Total internal reflexion

[Maurice]

Hello,
Can somebody help me?
Light enters an optical fiber(ideal glass optical fiber without cladding or buffer)at the normally cleaved end.

How calculate the greatest angle of incidence I that will result in total internal reflection of the light?

I know Snell's Law and refraction Law and I have found 21 or 42 degrees but that don't seem correct...

Thanks.

 

[swansont]

Snell's law is the right method. What is happening if you don't have total internal reflection? Maybe if you showed how you got 21 or 42 degrees, it would help?

[Maurice]

ni.sini=n2.sinr (I=incident,r=refracted)

so 1.sini=1.5.sinr

so sini/sinr=1.5 I>r

into glass 90-r=critical angle and then?

I think it is 50 degrees

No sorry 90-50=40 degrees

[swansont]

That works for if you are inside the glass and the outside is air or vacuum (and arcsin of 1/1.5 = 0.667 is 41.8 degrees) but the question you asked was for the angle of light incident into the fiber via the cleaved end. It looks to me like you need to apply snell's law twice — once for the light entering, and one for the TIR for not exiting the sides.

[Maurice]

Sorry I have some difficulties to clearly understand but it seems to be 28 degrees for the incident angle

[swansont]

A diagram usually helps in these situations. Similar to this http://www.uslasercorp.com/envoy/images/fobd.h1.gif

 

There's an acceptance angle for the cleaved surface, and TIR in the fiber itself

[Maurice]

Yes,I think that the solution is that the greatest angle of incidence is 14 degrees respect to the normal.