This is a continuation from the previous tutorial - weakly guiding fibers.

In a graded-index fiber, the index profile, n(r), in the core of the fiber is a function of the radial distance, r, from the center of the fiber, as shown in figure 7 below.

It starts at a value of n(0) = n1 at the center of the fiber and gradually decreases to a value of n(a) = n2 at the boundary between the core and the cladding.

The fiber V number defined in (1) [refer to the step-index fibers tutorial] can still be used, but the properties of a graded-index fiber are also determined by the specific functional dependence of n(r) on r. The numerical aperture, for example, is a function of radial position:

$\tag{67}NA(r)=\sqrt{n^2(r)-n_2^2}$

which decreases from $$NA(0)=\sqrt{n_1^2-n_2^2}$$ at the core center to $$NA(a)=0$$ at the core-cladding boundary.

Because the index profile is no longer piecewise constant, the electric susceptibility $$\epsilon(r)=\epsilon_0n^2(r)$$ is also a function of radial position. As a result, piecewise homogeneous wave equations for $$\mathcal{E}_z$$ and $$\mathcal{H}_z$$, such as those in (9) and (10) [refer to the step-index fibers tutorial], cannot be used.

For a graded-index fiber with $$\epsilon(r)$$, it can be shown by substitution of (3) and (4) [refer to the step-index fibers tutorial] in (24) and (25) [refer to the wave equations for optical waveguides tutorial], respectively, that

$\tag{68}\frac{\partial^2\mathcal{E}_z}{\partial r^2}+\frac{1}{r}\frac{\partial\mathcal{E}_z}{\partial r}+\frac{1}{r^2}\frac{\partial^2\mathcal{E}_z}{\partial \phi^2}+(k^2-\beta^2)\mathcal{E}_z=-\text{i}\beta\frac{\text{d}\text{ ln }\epsilon}{\text{d}r}\mathcal{E}_r$

$\tag{69}\frac{\partial^2\mathcal{H}_z}{\partial r^2}+\frac{1}{r}\frac{\partial\mathcal{H}_z}{\partial r}+\frac{1}{r^2}\frac{\partial^2\mathcal{H}_z}{\partial \phi^2}+(k^2-\beta^2)\mathcal{H}_z=\frac{\text{d ln }\epsilon}{\text{d}r}\left(-\text{i}\beta\mathcal{H}_r+\frac{\partial\mathcal{H}_z}{\partial r}\right)$

where $$k^2=\omega^2\mu_0\epsilon(r)=\omega^2n^2(r)/c^2$$ is a function of r, and the relation

$\tag{70}\frac{\pmb{\nabla}\epsilon}{\epsilon}=\frac{\text{d ln }\epsilon}{\text{d}r}\hat{r}$

is used.

The longitudinal field components $$\mathcal{E}_z$$ and $$\mathcal{H}_z$$ cannot be solved without solving the transverse field components simultaneously. This is a manifestation of the complicated vectorial nature of the mode fields in a fiber, which is caused by the geometry and structure of the dielectric fiber waveguide.

In general, the problem has to be solved numerically with vectorial wave equations. However, some approximate analytic approaches exist that allow us to gain much understanding of the key characteristics of a graded-index fiber without numerical solutions.

All graded-index optical fibers in practical applications are weakly guiding fibers with very small index changes satisfying the condition in (48) [refer to the weakly guiding fibers tutorial]. In addition, the index profiles are usually quite smooth so that the index gradients are small. Under these assumptions, $$\pmb{\nabla}\epsilon/\epsilon$$ is very small, and an approximation can be made to neglect the terms on the right-hand sides of (68) and (69), resulting in the following approximate homogeneous equations for $$\mathcal{E}_z$$ and $$\mathcal{H}_z$$:

$\tag{71}\frac{\partial^2\mathcal{E}_z}{\partial r^2}+\frac{1}{r}\frac{\partial\mathcal{E}_z}{\partial r}+\frac{1}{r^2}\frac{\partial^2\mathcal{E}_z}{\partial \phi^2}+(k^2-\beta^2)\mathcal{E}_z\approx0$$\tag{72}\frac{\partial^2\mathcal{H}_z}{\partial r^2}+\frac{1}{r}\frac{\partial\mathcal{H}_z}{\partial r}+\frac{1}{r^2}\frac{\partial^2\mathcal{H}_z}{\partial \phi^2}+(k^2-\beta^2)\mathcal{H}_z\approx0$

By separation of variables, these equations yield the same $$\phi$$ dependence as that of the mode fields of a step-index fiber discussed in the step-index fibers tutorial and expressed in (23)-(26) [refer to the step-index fibers tutorial]. The r dependence is given by solution of

$\tag{73}\frac{\text{d}^2 R}{\text{d}r^2}+\frac{1}{r}\frac{\text{d}R}{\text{d}r}+\left(k^2-\beta^2-\frac{m^2}{r^2}\right)R=0$

Because the $$\phi$$ dependence of the mode fields of a graded-index fiber is exactly the same as that of the mode fields of a step-index fiber and is independent of its index profile, the classification of the normal modes into the basic TE, TM, HE, and EH types discussed in the step-index fibers tutorial and the concept of the LP modes as appropriate linear combinations of basic normal modes for a weakly guiding fiber discussed in the weakly guiding fibers tutorial are still valid.

