# Modes in optical fibers

I am trying to understand the modes in step-index optical fibers and I saw that they say the electric field distribution in the core and cladding is as bellow. my question is that which component of electric field is this?

"TE" means "Transverse Electrical". That means these modes have E-field perpendicular to the axis of the fiber. Because the fiber is circularly symmetric, the E field could point in any direction perpendicular to the axis (we say that the modes in different polarizations are degenerate).

However, circular dielectric waveguide like step-index optical fiber does not actually have TE modes. The actual modes in optical fiber are called LP (linearly polarized), HE, and HM (hybrid electric or magnetic) modes. These modes have E-field that is not exactly perpendicular to the fiber axis. They will behave approximately like TE modes, but not exactly. The approximation will be better, the stronger the index contrast between core and cladding.

• thanks for your answer but I mean is the z component of the electric field or r or phi components of the electric field? because I want to plot HE electric filed but I do not know which component I have to plot I just think that it is z component – sara Sep 15 '18 at 15:56
• If "z" is the fiber axis, then "perpendicular to the fiber axis" means there is no z component. The TE modes (if they exist) have zero E-field in the z-direction. The LP mode is linearly polarized: E-field is either in the x or y direction, depending which polarization. That means the $r$ and $\phi$ components will be different at different locations in the transverse plane. – The Photon Sep 15 '18 at 16:00
• sorry but I think I can not understand what you mean maybe that is because I have just started learning about optical fibers. suppose that I have HE11 (fundamental mode) and it has both x and y components of the electric field and I want to plot the field and I know that in the cladding it is decaying but I do not know this plot demonstrates which component @The Photon – sara Sep 15 '18 at 16:26
• The E-Field is circularly symmetric about the z-axis. So, it doesn't matter which way you slice it, or what you call 'x' and what you call 'y'. – JQK Sep 15 '18 at 17:06
• @The Photon thank you very much for your kind answers – sara Sep 15 '18 at 17:41

As The Photon already pointed out, "TE" means "Transverse Electrical", which is just telling you that there is no component along the direction of propagation - $z$, in this case. See here for the distinction.

Now, if you want to find the functional form of your electric field, you have to solve the wave equation.
Since your problem is cylindrically symmetric around $z$, the direction of propagation, your solution will exhibit the same symmetry.

These solutions have a name, the Laguerre-Gassian modes, which look like this:

$$E(r, \phi, z) \propto e^{-\frac{r^2}{w^2(z)}}\cdot e^{-il\phi}\cdot e^{-ikz}.$$

You can see that the $\phi$ dependence is in a pure phase factor.

What you actually see on a camera is the intenstiy $I \propto E^{\dagger}E$ so $$I \propto e^{-2\frac{r^2}{w^2(z)}}.$$

So only the radial part remains, and you can drop the $\phi$ depdendence from $E(r, \phi, z) \propto e^{-\frac{r^2}{w^2(z)}}$. Which is exactly what is plotted in your drawings. The $y$ axis is $r$.

(I have just written the fundamental mode, which is a gaussian, like your $TE_0$, on the Wiki link you can see that there is an $l$ dependent spatial part, which is what gives you the different shape of the other $TE$ modes: for details look at the Mathematica script below:

)

• I have seen that you are also active on Mathematica, so I have added a short script to reproduce your field profiles. – SuperCiocia Sep 15 '18 at 17:50
• Thanks for the Mathematica code. I tried to solve my problem as you said but I have a problem: after solving Maxwell's equations the r dependence of my electric field is like Bessel functions and I chose cos(lϕ) instead of exponential dependence. so in the core, I have BesselJ and in the cladding I have BesselK. and after plotting it I can not obtain the desired plot @SuperCiocia – sara Sep 15 '18 at 18:16
• Yeah this is the typical problem of when to take the real part of the electric field. Your way of taking the real part immediately is the correct one. But then you get a time dependent $\cos^2$ in the intensity. So what you do, is take the time average, $\langle \cos^2 \rangle = 1/2$. You can take the average because the oscillation happens at $100$s of THz frequencies so extremely fast. The "advantage" of working with complex field and taking the intensity as $E^{\dagger}E$ is that it does the time-averaging automatically. – SuperCiocia Sep 15 '18 at 18:24
• Actually sorry my comment is wrong, I thought you were referring to the $z$ phase factor. Let me think about it. Sorry. – SuperCiocia Sep 16 '18 at 0:10
• OKay. Here it is.Laguerre-Gauss modes are not the real solutions of the full wave equation, but of an approximation of the latter known as the paraxial Helmholtz equation. This equation is actually complex because there is a factor of $i$ appearing, so the complext exponential as a solution is not just a maths trick, it's the real solution. You cannot just take the real part. You have to keep both terms. However you can use my reasoning that the $\phi$ term does not give rise to an intensity dependence upon $E^{\dagger}E$, so you can drop it. – SuperCiocia Sep 16 '18 at 16:56