# Refraction through an optical fiber

Tags:
1. Feb 23, 2015

### SnowAnd38Below

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

Snell's Law: n1*sin(Ø1) = n2*sin(Ø2)

3. 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.

#### Attached Files:

• ###### Screen Shot 2015-02-23 at 3.19.11 PM.png
File size:
47.1 KB
Views:
105
2. Feb 23, 2015

### sk1105

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.

3. Feb 24, 2015

### SnowAnd38Below

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.

4. Feb 24, 2015

### ehild

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.