Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 6.4.3

Fibre optics

An optical fibre is given in the picture below. The refractive indices of the core and the cladding are $1.49$ and $1.48$ , respectively.

Calculate the critical angle of the total reflection phenomenon.
Find the largest angle a guided ray can make with the fibre axis.
Find the acceptance angle and the numerical aperture for light launched from air.
Over $L = 1 km$ , find the delay between the axial ray and the steepest guided ray.

Source: S. J. Orfanidis, Electromagnetic Waves and Antennas, Example 7.5.5.

Try it yourself first.

Strategy

A ray stays in the core if it meets the core-cladding wall beyond the critical angle. The angle $θ_{i}$ is measured from the normal of the wall, the angle $ϕ$ of the ray from the fibre axis, so $ϕ = 9 0^{\circ} - θ_{i}$ . Snell's law at the end face converts the largest $ϕ$ into an angle in air, and a ray at $ϕ$ travels the path $L / cos ϕ$ at the speed $c / n_{1}$ .

The ray goes from the denser core into the cladding, so total reflection is possible:

sin θ_{c} = \frac{n _{cladding}}{n _{core}} = \frac{1.48}{1.49} = 0.99329 ⟹ θ_{c} = 83.3 6^{\circ} .

Every ray with $θ_{i} > 83.3 6^{\circ}$ is trapped in the core.

$ϕ_{m a x} = 9 0^{\circ} - θ_{c} = 6.6 4^{\circ}$ : the guided rays are almost parallel to the axis.

At the end face $n_{0} sin θ_{a} = n_{1} sin ϕ_{m a x} = n_{1} cos θ_{c}$ , with $n_{0} = 1$ for air, and $n_{1} cos θ_{c} = n_{1} 1 - n_{2}^{2} / n_{1}^{2}$ :

NA = sin θ_{a} = n_{1}^{2} - n_{2}^{2} = 2.2201 - 2.1904 = 0.0297 = 0.172, θ_{a} = 9.9 2^{\circ} .

The axial ray needs $t_{0} = L n_{1} / c = 1 0^{3} \times 1.49/ (3 \times 1 0^{8}) = 4.967 μ s$ . The steepest guided ray travels $L / cos ϕ_{m a x} = L / sin θ_{c} = L n_{1} / n_{2}$ , so

Δ t = \frac{L n _{1}}{c} (\frac{n _{1}}{n _{2}} - 1) = 4.967 μ s \times 0.00676 = 33.6 ns .

Answer

θ_{c} = 83.3 6^{\circ}, ϕ_{m a x} = 6.6 4^{\circ}, θ_{a} = 9.9 2^{\circ}, NA = 0.172, Δ t = 33.6 ns/km

Check

$sin ϕ_{m a x} = sin 6.6 4^{\circ} = 0.1156$ and $1.49 \times 0.1156 = 0.172 = sin θ_{a}$ . For a small index step $Δ t \approx (L n_{1} / c) Δ n / n_{2}$ with $Δ n = n_{1} - n_{2} = 0.01$ : $4.967 μ s \times 0.01/1.48 = 33.6 ns$ .

(b) Delay spread and acceptance angle. — (a) The acceptance cone.

What the graphs say

A delay spread of $33.6 ns$ limits this step-index fibre to roughly $1/ (2Δ t) \approx 15 Mbit/s$ over a kilometre. Telecommunication fibre uses $Δ n \approx 0.005$ and a core so thin that only one mode propagates, and graded-index fibre equalises the delays of the rays.

Problem 6.4.3: Fibre optics