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I just learned from a previous question that light encounters bend loss moving around curves in optical cable. Why is this and how can I calculate it?

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Refer to Chapter 23 of Snyder and Love, "Optical Waveguide Theory", Chapman and Hall, 1983.

The model for bend loss is highly complicated. One can do a full coupling calculation between the propagating modes and the radiation field. This calculation is thoroughly loathesome; I show how this is done in my paper:

J. D. Love, R. W. C. Vance and A.Joblin, "Asymmetric, adiabatic multipronged planar splitters", Optical and Quantum Electronics 28(4):353-369

Unfortunately, this one has no pdf.

The Snyder and Love model works as follows. The bent waveguide is replaced by a straight one, and the deviation from straightness is replaced by a current density corresponding to a first order perturbation to Maxwell's equations. Then, the perturbing current density is thought of as an antenna, and the radiation loss from it is calculated.

For a step-index profile optical fiber, the amplitude attenuation co-efficient is:

$$\gamma = \sqrt{\frac{\pi\,\rho}{2\,R_b}}\,\frac{n_{co}^2-n_{cl}^2}{n_e^2-n_{cl}^2}\,k\,\sqrt{n_{co}^2-n_e^2}\,\exp\left(-\frac{8}{3}\,k\,n_{co}^2\,\sqrt{n_{co}^2-n_{cl}^2}\,R_b\right)$$

where $n_{co},\,n_{cl}$ are the core and cladding refractive indices, $n_e$ is the mode's effective index, $k$ is the freespace wavenumber, $\rho$ is the fiber core radius and $R_b$ the bend radius.

The effective index for most practical fibers is of the order of $0.9\,n_{co} + 0.1\,n_{cl}$. The above formula is a highly sensitive function of the bend radius and core index and moreover the effective index I have just given is only an estimate, so measurement is also recommended to check calculations. You can see these calculations worked through in my paper:

R. W. C. Vance, "Planar ring resonator realisation of symmetric 3×3 fibre coupler", Electronics Letters, August 1994

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