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

Problem 7.6.3

A single-mode fibre

A step-index silica fibre: core $n_{1} = 1.4500$ , cladding $n_{2} = 1.4450$ , core radius $a = 4.0 μ m$ , cladding diameter $125 μ m$ .

Find NA, the acceptance angle (launch from air) and the critical angle at the core–cladding boundary.
Find $V$ at $1310$ and $1550 nm$ . Single-mode? Largest single-mode core radius at $1310 nm$ ?
Find the cutoff wavelength. Which LP modes propagate at $850 nm$ ?
At $1550 nm$ solve for LP $_{01}$ : $n_{eff}$ and the core power fraction. Is the cladding thick enough?

Try it yourself first.

Strategy

NA and the two angles are plane-wave optics (Section 6.4). The mode content depends on $V = k_{0} a NA$ alone: LP $_{11}$ starts at the first zero of $J_{0}$ , $V = 2.405$ , LP $_{21}$ and LP $_{02}$ at $3.832$ . For LP $_{01}$ the eigenvalue equation $u J_{1} (u) / J_{0} (u) = w K_{1} (w) / K_{0} (w)$ with $u^{2} + w^{2} = V^{2}$ is solved numerically, and the power fraction in the core is $Γ = 1 - \frac{u ^{2}}{V ^{2}} (1 - \frac{K _{0}^{2} ( w )}{K _{1}^{2} ( w )})$ .

NA θ_{a} = 1.450 0^{2} - 1.445 0^{2} = 0.014475 = 0.1203, = arcsin 0.1203 = 6.9 1^{\circ},

$θ_{c} = arcsin (1.445/1.450) = 85.2 4^{\circ}$ , and $Δ \approx (n_{1} - n_{2}) / n_{1} = 0.345 %$ : a weakly guiding fibre.

$V = \frac{2 π \times 4.0}{λ _{0}} \times 0.1203 = \frac{3.024}{λ _{0} [ μ m ]}$ , so

V_{1310} = 2.308, V_{1550} = 1.951,

both below $2.405$ : the fibre is single-mode at both wavelengths. At $1310 nm$ the radius may grow until $V = 2.405$ :

a_{m a x} = \frac{2.405 \times 1.31 μ m}{2 π \times 0.1203} = 4.17 μ m .

$λ_{c} = \frac{2 π a NA}{2.405} = \frac{3.024}{2.405} = 1.257 μ m$ . At $850 nm$ , $V = 3.024/0.85 = 3.557$ : LP $_{01}$ and LP $_{11}$ propagate, while LP $_{21}$ and LP $_{02}$ need $3.832$ . Counting two polarisations, and two orientations of LP $_{11}$ , that is $2 + 4 = 6$ field patterns, close to the estimate $V^{2} /2 = 6.3$ .

At $V = 1.951$ the LP $_{01}$ equation gives $u = 1.511$ , $w = 1.234$ (both sides equal $1.672$ ). Then $b = w^{2} / V^{2} = 0.400$ and

n_{eff} = n_{2}^{2} + b NA^{2} = 1.445 0^{2} + 0.400 \times 0.014475 = 1.44700.

With $K_{0} (w) / K_{1} (w) = 0.738$ and $u^{2} / V^{2} = 0.600$ ,

Γ = 1 - 0.600 (1 - 0.544) = 0.727.

In the cladding the field is $J_{0} (u) K_{0} (w r / a) / K_{0} (w)$ times its value on the axis. It falls to $1 %$ at $r = 14.8 μ m$ and to $2 \times 1 0^{- 9}$ at the cladding surface, $r = 62.5 μ m$ . The cladding is thick enough by a wide margin.

Answer

NA = 0.120, θ_{a} = 6. 9^{\circ}, θ_{c} = 85. 2^{\circ} λ_{c} = 1257 nm; 850 nm : LP_{01}, LP_{11} 1550 nm : n_{eff} = 1.44700, Γ = 72.7 %

Check

$n_{2} < n_{eff} < n_{1}$ . $sin θ_{a} = n_{1} cos θ_{c} = 1.45 \times cos 85.2 4^{\circ} = 0.1203 = NA$ .

(a) LP_{01} field across the fibre. — (a) LP $_{01}$ field across the fibre.

(b) The fibre on the LP chart. — (a) LP $_{01}$ field across the fibre.

Remark

At $1550 nm$ $27 %$ of the power runs in the cladding. That is why bends cost more at $1550 nm$ than at $1310 nm$ , and why the cladding must be many decay lengths thick.

Problem 7.6.3: A single-mode fibre