Internal Reflection / Fiber Optics

Click For Summary
SUMMARY

The discussion focuses on determining the smallest inner radius of curvature, R, for a fiber optic cable with a diameter of 0.900 mm, where the core has a refractive index of n = 1.46 and the cladding has a refractive index of n = 1.32. The critical angle for internal reflection is calculated using the equation Sin(Θ)c = n1/n2, leading to a geometric approach for finding R. Participants emphasize starting the light ray tangent to the inner radius of the cable to simplify calculations and ensure accurate reflections.

PREREQUISITES
  • Understanding of Snell's Law and critical angles in optics
  • Basic geometry, particularly involving circles and right triangles
  • Familiarity with fiber optic cable structure and properties
  • Ability to visualize light paths and reflections in optical systems
NEXT STEPS
  • Calculate the critical angle for internal reflection in fiber optics using Snell's Law
  • Explore geometric methods for analyzing light paths in curved optical fibers
  • Research the effects of bending radius on signal loss in fiber optic cables
  • Study advanced topics in fiber optics, such as modal dispersion and waveguide theory
USEFUL FOR

Students and professionals in optics, physics, and engineering fields, particularly those working with fiber optic technology and seeking to understand light behavior in curved mediums.

jfulky
Messages
3
Reaction score
1

Homework Statement


3. A fiber optic cable is essentially a thin fiber of plastic with n = 1.46, surrounded by a protective material with n = 1.32. The light can travel around a bend in the fiber if it can internally reflect inside the fiber. If the diameter of the fiber is d = 0.900 mm find the smallest inner radius of curvature, R, into which the fiber can be bent that will allow light that is initially traveling parallel to the fiber to remain inside the fiber.

Homework Equations



Sin(Θ)c = n1/n2

The Attempt at a Solution


My work is in the thumbnail. I am not entirely sure how to approach this problem. My though process was to find the critical angle in which the light is internally reflected, and then use geometry to find the radius of curvature, since I am not sure of any more equations that would help me. I also though of using arc length in some way, to relate the distance the light traveled in the tube to the circumference of the circle.

Any help would be appreciated!
 

Attachments

  • Screenshot 2016-05-29 18.00.13.png
    Screenshot 2016-05-29 18.00.13.png
    2.4 KB · Views: 559
  • IMG_20160529_181234.jpg
    IMG_20160529_181234.jpg
    16.6 KB · Views: 552
Physics news on Phys.org
jfulky said:

Homework Statement


3. A fiber optic cable is essentially a thin fiber of plastic with n = 1.46, surrounded by a protective material with n = 1.32. The light can travel around a bend in the fiber if it can internally reflect inside the fiber. If the diameter of the fiber is d = 0.900 mm find the smallest inner radius of curvature, R, into which the fiber can be bent that will allow light that is initially traveling parallel to the fiber to remain inside the fiber.

Homework Equations



Sin(Θ)c = n1/n2

The Attempt at a Solution


My work is in the thumbnail. I am not entirely sure how to approach this problem. My though process was to find the critical angle in which the light is internally reflected, and then use geometry to find the radius of curvature, since I am not sure of any more equations that would help me. I also though of using arc length in some way, to relate the distance the light traveled in the tube to the circumference of the circle.

Any help would be appreciated!
A couple of comments on your work so far:
  • You analyzed a light ray starting in the middle of the fiber optic cable. I do not recommend using a a light ray starting in the middle of the cable. Instead, let the light ray start out tangent to the inner radius of the bent cable -- not in the middle of the cable, but rather just touching (and tangent to) the inner side of the cable.
  • In your diagram, you have the entire 90 deg bend accomplished by a single reflection; in other words you drew \theta_C = 45^\circ. This is not only unrealistic, but it will add confusion later down the road. In a realistic cable, there will be many reflections down the road to achieve a total 90^\circ bend. Also, as part of the process of solving this, there will be (depending on your approach) a different right triangle involved, even if it's only a small part of the overall bend. I don't want to confuse this right triangle with any other possible right triangles.

Start with drawing a small section of the cable such that there are parts of two concentric circles. The light ray starts out vertical, as you have already drawn it, except it starts out tangent to the inner radius. Then let the light ray follow a straight line until it reflects off of the outer radius, and then comes back to being tangent once again with the inner radius.

But don't force \theta_C = 45^\circ. It's okay to allow it to be a larger angle than that.

Now you can use geometry to solve for one of the radii. (And if you solved it the way I did, there will be a right triangle that fits in there somewhere.) :smile:

Edit: Just to be clear, start with a figure like something along these lines:

FiberOptic.png
 
Last edited:
Thank you for the explanation, the problem is actually extremely simple, I don't know why I missed that insight! As a side note: when dealing with light ray problems, is it reasonable to align the light ray in any way you want that simplifies the problem? I sometimes struggle with what to assume the light ray's path is.
 
  • Like
Likes   Reactions: collinsmark
jfulky said:
Thank you for the explanation, the problem is actually extremely simple, I don't know why I missed that insight! As a side note: when dealing with light ray problems, is it reasonable to align the light ray in any way you want that simplifies the problem? I sometimes struggle with what to assume the light ray's path is.
I suggested aligning the light ray tangent to the inner radius because it is the extremum (i.e., it is an extreme case).

Whenever the problem asks you to look for "maximum <something>" or "minimum <something>", it's an indication that you might want to consider aligning the light ray in one of its exreme cases. :smile: [Edit: that is: an extremum that makes logical sense; you don't have to worry about aligning the light ray into physically impossible configurations, but you do want to look for cases at the physical limits.]
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
8K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
15K
Replies
10
Views
6K