# Why is it possible to couple multiple wavelengths into a single mode optical fiber?

I'm quite new to the mode theory, but as I understand, single mode fiber should only allow a single pattern of wavelength + polarization.

I'm assuming a non-modulated non-coherent light (a white LED, for example) coupled into single-mode fiber.

According to brief info I found, including this post, multiple modes of light can propagate through a single mode fiber but will experience losses, Will energy of all coupled modes transfer into heat inside of a single-mode core due to interference? Or loss occurs only in context of data transfer, when light is modulated?

• Often when wavelengths are close enough together we can treat them all the same Commented Aug 25, 2021 at 1:00
• There may be slightly different attenuation rates/km, but as @user253751 wrote, there's no steep cliff for supporting single-mode vs. wavelength. Polarization is usually controlled by making the core elliptical. BTW, most loss ends up in the cladding or exiting the fiber rather than being absorbed. Commented Aug 25, 2021 at 14:35
• @CarlWitthoft so this means single mode fiber will just gradually radiate away light until only one mode is left? Commented Aug 27, 2021 at 20:44

Single-mode fibers are optical fibers that for a specific wavelength range of interest allow a "single" mode (actually two modes, given there's always two orthogonal polarization modes). For wavelengths above a threshold, there is no posible propagation through the fiber. Meanwhile, for a lower wavelength threshold, two propagating modes start to be possible, then a third, and so on... (x2 because of polarization).

This means that to say a fiber is single-mode you should specify for what wavelength, and one could argue that all fibers are single-mode for a certain wavelength range. The single-mode range is mainly dependent on the core size.

The wavelengths that cannot propagate still get diffracted by the fiber but never couple into the fiber nor reach the other end. So, in the case of white light, depending on the fiber, you can have a combination of wavelengths lost (not coupled), wavelengths propagating in a single mode and wavelengths propagating in more than one mode.

• Thanks for your answer, but what happens to the "excessive" energy, if I, for example, am trying to couple 2 or more modes of same wavelength and polarization into a fiber that allows only 1 of such? Will they both get refracted, maybe turn into heat? Commented Aug 3, 2022 at 13:40
• You can define modes of propagation inside a fiber and count them (for a given wavelength) because they are determined by the indices of refraction and geometry of the fiber. The light that is coming from free space will not be in one or another predefined modes, but it will be a superposition of incoherent waves. If you decompose it in the modes of the fiber (i.e. the modes that fulfill a wave eq. inside the fiber) you will still have extra components that don't propagate. This means they fulfill an equation of exponential decay. Their energy dissipates/scatters into infinity Commented Aug 3, 2022 at 18:52
• I think that, more realistically, some light might couple into modes supported by the cladding and still propagate along the fiber some distance. This would depend also on curvature. Dissipation as heat along the fiber happens but it is very small (losses are seen in km ranges), and it's not what happens to the rejected modes. Commented Aug 3, 2022 at 18:55
• Thanks, could you possibly suggest some reading or course on the topic (light modes)? Commented Aug 5, 2022 at 10:42

The single mode operation of optical fiber depends on V-number that relates to diameter, wavelength, and refractive index.

\begin{align} \ V^2 &= \frac{2𝜋𝑑}{𝜆} (𝑛_1^2−𝑛_2^2)\\\\ \end{align}

The optical fiber works best at 850nm and 1310-1550nm of spectrum because of minimum loss. All optical network compnents (amplifiers, couplers) are designed around that spectrum.

You can launch multiple modes of UV waves (small wavelength) into a single mode fiber but attenuation would be too high becuase of rayleigh scattering and UV absorption.