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!
[Physics] The core diameter of the optical fiber is equal to the wavelength of light
fiber opticsgeometric-opticsoptics
Related Solutions
Yes, but.
The quality of the collimated beam may not be fantastic. And it depends on how the light is launched into the fiber. If the incident light has a narrow angular dispersion, and the fiber is short and/or of very good quality with few bends, then the light coming out will have a low angular dispersion. But if the incident light is converging at an angle close to the NA of the fiber, then your analysis should be ok.
The classic, definitive reference for this stuff is:
Snyder & Love, "Optical Waveguide Theory"
My edition is the Chapman and Hall one.
The ratio you seek naturally depends on the fibre's refractive index profile. Snyder and Love give the formula for this ratio for many, many different situations in the book. The two most useful are:
Weak Guidance Step Index Profile Approximation:
The weak guidance approximation to the step-refractive-index profile fibre (S&L, §14.6, p313 in the book I have linked), whose core power fraction is given by:
$$\eta = \frac{W^2}{V^2} + \left(\frac{U\,K_0(W)}{V\,K_1(W)}\right)^2\tag{1}$$
where:
$$U^2+W^2 = V^2\tag{2}$$
and the eigenvalue equation for this fibre is:
$$U\frac{J_1(U)}{J_0(U)}=W\frac{K_1(W)}{K_0(W)}\tag{3}$$
and, as you know:
$$V = k\,\rho\,\sqrt{n_{co}^2-n_{cl}^2}\tag{4}$$
where $\rho$ is core radius, $k$ the freespace wavenumber and $n_{co}, \, n_{cl}$ the core and cladding indices respectively. $J_n$ is the $n^{th}$ order Bessel function of the first kind (the nonsingular one) and $K_n$ the the $n^{th}$ modified Bessel function of the second kind.
What you must do is, given $V$, solve (3) subject to (2) (there will be more than one solution if $V>\omega_{0,1}\approx 2.405$ where $\omega_{0,1}\approx 2.405$ is the first zero $J_0(z)$.
This is all a bit unwieldy and you'll need to build something like a Mathematica notebook to calculate with. However, there is an approximation to this one that works well:
Gaussian Approximation to the Weak Guidance Step Index Profile Approximation
For many if not most practical fibres, the modes look very much like generalised Gaussian modes (in particular Hypergeometric Gaussian modes) whose spotsizes do not vary with propagation (i.e. the optical fibre yields the right amount of focussing to exactly cancel diffraction). One of the major techniques in S&L is the use of variational principles to find best fit Gaussian modes for different fibres. For the step refractive index fibre, the stationary-Lagrangian Gaussian spotsize for the lowest order fibre mode is defined by:
$$\sigma = \frac{\rho}{\sqrt{2\,\log V}}\tag{5}$$
so that the amplitude profile is:
$$E(r) = \exp\left(-\frac{r^2}{2\,\sigma^2}\right) = \exp\left(-\log(V)\,\frac{r^2}{\rho^2}\right)\tag{6}$$
and intensity profile is:
$$I(r) = \exp\left(-2\,\log(V)\,\frac{r^2}{\rho^2}\right)\tag{7}$$
where $\rho$ is the core radius. From these formulas, the fraction of power within a core of radius $\rho$ is:
$$\eta = 1 - \exp\left(-2\,\log(V)\right) = 1-V^{-2}\tag{8}$$
You need to be a little careful with radius and mode field definitions: there are many around. Most often, the quantity called the "mode field diameter" is the Petermann II diameter. For a Gaussian beam, this is the width of the beam wherein the intensity is greater than $1/e^2$ times its peak value, although the Petermann II concept is rather more general than this idea and it has to do with diffracted beamwidths. From (6), the intensity has dropped to $1/e^2$ times the peak value at a radius of $\sqrt{2}\,\sigma$ from the core, so the Petermann II mode field diameter is $2\,\sqrt{2}\,\sigma$. The intensity profile is $I(r) \propto \exp\left(-\frac{r^2}{\sigma^2}\right)$ so that the fraction of power in the core or radius $\rho$ is:
$$\eta = \frac{\int_0^\rho\,r\,\exp\left(-\frac{r^2}{\sigma^2}\right)\,{\rm d}\,r}{\int_0^\infty\,r\,\exp\left(-\frac{r^2}{\sigma^2}\right)\,{\rm d}\,r} = 1-\exp\left(-\frac{\rho^2}{\sigma^2}\right) = 1-\exp\left(-\frac{8\, \rho^2}{\mathscr{P}^2}\right)\tag{9}$$
where $\mathscr{P}$ is the Petermann II diameter or "mode field diameter". So your formula is correct if $w$ stands for the mode field radius NOT diameter. For English speakers, $w$, being the first letter of "width", bespeaks a diameter, so this is confusing. It has to do with the standard notation for the waist parameter of a Gaussian beam, which is the mode field diameter divided by 2. Be careful. the $1-V^{-2}$ formula for the power fraction is much less confusing.
Another formula, very near in its calculated values to (5) for $1.2<V<\infty$ is Marcuse's Petermann II Radius formula:
$$\frac{w}{\rho}=\frac{\mathscr{P}}{2\,\rho}=0.65+\frac{1.619}{V^{\frac{3}{2}}}+\frac{2.879}{V^6}\tag{10}$$
and sometimes this is refined further to:
$$\frac{w}{\rho}=\frac{\mathscr{P}}{2\,\rho}=0.65+\frac{1.619}{V^{\frac{3}{2}}}+\frac{2.879}{V^6}-\left(0.016+\frac{1.561}{V^7}\right)\tag{10}$$
Further Words of Warning
Although the above formulas work for many situations aside from the step index profile if one sticks to $V$ parameters instead of using the core radius directly. That is, the use of the above formulas gives an "effective" core radius for non step index profiles. Mostly, the formulas are extremely insensitive to the exact profiles, so your formulas are adequate. There is one situation in particular where all this can fail spectacularly and that is for dispersion shifted communications fibres. Here the refractive index profile is tailored to shift the zero dispersion point (i.e. wavelength where the derivative of group velocity with respect to wavelength) is in the middle of the 1550nm communications band, whereas it naturally tends to fall around 1300nm for silica fibres. A complicated refractive index profile with several concentric regions of different index is used to achieve the dispersion shift. In this case, you need to check the literature for more appropriate formulas.
Best 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.