# Tag Info

22

Yes easily. Fiber-fed 5-10kW Nd-Yag lasers are commonly used for cutting metal in machine shops. Fibers are so transparent, especially when designed for a single wavelength laser, that the power loss and so heating in the fiber is very small. It's generally less than an optically fed laser where dirt accumulates on the lenses. Many systems have a thin ...

7

The fibres are too small (10um) for light to behave like a ray, you need the wave equation. Ray theory works when: (size of all dimensions in system)*(relative error tolerance) >> (wavelength of light). For macroscopic objects this is usually true. In a double slit experiment/demonstration, where ray theory fails, we combine a smallish slit size with a ...

7

My bet is that your fiber is very short (something like one meter or so) and that the fluctuations you see on the output mode are due to cladding modes, i.e. a part of the injected light propagating into the cladding of the fiber instead of the core. The resulting fluctuations are due to external perturbations of the fiber (thermal fluctuations or you ...

6

What you propose is a version of concentrated PV. Concentrated PV has been in development for some time. The big issue with PV, is lifetime cost per unit energy. There are no other constraints really - there are plenty of alternative commercial panels that material shortages aren't an issue; we know we can manage their exogenously variable power ...

6

I don't think "the general solution to this equation is a Gaussian beam" - this is a linear equation, so a superposition of solutions is also a solution, and an arbitrary superposition of Gaussian beams is not always a Gaussian beam.

6

The translucent sheets and optical fibers are being lit from below. Light enters one end with a relative small angle incidence. Once inside the sheet/fiber the light experiences total internal reflections multiple times because it hits the sides of the sheet/fiber with a large angle incidence. It's not until the light gets to the far end of the sheet/...

5

This may seem an simple question, but a truly correct answer is actually a bit more complicated. This is because there are different speeds and the question is which one is relevant. There is phase velocity (this is the "ordinary" speed of light, and the one that is described with the refractive index). BUT this is not the speed at which information ...

5

This sounds as though the aberration in the laser's output could be fluctuating owing to "mode hopping" (where several of the laser's cavity modes are active and playing a time varying role) so that, even at a constant output power, the aberration of the output beam varies with time. Wavefront aberration is roughly the Fourier-dual of Strehl ratio. This ...

5

As usual, whenever you have a complex-valued amplitude $\tilde{\mathbf{E}}(\mathbf r)$, it's a stand-in for a physical field that's obtained as its real part, i.e. $$\mathbf E(\mathbf r,t) = \mathrm{Re}\left( \tilde{\mathbf{E}}(\mathbf r)e^{-i\omega t}\right).$$ The presence of nontrivial imaginary parts of $\tilde{\mathbf{E}}(\mathbf r)$ signifies ...

4

Ok, first of all, if you are dealing with fiber optics things are little bit more difficult than just shine with bulb on fiber frontface. Talking about standard single mode telecommunication fiber, core diameter (e.g. this part of fiber, where is the light guided) is around 9 micrometers. If you have multimode, core will be like 50 micrometers, so this is ...

4

A $1\text{fs}$ pulse is extremely short, and unlikely to be sent down an optical fiber for any distance, let alone 50 kilometers, and often femotosecond pulses are produced at high enough powers that it can result in nonlinear effects when in contact with matter. Ignoring these concerns for the moment and assuming linear behavior, the dispersion of the ...

4

The general solution to the paraxial wave equation, $$\left[ \frac{\partial^2 }{\partial^2 x} + \frac{\partial^2 }{\partial^2 y} - 2ik\frac{\partial}{\partial z} \right]\Psi(x,y,z)=0,$$ is not a gaussian beam. To see this, you should note that this equation is now of parabolic type, which essentially means that you can take any plane at $z=z_0=\text{const}$...

4

In standard single-mode fiber, the polarization will tend to drift as the signal propagates (due to slight and varying birefringence of the glass, possibly stress-induced, coupling one polarization to the other). For short lenths (1 m or so) polarization is typically maintained fairly well. For long distances (1 km or more, I'd guess, but I don't have a ...

4

The problem is that if you have air as clad, then nothing may touch your fiber. This means that you have to suspend the fiber in air - at any point where you touch it, you would change the reflective properties. This would result to dispersion and losses in the signal. By ensuring that the reflection takes place on an inner boundary, you make the fiber ...

