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A step index fiber has a numerical aperture of 0.16, a core refractive index of 1.45 and core diameter of 90 mm, calculate (i) The acceptance angle . (ii) the refractive index of cladding.

written 5.9 years ago by teamques10 ★ 68k

•   modified 4.7 years ago

advanced communication system

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written 5.8 years ago by teamques10 ★ 68k

•   modified 5.8 years ago

Given :

N.A = 0.16 , n_1 = 1.45

A) N.A = $\sqrt{(n_1)^2-(n_2)^2}$

0.16 = $\sqrt{(1.45)^2-(n_2)^2}$

$(n_2)^2 = (1.45)^2 - 0.0256$

$n_2$ = 1.441

Refractive Index of cladding = 1.441

B) $sin \theta_a = \frac{N.A}{\eta_o}$

$sin \theta_a = \frac{0.16}{1}$

$\theta_a = sin^{-1} (0.16)$

$\theta_a = 9^o 12'$

Acceptace angle $\theta_a = 9^o 12'$

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