## Related questions with answers

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

Solution

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

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

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

$\begin{equation} \lambda=\dfrac{\pi \, R\, NA}{2.405} \end{equation}$

where we need to compute the numerical aperature

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

Plugging all the numbers into the equation (1)

$\boxed{\lambda=\dfrac{\pi \, 3.5\cdot 0.13 }{2.405}=594.4 \, \, \text{nm}}$

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