A glass-clad fiber is made with a core glass of a refractive index of 1.5 and the cladding is doped to give an index difference of 0.0005. Determine,

(a) the cladding refractive index,

(b) the critical reflection angle,

(c) the critical acceptance angle and

(d) the numerical aperture.

Solution:

Given,

$n_1=1.5$ and

$\Delta=0.0005$

(a) We know that,

$ \Delta=\frac{n_1-n_2}{n_1}\\ $

$\therefore \quad n_2=1.49925$

(b) The critical angle is given by,

$ \begin{aligned} \phi_c & =\sin ^{-1}\left(\frac{n_2}{n_1}\right) \\\\ \therefore \quad \phi_c & =\sin ^{-1}\left(\frac{1.49925}{1.5}\right)=88.2^{\circ}\\ \end{aligned} $

(c) The critical acceptance angle $\alpha$ is given by,

$ \begin{aligned}\\ \alpha & =\sin ^{-1} \sqrt{n_1^2-n_2^2}=\sin ^{-1} \sqrt{(1.5)^2-(1.49925)^2} \\\\ & =\sin ^{-1}(0.0474)=2.72^{\circ} \text { Ans. }\\ \end{aligned} $

(d) The numerical aperture NA is given by,

$ N A=\sin \alpha=\sin 2.72=0.0474 $