Fiber Optic Refraction: Calculating Light Ray Distance and Reflections

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Homework Help Overview

The discussion revolves around a problem involving the refraction of light in a fiber optic cable. The original poster seeks to calculate the distance a light ray travels between successive reflections within a fiber that has a specific index of refraction and diameter.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the use of Snell's law and the relationship between angles of incidence and refraction. There are questions about the correct application of formulas and the interpretation of the problem statement. Some participants express confusion regarding the angles involved and the calculations needed to arrive at the distance.

Discussion Status

The discussion is active, with participants sharing their thoughts and calculations. Some have proposed potential methods for solving the problem, while others are still grappling with the concepts and seeking clarification. There is no explicit consensus on the approach, but several lines of reasoning are being explored.

Contextual Notes

Participants note the challenge of understanding the problem's requirements and the implications of the given parameters, such as the index of refraction and the fiber's diameter. There is also mention of potential errors in previous calculations and the need for careful application of trigonometric principles.

kelvin_ng
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Question:

A ray of light enters a light fiber at an angle of 15degree with the long axis of the fiber. Calculate the distance the light ray travels between succesive reflections off the sides of the fiber has an index of refraction 1.6 and is 10^{-4}mm in diameter.
 
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the answer given is L = 3.9 x 10^{-4} m.
 
what is the solution to this question?
I just can't get to the answer.
 
is it using n = sine i/sine r or n = 1 / sine c?
and is it i = 90degree - 15degree = 75degree?
i don't know actually what the question mean by.
Can someone help me?
 
kelvin_ng said:
is it using n = sine i/sine r or n = 1 / sine c?
and is it i = 90degree - 15degree = 75degree?
i don't know actually what the question mean by.
Can someone help me?

I think index of refraction = sin(angle of incidence)/sin(angle of refraction)
 
kelvin_ng said:
is it using n = sine i/sine r or n = 1 / sine c?
and is it i = 90degree - 15degree = 75degree?
i don't know actually what the question mean by.
Can someone help me?

I think you need to use 75 degrees.
 
but then i get the angle of refraction is 37.14 degree.
then what else i could do to get L = 3.9 x 10^{-4} m ?
 
I don't think the angle of reflection has anything to do with the refractive index, if total internal reflection takes places.

"A ray of light enters a light fiber at an angle of 15degree with the long axis of the fiber. Calculate the distance the light ray travels between succesive reflections off the sides of the fiber has an index of refraction 1.6 and is 10^-4mm in diameter."

I get an answer 3.7320508075688776*10^-4 using a computer.

We find that tan(incidence)=diameter/(distance between each reflection)

Therefore tan(pi/12)=10^-4/x

Regards,
Sleek.
 
Last edited:
lolx, i know what to do now.
I just get the answer.
Just by using trigonometri. 10^-4 / sin 15 to find the hipotenus.
 
  • #10
Ops, I think i blundered. I found the length of fiber traveled by light instead. Yes, you have to use sine. Answer come out to be 3.86*10^-4 ~ 3.9*10^-4mm. Sorry about that.
 
  • #11
No need to sorry.. ^^
Is very nice that u guyz is tryin to help.
 

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