- small glass spheres to couple to fiber optic cables
  1. Fiber Optics Materials (Replies: 3)

Loading...