Refraction through an optical fiber

Click For Summary
SUMMARY

The discussion focuses on calculating the maximum angle of incidence (Ø1) for a laser beam entering a new type of optical fiber with an index of refraction (n) of 1.23, while the refractive index of air (nair) is 1.00. The critical angle for total internal reflection is determined using Snell's Law, specifically the equation n1*sin(Ø1) = n2*sin(Ø2). The solution confirms that Ø1 must be 45.7° to prevent the beam from escaping the fiber, emphasizing the importance of understanding the critical angle and its application in optical fibers.

PREREQUISITES
  • Understanding of Snell's Law in optics
  • Knowledge of refractive indices (e.g., n = 1.23 for fiber, nair = 1.00)
  • Concept of critical angle and total internal reflection
  • Familiarity with the behavior of light in optical fibers
NEXT STEPS
  • Calculate the critical angle for different materials using Snell's Law
  • Explore the principles of total internal reflection in optical fibers
  • Investigate the impact of varying refractive indices on light propagation
  • Learn about the design and applications of optical fibers in telecommunications
USEFUL FOR

This discussion is beneficial for physics students, optical engineers, and anyone interested in the principles of light behavior in optical fibers and their applications in technology.

SnowAnd38Below
Messages
2
Reaction score
0

Homework Statement



Given a "new type" of optical fiber (index of refraction n = 1.23), a laser beam is incident on the flat end of a straight fiber in air. Assume nair = 1.00. What is the maximum angle of incidence Ø1 if the beam is not to escape from the fiber? (See attached file for diagram).

Homework Equations



Snell's Law: n1*sin(Ø1) = n2*sin(Ø2)

The Attempt at a Solution



I know the answer is 45.7°, I just can't seem to generate it. For the beam not to escape from the fiber, the angle of incidence must be the critical angle, such that Ø2=90° (or sin(Ø2)=1). But I don't know if I'm supposed to try to find the critical angle, then work backwards to find the initial angle of incidence from the air into the fiber or not. There's no cladding to consider in this problem, so I'm just very confused.
 

Attachments

  • Screen Shot 2015-02-23 at 3.19.11 PM.png
    Screen Shot 2015-02-23 at 3.19.11 PM.png
    42.6 KB · Views: 867
Physics news on Phys.org
Assuming you've defined ##θ_2## as the angle between the normal and the beam inside the fibre, this doesn't have to be 90°, as the beam can be reflected off the inside edge of the fibre without escaping. Indeed, if it did have to be 90°, fibres wouldn't work round corners. Revise the definition of critical angle in optics, and it should be clearer how to get the right answer.
 
I guess I didn't explain my attempt very well. I was trying to treat the problem with Snell's Law twice; once for the beam entering the fiber from the air, and a second time for refracting in the fiber such that the second angle of incidence is the critical angle, guaranteeing full reflection of the beam back into the fiber.
 
SnowAnd38Below said:
I guess I didn't explain my attempt very well. I was trying to treat the problem with Snell's Law twice; once for the beam entering the fiber from the air, and a second time for refracting in the fiber such that the second angle of incidence is the critical angle, guaranteeing full reflection of the beam back into the fiber.
Yes, it will do. Find the critical angle for total reflection for the interface fibre-air, then backwards the angle of reflection at the front surface and from that, the angle of incidence.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 1 ·
Replies
1
Views
15K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
8K