# Why does a single mode fibre have a cutoff wavelength?

I earlier had a doubt based on ray theory that light must reflect and so all wavelengths should propagate through SMF. This Q/A Single-mode fibers and ray-theory of light does specify that we have to consider wave theory for propagation of wave in SMF, but i still don't get why a SMF has a cutoff wavelength.

Any optical fibre - and any optical waveguide in general - has a cutoff wavelength (and therefore a cutoff frequency or a cutoff energy) because wether light is confined in the film region (guided modes) or escapes to the substrate (substrate-radiation modes) depends on the propagation constant, $\beta$, which is related to frequency trough the dispersion relation. $\beta$ is just the quantity introduced to account for the propagation of the wave along the optical axis of the waveguide or fibre, and it appears in the wave equation. Since ray theory doesn't consider this quantity nor the effect of dispersion, it fails to account for the existence of a cutoff wavelength.
This figure shows how $\beta$ and $\omega$ relate in an optical fibre or waveguide (taken from Tamir et al: Integrated Optics - Springer Verlag 1979). The $n_s k$ line corresponds to the cutoff region (the cutoff condition is precisely $N_\nu = n_s$ and therefore $\beta_\nu = k n_s$, where $N_\nu$ is the effective index and $n_s$ the refraction index of the substrate).