about fiber optic refraction

kelvin_ng

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[tex]^{-4}[/tex]mm in diameter.

kelvin_ng

the answer given is L = 3.9 x 10[tex]^{-4}[/tex] m.

kelvin_ng

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

kelvin_ng

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?

learningphysics

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

learningphysics

I think you need to use 75 degrees.

kelvin_ng

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[tex]^{-4}[/tex] m ?

Sleek

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.

kelvin_ng

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

Sleek

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.

kelvin_ng

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