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A glass clad fiber is made with core glass of refractive index 1.5 and the cladding is dopped to give fractional index difference of 0.0005.

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

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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

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