What happens if the core diameter of the optical fiber is equal to the wavelength of light. For example: There is an optical fiber (core diameter = 1.5 μm) and a light source (λ = 1.5 μm). What happens to this light if it is transmitted in the optical fiber. Help me. Thank you very much!
Answer
Probably nothing too unusual. The modal field will tend to be a little larger than the core. If the core cladding index difference is small, then the modal field spreads out a long way into the cladding.
One cannot know what will happen without knowing the index difference between the core and cladding. If the core-cladding index difference is very large, the field tends to be very strongly confined to the core. But Chapter 12 of "Optical Waveguide Theory" by Alan Snyder and John Love gives exact expressions for the field vectors and eigenvalue equations in a step index profile round fiber. There are always solutions to the eigenvalue equation and you will always get a propagating field in a dielectric waveguide; it may simply be that even the fundamental mode reaches far into the cladding.
No matter how big the core cladding index difference may be, there is a limit to how tightly the field can be confined. If we replace the core-cladding interface with a perfect conductor to force guiding (or as a common approximation to a fiber with a very big core cladding index difference), then 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 microwaves from getting through the little holes in the mesh screen on your microwave door, for example.
If we have a pure glass fiber and we draw it down to less than this diameter, we have a photonic wire waveguide whose fields penetrate the air around the fiber significantly. Such a device makes a sensitive gas absorption spectroscopy transducer.
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