1. ### kelvin_ng

7
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.

2. ### kelvin_ng

7
the answer given is L = 3.9 x 10$$^{-4}$$ m.

3. ### kelvin_ng

7
what is the solution to this question?
I just can't get to the answer.

4. ### kelvin_ng

7
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?

5. ### learningphysics

4,124
I think index of refraction = sin(angle of incidence)/sin(angle of refraction)

6. ### learningphysics

4,124
I think you need to use 75 degrees.

7. ### kelvin_ng

7
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 ?

8. ### Sleek

59
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: Sep 13, 2007
9. ### kelvin_ng

7
lolx, i know what to do now.
Just by using trigonometri. 10^-4 / sin 15 to find the hipotenus.

10. ### Sleek

59
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. ### kelvin_ng

7
No need to sorry.. ^^
Is very nice that u guyz is tryin to help.