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A silica optical fiber with a core diameter large enough to be considered by ray theory analysis has a core reflective index of 1.50 and a cladding refractive index of 1.47.
written 5.9 years ago by gravatar for teamques10 teamques10 ★ 68k •   modified 4.7 years ago

Calculate (i) Critical angle, (ii) NA of fiber, (iii) Acceptance angle in air for fiber.
advanced communication system
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written 5.8 years ago by gravatar for teamques10 teamques10 ★ 68k

Given: $n_1 = 1.5 \\n_2 = 1.47 $

To find:

  1. Critical angle

  2. Numerical Aperture (NA)

  3. Acceptance angle

Solution:

$ \phi _c = sin^{-1} (\frac{n2}{n1}) = sin^{-1} (\frac{1.47}{1.5}) \\ \phi _c = 78.52^o \\ Critical Angle \ \phi _c = 78.52^o \\ N.A = \sqrt{n_1^2-n_2^2} \\ = \sqrt{1.5^2-1.47^2} \\ = 0.2984 \\ Numerical Aperature = 0.2984 \\ N.A = sin \phi \\ 0.2924 = sin \phi \\ \phi_a = 17.36^o \\ Accepatance angle = 17.36^o $
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