# Incident Angle Limitation Derivation in Optical Fibre

#### 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.

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#### haruspex

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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
First (for those of us unfamiliar with this topic), please describe the path of the light rays of interest. Better still, provide a diagram.
Snells Law:

n1sinθ1=n2sinθ2
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.

#### Eats Dirt

First (for those of us unfamiliar with this topic), please describe the path of the light rays of interest. Better still, provide a diagram.

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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#### 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 $$\cos(\arcsin(x))=(1-x^2)^\frac{1}{2}\\ n_1\sin(\theta_i)=n_2(1-(\frac{n_3}{n_2})^2)^\frac{1}{2}\\ \sin(\theta_i)=\frac{n_2}{n_1}(1-(\frac{n_3}{n_2})^2)^\frac{1}{2}\\ \sin(\theta_i)=\frac{1}{n_1}((n_2)^2-(n_3)^2)^\frac{1}{2}$$

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Gold Member
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 $\cos(\arcsin(x))=(1-x^2)^\frac{1}{2}$
$n_1\sin(\theta_i)=n_2(1-(\frac{n_3}{n_2})^2)$
$\sin(\theta_i)=\frac{n_2}{n_1}*(1-(\frac{n_3}{n_2})^2)$

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

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#### Eats Dirt

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

"Incident Angle Limitation Derivation in Optical Fibre"

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