Question

Consider a step-index fiber for which n1 = 1.475, n2 = 1.460 and a = 25µm. What is the maximum value of θ for which the ray ill be guided by the fiber. 

Answer 1

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

$\sin\theta=\sqrt{1.475^2-1.46^2}=0.209,$

$\theta=12.1°.$