Answer to Question #277086 in Optics for Ali

Question

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.

Answer 1

N=742

NA = 0.3

d = 70 $\mu$

Number of modes of propagation in graded index fiber is

$N = 2.45(\frac{d \times NA}{λ})^2 \\ 742 = \frac{2.45(70 \times 10^{-6} \times 0.3)^2}{λ^2} \\ λ^2 = \frac{2.45(21 \times 10^{-6})^2}{742} \\ λ = \sqrt{1.456 \times 10^{-12}} \\ λ = 1.2067 \times 10^{-6} \; \mu m$

For single mode operation:

$V≤2.405 \\ \frac{2 \pi \times a \times NA}{λ } ≤ 2.405 \\ a ≤ \frac{2.405 \times λ }{2 \pi \times NA} \\ a ≤ \frac{2.405 \times 1.2067 \times 10^{-6}}{2 \pi \times 0.3} \\ a ≤ 1.54 \times 10^{-6} \; m$

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

$d ≤ 2 \times 1.54 \; \mu m \\ d ≤ 3.08 \; \mu m$

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