Question #277086

A graded index fiber with a parabolic index profile supports the propagation of




742 guided modes. The fiber has a numerical aperture in air of 0.3 and a core diameter




of 70 μm. Determine the wavelength of the light propagating in the fiber.




Further estimate the maximum diameter of the fiber which gives single-mode




operation at the same wavelength.

1
Expert's answer
2021-12-08T10:00:50-0500

N=742

NA = 0.3

d = 70 μ\mu

Number of modes of propagation in graded index fiber is

N=2.45(d×NAλ)2742=2.45(70×106×0.3)2λ2λ2=2.45(21×106)2742λ=1.456×1012λ=1.2067×106  μmN = 2.45(\frac{d \times NA}{λ})^2 \\ 742 = \frac{2.45(70 \times 10^{-6} \times 0.3)^2}{λ^2} \\ λ^2 = \frac{2.45(21 \times 10^{-6})^2}{742} \\ λ = \sqrt{1.456 \times 10^{-12}} \\ λ = 1.2067 \times 10^{-6} \; \mu m

For single mode operation:

V2.4052π×a×NAλ2.405a2.405×λ2π×NAa2.405×1.2067×1062π×0.3a1.54×106  mV≤2.405 \\ \frac{2 \pi \times a \times NA}{λ } ≤ 2.405 \\ a ≤ \frac{2.405 \times λ }{2 \pi \times NA} \\ a ≤ \frac{2.405 \times 1.2067 \times 10^{-6}}{2 \pi \times 0.3} \\ a ≤ 1.54 \times 10^{-6} \; m

a-radius ≤ 1.54 μ\mu m and diameter d = 2a

d2×1.54  μmd3.08  μmd ≤ 2 \times 1.54 \; \mu m \\ d ≤ 3.08 \; \mu m

The maximum diameter d=3.08  μmd = 3.08 \; \mu m


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS