Question

A stepped-index fiber has indices of 1.451 and 1.457. If the core radius is 3.5μm3.5 \mu \mathrm{m}, determine the cut-off wavelength above which the fiber will sustain only the fundamental mode.

Solution

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Starting with the information we got from the text of the exercise

nf=1.457nc=1.451R=3.5μm\begin{align*} n_f&=1.457\\ n_c&=1.451 \\ R&=3.5 \, \, \mu \text{m} \end{align*}

We will compute the cut-off wavelength with the help with help of the equation

λ=πRNA2.405\begin{equation} \lambda=\dfrac{\pi \, R\, NA}{2.405} \end{equation}

where we need to compute the numerical aperature

NA=(nf2nc2)1/2=(1.45721.4512)1/20.13\begin{equation} \text{NA}=\left(n_f^2-n_c^2 \right)^{1/2}=\left( 1.457^2-1.451^2 \right)^{1/2}\approx 0.13 \end{equation}

Plugging all the numbers into the equation (1)

λ=π3.50.132.405=594.4nm\boxed{\lambda=\dfrac{\pi \, 3.5\cdot 0.13 }{2.405}=594.4 \, \, \text{nm}}

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