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Graded Index Fiber's Optimal Grade Profile Parameter P = 2

  1. Jun 4, 2013 #1
    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:

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

    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 [itex]β_q[/itex] of mode "q" as:

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

    where M is the grand total number of modes.

    We know the group velocity is expressed as:

    [itex]v_q=\frac{dω}{dβ_q}[/itex]

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

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

    You can evaluate the inverse of the expression for group velocity: [itex]\frac{dβ_q}{dω}[/itex]

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

    [itex](\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}[/itex]

    Approximate by: [itex](1+δ)^{-1}≈1-δ[/itex] for [itex]|δ|≪1[/itex]

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

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