However, the r dependence of the mode fields is no longer simply described by the Bessel functions. It depends on the specific functional form of the index profile n(r).

Approximate solutions of (73) can be obtained for the r dependence of the guided mode fields of a graded-index fiber using the WKB approximation in a manner similar to, but somewhat more complicated than, that outlined in the graded-index planar waveguides tutorial for obtaining the approximate solutions for a graded-index planar waveguide. Here we summarize the results without going through the details.

The existence and the characteristics of guided modes in a graded-index fiber depend on the sign of the following function:

$\tag{74}p^2(r)=k^2(r)-\frac{m^2-1/4}{r^2}-\beta^2$

For any given guided mode, there exist two turning points r = r1 and r = r2, with 0 < r1 < r2 < a, where p(r) = 0. For the mode field to exist and be guided, it is necessary that p2(r) > 0 within the radial range r1 < rr2, but p2(r) < 0 for r <  r1  and r > r2. The eigenvalue equation for the guided modes of a graded-index fiber is given in terms of the function p(r):

$\tag{75}\displaystyle\int\limits_{r_1}^{r_2}p(r)\text{d}r=\displaystyle\int\limits_{r_1}^{r_2}\left[k^2(r)-\frac{m^2-1/4}{r^2}-\beta_{mn}^2\right]^{1/2}\text{d}r=\left(n-\frac{1}{2}\right)\pi, \qquad n=1,2,3...$

The allowed values of β can be obtained by solving this equation with integral values of m and n. Evidently, the solution depends on the precise form of n(r). For a given azimuthal index m, there exists only a finite number of discrete values of βmn that are allowed for the guided modes.

While the $$\phi$$ dependence of the mode fields remains the same as that of the mode fields discussed in the step-index fibers tutorial, the radial variations can be approximated with the following asymptotic form at radial locations away from the immediate vicinity of the turning points where p(r) = 0:

\tag{76}R(r)\sim\left\lbrace\begin{align}&\frac{1}{\sqrt{r|p(r)|}}\exp\left[-\displaystyle\int\limits_r^{r_1}|p(r')|\text{d}r'\right],&&r\lt r_1\\&\frac{2}{\sqrt{rp(r)}}\cos\left[\displaystyle\int\limits_{r_1}^rp(r')\text{d}r'-\frac{\pi}{4}\right],&&r_1\lt r\lt r_2\\&\frac{(-1)^{n-1}}{\sqrt{r|p(r)|}}\exp\left[-\displaystyle\int\limits_{r_2}^r|p(r')\text{d}r'\right],&&r\gt r_2\end{align}\right.

where the factor of (-1)n-1 is used for the correct phase at r = r2.

For a graded-index fiber, this radial solution replaces the Bessel functions $$J_m(hr)$$ and $$K_m(hr)$$ in (23)-(26) of the mode fields of a step-index fiber [refer to the step-index fibers tutorial]. Note the similarity between the form of R(r) in oscillatory and evanescent regions and that of the first terms in (18) and (19), respectively, for the asymptotic behavior of $$J_m(x)$$ and $$K_m(x)$$ at large values of x.

Number of modes

Graded-index fibers are primarily used as low-dispersion multimode fibers. The total number of modes supported by a graded-index multimode fiber can be estimated using the eigenvalue equation in (75). It can be seen from (75) that for a given azimuthal mode index m, the largest number, n(m), for the radial mode index is obtained when β has a minimum value. Since β > k2 for a guided mode, the minimum value of β can be approximated by k2 for a fiber that has a large number of modes. Therefore, n(m) is approximately given by

$\tag{77}n(m)\approx\frac{1}{\pi}\displaystyle\int\limits_{r_1}^{r_2}\left[k^2(r)-\frac{m^2-1/4}{r^2}-k_2^2\right]^{1/2}\text{d}r\approx\frac{1}{\pi}\displaystyle\int\limits_{r_1}^{r_2}\left[k^2(r)-\frac{m^2}{r^2}-k_2^2\right]^{1/2}\text{d}r$

Meanwhile, for guided modes, it is necessary that p(r) > 0 for r1 < r < r2, as discussed above. Therefore, the largest number, mmax, of the azimuthal mode index for guided modes is

$\tag{78}m_{max}=r\sqrt{k^2(r)-k_2^2}$

The total number of guided modes can then be estimated as

$\tag{79}M=4\sum_{m=0}^{m_{max}}n(m)\approx4\displaystyle\int\limits_{0}^{m_{max}}n(m)\text{d}m$

where the factor 4 accounts for the four-fold degeneracy of most high-order guided modes as discussed in the preceding section, and the summation over m is replaced by an integral for a fiber of a large number of densely spaced modes.

