Question

Consider a step-index fiber for which n1 = 1.5, n2 = 1.30 and a = 25µm. What is the maximum value of numerical aperture.

Answer 1

Given, $n_1=1.5,n_2=1.30$

The numerical aperture is-

$NA=\sqrt{n_1^2-n_2^2}$

$=\sqrt{(1.51)^2-(1.49)^2}=\sqrt{2.28-2.22}=\sqrt{0.06}=0.245$