Question

step index fibre has a core diameter of 29×10-6 m .The R.I. of core & cladding are 1.52 &1.5189 respectively. If the light of wavelength 1.3 µm is transmitted through the fibre. Determine normalized frequency of fibre & number of modes the fibre will support.

Answer 1

For the given step-index fiber, the normalized frequency is 4.05, and the number of modes the fiber will support is about 8.

Given,

For a step-index fiber:

core diameter = 29 × 10⁻⁶ m,

Refractive index of core = 1.52,

Refractive index of cladding = 1.52, and

Wavelength of light = 1.3 μm.

To find,

Normalized frequency,

The number of modes the fiber will support.

Solution,

Firstly, we know that for optical fiber, the normalized frequency, which is denoted by V, is given as

$V=\frac{2\pi a}{\lambda} \sqrt{n_1^2-n_2^2} \hfill ...(1)$

where,

a = core radius,

λ = light wavelength,

n₁ = refractive index of core,

n₂ = refractive index of cladding.

Further, for a step-index fiber, the number of modes, which is also called the mode volume, is given by

$\frac{V^{2} }{2} \hfill ...(2)$

Here, it is given that,

core diameter = 29 × 10⁻⁶ m

⇒ core radius = a = 14.5 × 10⁻⁶ m,

λ = 1.3 × 10⁻⁶ m,

n₁ = 1.52,

n₂ = 1.5189.

• Thus, for the given fiber, using (1), the normalized frequency can be determined as follows.

$V=\frac{2(3.14)(14.5 \times 10^{-6})}{1.3 \times 10^{-6}} \times \sqrt{(1.52)^2-(1.5189)^2}$

$\implies V=\frac{91.06}{1.3} \times \sqrt{(2.3104)-(2.307057)}$

$\implies V= 70.04615 \times \sqrt{0.003343}$

$\implies V= 70.04615 \times 0.057819$

⇒ V = 4.049998

⇒ V ≈ 4.05.

• Now, using (2), the number of modes will be

$= \frac{(4.05)^{2} }{2}$

$=\frac{16.4025}{2}$

= 8.20125.

≈ 8.

⇒ the number of modes ≈ 8.

Therefore, for the given step-index fiber, the normalized frequency is 4.05, and the number of modes the fiber will support is about 8.

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