# Incident Angle Limitation Derivation in Optical Fibre

1. Feb 21, 2014

### Eats Dirt

1. The problem statement, all variables and given/known data
derive the angle of limitation sinθ=[(n2^2-n3^2)^1/2]/n1

n1 is the air out side of the fibre
n2 is inside of the fibre
n3 is the fibre wall

2. Relevant equations

Snells Law:

n1sinθ1=n2sinθ2

3. The attempt at a solution

I'm pretty stuck and don't really know where to go after getting the angles from geometry. I have pi/2-θ2 for the angle of reflection off of the optical wall. So I put this angle into Snells Law then just kind of get stuck.

2. Feb 22, 2014

### haruspex

First (for those of us unfamiliar with this topic), please describe the path of the light rays of interest. Better still, provide a diagram.
Not quite.
When you have that corrected, use it at both the point of entry and at the point of internal reflection. (What is the minimum angle of incidence to get internal reflection? )
Connect the two using cos2 = 1 - sin2.

3. Feb 22, 2014

### Eats Dirt

Snells Law: n1Sin(θi)=n2Sin(θt) where t is the transmitted and i is the incident ray and their angles are measured relative to the normal of the surface.

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4. Feb 22, 2014

### Eats Dirt

Ok I think I've got it,

$$\sin\theta$$

$$n_1\sin\theta_i=n_2\sin\theta_t$$

Known that the critical angle is $$\arcsin(\frac{n_2}{n_1})$$
$$\arcsin(n3/n2)=\frac{\pi}{2}-\theta_2$$
$$\theta_2=\frac{pi}{2}-\arcsin(\frac{n3}{n2})$$

$$n_1\sin(\theta_i)=n_2\sin(\theta_2)\\ n_1\sin(\theta_i)=n_2\sin(\frac{pi}{2}-\arcsin(\frac{n_3}{n_2}))\\$$

use the identity $$\sin(A-B)=sinAcosB-cosAsinB$$

$$n_1\sin(\theta_i)=n_2\sin(\frac{pi}{2})\cos(\arcsin(\frac{n3}{n2}))$$
use the identity $$\cos(\arcsin(x))=(1-x^2)^\frac{1}{2}\\ n_1\sin(\theta_i)=n_2(1-(\frac{n_3}{n_2})^2)^\frac{1}{2}\\ \sin(\theta_i)=\frac{n_2}{n_1}(1-(\frac{n_3}{n_2})^2)^\frac{1}{2}\\ \sin(\theta_i)=\frac{1}{n_1}((n_2)^2-(n_3)^2)^\frac{1}{2}$$

Last edited: Feb 22, 2014
5. Feb 22, 2014

Converted to LaTeX for easier reading
$n_1\sin(\theta_i)=n_2\sin(\theta_t)$

Known that the critical angle is $\arcsin(\frac{n_2}{n_1})$[general case]
so $\arcsin(\frac{n_3}{n_2})=\frac{\pi}{2}-\theta_2$
$\theta_2=\frac{pi}{2}-\arcsin(n_3/n_2)$

$n_1\sin(\theta_i)=n_2\sin(\theta_2)$
$n_1\sin(\theta_i)=n_2\sin(\frac{\pi}{2}-\arcsin(\frac{n_3}{n_2}))$
use the identity $\sin(A-B)$$=$$sinAcosB-cosAsinB$

$n_1\sin(\theta_i)=n_2\sin\frac{\pi}{2}\cos(\arcsin(\frac{n_3}{n_2})$
use the identity $\cos(\arcsin(x))=(1-x^2)^\frac{1}{2}$
$n_1\sin(\theta_i)=n_2(1-(\frac{n_3}{n_2})^2)$
$\sin(\theta_i)=\frac{n_2}{n_1}*(1-(\frac{n_3}{n_2})^2)$

$\sin(\theta_i)=\frac{1}{n_1}*((n_2)^2-(n_3)^2)^\frac{1}{2}$

Last edited: Feb 22, 2014
6. Feb 22, 2014

### Eats Dirt

I edited my earlier message to convert it to latex :) took me a while as I have only used it a few times prior.