Substituting (77) and (78) in (79) and noting that the minimum value of r1 is r = 0 while the maximum value of r2 is r = a, we have

$\tag{80}M\approx\frac{4}{\pi}\displaystyle\int\limits_0^a\int\limits_0^{r\sqrt{k^2(r)-k_2^2}}\left[k^2(r)-k_2^2-\frac{m^2}{r^2}\right]^{1/2}\text{d}m\text{d}r=\displaystyle\int\limits_0^a[k^2(r)-k_2^2]r\text{d}r$

In terms of the index profile of the fiber, the total number of modes is

$\tag{81}M\approx\frac{\omega^2}{c^2}\displaystyle\int\limits_0^a[n^2(r)-n_2^2]r\text{d}r$

Following the line of argument leading to (80), we can find the number Mβ of guided modes that have propagation constants larger than β to be

$\tag{82}M_\beta=\displaystyle\int\limits_0^{r_2(\beta)}[k^2(r)-\beta^2]r\text{d}r$

where r2(β) is determined by k(r2) = β.

Power-law index profiles

We consider here the following power-law index profile:

\tag{83}n(r)=\left\lbrace\begin{align}&n_1\left[1-2\Delta\left(\frac{r}{a}\right)^{\alpha}\,\right]^{1/2}, &&0\le r\le a\\&n_2,&&r\gt a\end{align}\right.

where

$\tag{84}\Delta=\frac{n_1^2-n_2^2}{2n_1^2}\approx\frac{n_1-n_2}{n_1}$

with Δ ≪ 1, the fiber core has a linear index profile for $$\alpha=1$$. It becomes a step-index fiber for $$\alpha=\infty$$. In terms of Δ, the V number of a fiber is

$\tag{85}V=\frac{\omega}{c}a\sqrt{n_1^2-n_2^2}=\frac{\omega}{c}an_1\sqrt{2\Delta}$

Substituting (83) in (81) and using (85), the total number of modes can be obtained:

$\tag{86}M=\frac{\alpha}{\alpha+2}\frac{V^2}{2}$

From (82), the number of modes with propagation constants larger than β is found to be

$\tag{87}M_\beta=M\left(\frac{1-\beta^2/k_1^2}{2\Delta}\right)^{(\alpha+2)/\alpha}$

Therefore, the propagation constant can be written

$\tag{88}\beta=k_1\left[1-2\Delta\left(\frac{M_\beta}{M}\right)^{\alpha/(\alpha+2)}\right]^{1/2}$

The relation in (86) between M and V for a graded-index fiber applies only when M is a large number. It fails for a fiber that supports only a few modes and clearly is not applicable to single-mode fibers.

A single-mode graded-index fiber is also determined by a cutoff V number similar to (47) [refer to the step-index fibers tutorial] for a step-index fiber. However, the cutoff V number for a given graded-index fiber depends on the particular index profile of the fiber. Specifically, the condition for a fiber with a quadratic index profile of $$\alpha=2$$ to be single moded is V < 3.53. For other power-law profiles, the condition is approximately $$V\lt2.405\sqrt{1+2/\alpha}$$.

Example

If a graded-index fiber has all of the parameters of the step-index fiber designed in the weakly guiding fibers tutorial example, except for a quadratic index profile of $$\alpha=2$$, how many guided modes does it support at 850 nm? What is the propagation constant of its 200th mode? What should its core diameter be for the graded-index fiber to support at least 1000 modes at 850 nm if its index parameters and profile remain unchanged?

The fiber designed in the weakly guiding fibers example has V = 50.05 and supports 1015 modes at 850 nm wavelength. A graded-index fiber with the same parameters but with $$\alpha=2$$ also has the same V number according to (85), but its mode number is given by (86). Therefore, the number of modes it supports is

$M=\frac{2}{2+2}\times\frac{50.05^2}{2}=626$

which is much smaller than that of the step-index fiber. With n1 = 1.48, we have k1 = 10.94 μm-1. From the weakly guiding fibers example, we know that Δ = 0.0169. The propagation constant for the 200th mode can then be found using (88) by taking Mβ = 200 to be β = 0.99k1 = 10.83 μm-1, which is only 1% below k1 because Δ is only 1.69%.

For M > 1000, we find that V > 63.25 is required by using (86) with $$\alpha=2$$. With n1 = 1.48, n2 = 1.455, and λ = 850 nm, we find from (85) that a > 31.6 μm. Therefore, the minimum core diameter is 63.2 μm, which is clearly larger than the 50 μm core of the step-index fiber that supports the same number of modes.

The next part continues with the Attenuation in Fibers tutorial.