3

First, be careful about mixing rays and modes. Your picture shows a ray-optics view of multimode fiber, but your text talks about modes. Some of your questions are best answered in terms of rays and some are best answered in terms of modes. Generally, fiber is characterized by a numerical aperture, which you'll find listed on its datasheet. The numerical ...

3

Two extreme cases: For a very thick multi-mode fiber, the angles are somewhat preserved I think. The opposite extreme is a thin single-mode fiber, in which the angles are most definitely not preserved. The light enters the fiber or it doesn't, there is no other spatial or angular information. So the answer to your question is: it depends on the fiber!

3

Light travels through an optical fiber in a set of discreet "modes". Each mode correspond to a specific angle at which the light field propagates, bouncing up and down in the fiber. Depending on the wavelength and the type of fiber (diameter, refractive index, etc.) there may exist one or several propagation modes. Then the method of coupling light into the ...

3

Well, ya gotta be careful. As you noted, there's an acceptance cone angle. Now, consider an idealized case where the fiber is perfectly straight and perfectly cylindrical. Then, even for skew rays (per wikipedia, "ray that does not propagate in a plane that contains both the object point and the optical axis. Such rays do not cross the optical axis ...

3

"White" light is light of multiple frequencies, usually a broad band of different frequencies ranging from ~($390nm$ to $700nm$) though white on your computer monitor consists of only three different frequencies (red green and blue). It is possible to send multiple frequencies down an optical fiber. However the fiber will act like a prism and diffract the ...

3

In fibre optics ,a optical phenomenon known as total internal reflection is used to transmit light rays. In case of the simplest form of optical fiber, light entering one end of the fiber strikes the boundary of the fiber and is reflected inward. The light travels through the fiber in a succession of zigzag reflections until it exits from the other end of ...

3

How much light couples into the fiber depends on the NA (numerical aperture) of the fiber, the diameter of the beam of light entering the collimating lens, and the focal length of the lens. Typical fibers accept light only from a limited angle about the axis of the fiber. This is referred to as the NA of the fiber. If the NA is large (e.g. 0.7) the fiber ...

3

The manufacturer generally specifies a cutoff wavelength for single mode optical fibre designed for specific wavelength. For example, a single mode fibre desgined for 700nm would have a cutoff wavelength few tens of nanometers below 700, lets say 680nm. This means that for wavelengths longer than 680nm, the Normalized Frequency or the V-number of the fibre ...

3

Since your question is purely theoretical I assume you are referring to the intrinsic noise limits. This limit is called shot noise limit. Generally photon detection is complicated by the fact that photons arrive (are generated) in a random fashion. This is because the photon emission/absorption events occur independently and there is no communication ...

3

It is certainly possible. It depends, as others have said, on the detector. But it also depends on the detection electronics, and the techniques used to do the measurement. Common sources of noise are shot noise, dark current noise, statistical fluctuations in the detection mechanism, and thermal noise in the detection electronics. Which of these are ...

3

It depends. The first challenge is damage at the surface. Is your fiber single-mode or multi-mode? What is the mode-field diameter? If it is single-mode then the mode diameter will probably ~10um, or so, and the fluence will be very large. If you are using multi-mode fiber the mode could be quite large, or not, depending on the fibre. The second ...

3

In a mirror the thin layer of metal at the back provides the reflection. If this metal is exposed to air it will 'tarnish' - ie become dull due to formation of oxides on the surface. It can also be scratched quite easily. Both these reduce the quality of reflection. Glass provides a layer of protection for the metal. Glass is transparent and also hard,...

3

No, a mode does not have a specific frequency, but a specific field distribution. For each mode there is a minimum frequency that can be propagated (cut-off frequency) but above the cut-off frequency any frequency can be transmitted. In a single-mode fiber, the diameter of the core is sufficiently small to have just one mode above cut-off in the range of ...

3

I have in my understanding that "Mode" is just a path of light propagation in a fiber. That's a very rough way of understanding it. A better way to describe it is that a mode is a way for the EM field to satisfy the boundary conditions imposed by the physical structure of the fiber, such that the field patterns at different positions along the fiber's ...

3

Emilio Pisanty answers this most succinctly with: "Who says that modes can't share eigenvalues?" So yes, degenerate modes share the same eigenvalue. This is the fundamental definition of "degeneracy". Degenerate modes share eigenvalues of the linear operator $\mathscr{L} = \nabla^2 + k^2 n^2$, which is the operator whose eigenvalue problem we solve for ...

3

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, ...

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