Find (i) The cladding index, (ii) The critical internal reflection angle, (iii) The Numerical aperture

Solution :

Given , $n_1 = 1.5 , \triangle = 0.0005$

i) Let the refractive index of cladding be $n_2$ . So we have

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

0.0005 = $\frac{1.5 - n_2}{1.5}$

$n_2$ = 1.5 - 1.5 x 0.005 = 1.4925

ii) $sin \theta_c = n2/n1$

$\theta_c = sin^{-1} (n2/n1)$

$\theta_c = sin^{-1} (1.4925/1.5)$

$\theta_c = 84.268^o$

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

NA = ${\sqrt{(1.5)^2-(1.4925)^2}}$

NA = ${\sqrt{ 2.25 - 2.2275}}$

NA = ${\sqrt{0.02244}}$

NA = 0.1498

NA = 0.15