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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.
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written 5.2 years ago by
teamques10
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• modified 5.2 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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