Question

A multimode step index fiber has a relative refractive index difference of 1%, and a core refractive index of 1.5. The number of modes propagating at a wavelength of 1.3um is 1100. Estimate the diameter of fiber core

Answer 1

Answer on Question #70230, Physics / Optics

A multimode step index fiber has a relative refractive index difference of $1\%$, and a core refractive index of 1.5. The number of modes propagating at a wavelength of $1.3\,\mu\mathrm{m}$ is 1100. Estimate the diameter of fiber core

Solution:

**Given:**

- Relative index difference, $\Delta = 1.0\% = 0.01$

- Wavelength, $\lambda = 1.3\,\mu\mathrm{m}$

- Core refractive index, $n_1 = 1.5$

- Number of modes, $M = 1100$

- diameter of fiber core, $d = ?$

The total number of modes in a multimode fiber is given by

$$M = \frac{1}{2} \left[ \frac{\pi d}{\lambda} \cdot N A \right]^2$$

where

- $d$ is core diameter

$$d = \frac{\lambda \sqrt{2M}}{\pi N A}$$

NA is numerical aperture

$$N A = n_1 (2 \Delta)^{\frac{1}{2}}$$ $$N A = 1.5 \times (2 \times 0.01)^{\frac{1}{2}} = 0.21213$$

So,

$$d = \frac{1.3 \times 10^{-6} \sqrt{2 \times 1100}}{\pi \times 0.21213} = 91.5 \times 10^{-6} \, \mathrm{m} = 91.5\,\mu\mathrm{m}$$

**Answer:** $91.5\,\mu\mathrm{m}$

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