A manufacturer wishes to make a silica core, step-index fiber with $\mathrm{V}=75$ and a numerical aperture $\mathrm{NA}=0.30$ to be used at $820 \mathrm{~nm}$. If $n_1=1.458$, what should the core size a and cladding index be?

Solution:

Solution: We know that the numerical aperture is given by,,

Given

$\begin{aligned} & N A=\sqrt{n_1^2-n_2^2} \\\\ & N A=0.3, n_1=1.458 \\\\ & \therefore \quad n_2^2=n_1^2-(N A)^2=(1.458)^2-(0.3)^2 \text { or } n_2=1.426 \\\\ & \end{aligned}\\ $

Also, normalized frequency is,

$V=\frac{2 \pi a}{\lambda}\left(n_1^2-n_2^2\right)^{1 / 2}=\frac{2 \pi a}{\lambda}(N A)\\ $

Given,

$\quad \lambda=820 \mathrm{~nm}$ and $V=75$

$\therefore \quad a=\frac{V \lambda}{2 \pi(N A)}=\frac{75 \times 820}{2 \pi \times 0.3}=32.6 \mu \mathrm{m} \quad$