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)\\ …