Question

Consider a step-index fiber for which n1 = 1.45, n2 = 1.24 and a = 24µm. What is the maximum value of numerical aperture.

Answer 1

$\sin \theta=\sqrt{n_1^2-n_2^2},$

$\sin\theta=\sqrt{1.45^2-1.24^2}=0.752,$

$\theta=48.7°.$