example_problem

example_problem - kHz 284.4 s 1.23 0.35 0.35 BW r = = = t...

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EXAMPLE Compare the acceptance cone angle and dispersion-limited bandwidth for two different optical fibers. Both fibers are 1 km long. Fiber A has a core with a refractive index of 1.6 and cladding with an index of 1.3. Fiber B has core and cladding indices of 1.6 and 1.5, respectively. SOLUTION Fiber A: The numerical aperture must first be calculated. NA = 0.933 1.3 - 1.6 n - n 2 2 2 2 2 1 = = The acceptance cone can be directly calculated as: θ accept = sin -1 (NA) = sin -1 (0.933) = 68.87˚ ( 29 ( 29 s 1.23 1 - 1.3 1.6 m/s 10 3 m 1000 1.6 1 - n n L n 8 2 1 1 r μ = × = = c t t Knowing this, we can estimate the bandwidth by
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Unformatted text preview: kHz 284.4 s 1.23 0.35 0.35 BW r = = = t Fiber B: The numerical aperture must first be calculated. NA = 0.557 1.5-1.6 n-n 2 2 2 2 2 1 = = The acceptance cone can be directly calculated as: accept = sin-1 (NA) = sin-1 (0.557) = 33.83 ( 29 ( 29 s 0.35 1-1.5 1.6 m/s 10 3 m 1000 1.6 1-n n L n 8 2 1 1 r = = = c t t Knowing this, we can estimate the bandwidth by kHz 984.4 s 0.35 0.35 0.35 BW r = = = t...
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This note was uploaded on 11/17/2011 for the course ECT 350 taught by Professor Brom during the Summer '11 term at N.C. A&T.

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