A step-index fiber has normalized frequency $V=266$ at $1300 \mathrm{~nm}$ wavelength. If the core radius is $25 \mu \mathrm{m}$, calculate the numerical aperture.

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prajapatijaimin • 3.3k

• modified 3.0 years ago

optical fiber communication

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written 3.0 years ago by

prajapatijaimin • 3.3k

Solution:

We know that the normalized frequency is given by,

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

Also,

$ N A=\sqrt{n_1^2-n_2^2} \quad \therefore V=\frac{2 \pi a}{\lambda} N A\\ $

Given

$V=26.6, a=25 \mu \mathrm{m}, \lambda=1300 \mathrm{~nm}=1.3 \mu \mathrm{m}$

$ N A=\frac{V \lambda}{2 \pi a}\\ $

$ =\frac{26.6 \times 1.3}{2 \pi \times 25} $

$ =0.22 $

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