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I would love to know about propagation modes in a subwavelength-diameter optical fiber. For example: There is an optical fiber (core diameter = 0.7 μm), the refractive index of the core (n1 = 2.1), the cladding is air (refractive index n2 = 1). What happens if light (λ = 1.5 μm) is transmitted in this fiber ? How many propagation modes ? Is optical energy lost between the output and the input of this fiber ? Thank you very much!

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The answer is given as part of my answer to your other similar question here.

In a dielectric waveguide, nothing special happens: the field spreads out so that it is significantly different from zero over a diameter that is much greater than the core diameter. There is still no loss for a straight fiber even if the core is arbitrarily small. However, the mode becomes more and more weakly guided: theoretically lossless for a straight fiber but more and more subject to loss, through coupling into the radiation field, when the fiber is bent. For a very small core, even the mildest bend curvatures lead to significant loss.

Chapter 23 of Snyder and Love, "Optical Waveguide Theory" describes this phenomenon in detail. The propagation constant for a bend step index fiber has a real (attenuating) part with a term given by (See Eq. 23-12 in Snyder and Love):

$$\exp\left(-\frac{4\,R_c\,\Delta\,W^3}{3\,\rho\,V^2}\right)$$

where $\rho$ is the core radius, $R_c$ the bend radius, $\Delta$ and $V$ are the waveguide parameters and $W$ the cladding eigenvalue for the fundamental mode. This quantity increases swiftly with decreasing bend radius, and the ratio $\Delta\,W^3/V^2$ measures the guiding strength: for small cores, this is large which means even mild bends lead to high loss. For large cores and large core-cladding index differences, this ratio is small, and it takes a very tight bend to beget significant loss.

In a metal clad waveguide, as in my other answer, the field vectors for the fundamental mode vary with radial position $r$ and axial position $z$ within the fiber like $J_0(k_\perp\,r)\,\exp\left(i\,\sqrt{k^2\,n^2-k_\perp^2}\,z\right)$ and we must have $k_\perp\,r_0 = \omega_{0\,1}\approx 2.405$, where $\omega_{0\,1}$ is the first zero of the Bessel function and $r_0$ the core radius, to fulfill the field continuity boundary conditions. This sets a minimum core radius $r_0$ that one can have for a propagating field; this minimum radius is $r_{min}\approx 2.405\,\lambda/(2\,\pi\,n)$. For $1.5{\rm \mu m}$ wavelength (in freespace) light and a core refractive index of 1.5, the minimum possible mode field diameter is about 770nm. If the core is any smaller, then $\sqrt{k^2\,n^2-k_\perp^2}$ is imaginary, all modes of the metal clad waveguide are cut off and propagation becomes evanescent, i.e. non power transporting. This is the mechanism that stops EM waves getting through subwavelength holes in conductive materials of any significant thickness: this is how the mesh on a microwave oven door stops leakage even though we can see through the mesh.

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