1. Jun 4, 2013

### SpecialK0

This was something I was thinking about yesterday regarding graded index optical fibers. So since most graded index fibers adhere to the power-law function for its refractive index profile of the core shown below:

$n^2 (r)=n_1^2 [1-2(\frac{r}{a})^p ∆], r≤a,$ where $∆ =\frac{n_1^2-n_2^2}{2n_1^2}≈\frac{n_1-n_2}{n_1}$

Why does the quadratic grade profile parameter of p = 2 so well suited for such a fiber? I understand the mathematics of it and the rigorous explanation (thank you Born & Wolf optics textbook), but intuitively, I'm just unsure what the quadratic dependency is compensating for so that the group velocity of any given mode is roughly equalized when p = 2.

The rigorous explanation is as follows:

So you have the propagation constant $β_q$ of mode "q" as:

$β_q≈n_1 k_0 [1-(\frac{q}{M})^\frac{p}{p+2}∆]$

where M is the grand total number of modes.

We know the group velocity is expressed as:

$v_q=\frac{dω}{dβ_q}$

To essentially rewrite $β_q$ in terms of ω, we first substitute $n_1 k_0=\frac{ω}{c_1}$ into the $β_q$ equation:

$β_q≈\frac{ω}{c_1} [1-(\frac{q}{M})^\frac{p}{p+2} ∆]$

You can evaluate the inverse of the expression for group velocity: $\frac{dβ_q}{dω}$

$\frac{dβ_q}{dω}=\frac{1}{c_1} [1+\frac{p-2}{p+2} (\frac{q}{M})^\frac{p}{p+2} ∆]$

$(\frac{dβ_q}{dω})^{-1}=\frac{dω}{dβ_q}=c_1 [1+\frac{p-2}{p+2} (\frac{q}{M})^\frac{p}{p+2} ∆]^{-1}$

Approximate by: $(1+δ)^{-1}≈1-δ$ for $|δ|≪1$

$\frac{dω}{dβ_q}=c_1 [1-\frac{p-2}{p+2} (\frac{q}{M})^\frac{p}{p+2} ∆]=v_q$

Now, plug in $p = 2$. $v_q≈c_1$, thus, for all modes "q", the respective group velocity is roughly all traveling at the same velocity $c_1$.