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

Given,

$n_1=1.475\\n_2=1.460$

$a=25 \ \mu m$

Numerical aperture (NA) =

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

$=\sqrt{(1.475)^2-(1.460)^2}$

$=\sqrt{2.25-2.19}=0.210$

therefore,

NA= 0.210

now,

we calculate $\theta_{max}$

$\theta_{max}=sin^{-1}(0.210)=12.12\degree$

Answer=12.12⁰