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I am trying to understand the theory of inter-modal dispersion in optical fibers. It seems quite obvious that if higher modes have a greater angle of incidence in the fiber than lower modes, the path length of higher modes through the fiber is larger. This is because higher modes undergo more reflections, but they also have a greater part of the light wave traveling is the cladding. Here the speed of light is a little higher than in the core and therefore the higher modes are moving faster than lower modes. The theory of a part of the light wave traveling in the cladding has to do with the evanescent field I think, but why do higher modes have a greater part of the wave traveling in the cladding in comparison with lower modes?

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At a smaller angle of incidence on the boundary (higher mode), the field of the evanescent wave penetrates more deeply into the optically rarer medium (cladding). See the derivation here, in which the characteristic depth of penetration of the evanescent electric field is given by $$ \frac{c}{\omega}\left((n_1\sin(\theta_I))^2-n_{2}^{2}\right)^{-1/2}. $$

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  • $\begingroup$ I can understand what you mean. Thank you for your answer. But if I may ask another question, if a third medium is brought close by, the evanescent field can (if powerful enough of course) cause an energy flow in this medium (evanescent coupling). Now a light wave (EM-wave) can occur in this third medium. But how is this possible since only the electric field is crossing the border to the cladding and subsequently to the third medium? Does the electric field causing a magnetic field in the third medium by polarizing the electrons or do I miss something important here? $\endgroup$ – Gregory Jun 2 '15 at 17:21
  • $\begingroup$ Crucially, there is a magnetic field associated with the evanescent wave. Neither the electric nor the magnetic field are allowed to be discontinuous at the boundary, and this applies no matter how many boundaries you have. So to address your proposed situation, an evanescent wave could extend into a third, adjacent medium. If the refractive index is greater than $n_2$, this evanescent wave could even give rise to a propagating wave in the third medium (the 'resonant tunneling' process). $\endgroup$ – tok3rat0r Jun 2 '15 at 19:36
  • $\begingroup$ Thank you for this explanation. But an evanescent field itself is invisible (like we can't detect it with our eyes) as far as I know. But if the electric and magnetic are continues across the border, why is this not visible, but it is when it is propagating in the third medium? $\endgroup$ – Gregory Jun 5 '15 at 19:26
  • $\begingroup$ We can't detect it with our eyes because the evanescent wave is a near field effect, localised within about half a wavelength (which for visible light is <0.5 micron) of the interface. To transmit the light to our eyes, or to any conventional detector, we need to convert it back into a propagating wave which can extend into the far field region. This doesn't mean evanescent waves are undetectable however, just that you need a detector which is very small, and very close to the surface $\endgroup$ – tok3rat0r Jun 5 '15 at 20:05
  • $\begingroup$ It's worth emphasising that no energy transfer across the boundary occurs in an evanescent wave. The fields induced by the incident wave, when averaged over the full cycle, recover all their energy from the fields induced by the evanescent component. This changes when a third, optically denser medium is brought close by. Now, the induced fields can couple to propagating (far-field) modes in the third medium. This causes energy to 'leak' across the boundary. $\endgroup$ – tok3rat0r Jun 5 '15 at 20:40
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For optical fibers with cylinder geometry there are an orbital angular momentum (OAM) mode number $\ell\in\mathbb{Z}$, and a radial mode number $m\in\mathbb{N}$. The average radial position grows with $m$. A fixed OAM $\ell$ leads to a centrifugal term in the radial equation, which diminishes the number of radial modes. See e.g. the RP photonic encyclopedia.

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