# Tag Info

5

There are two parts to those lights: the reflector, which gathers the bulb's output and creates as focused a beam as possible the lens, which modifies that beam as desired. A focused beam makes a lousy headlight. You only see a small patch way in front. The extra ridges in the pictured lenses are vertical, which means they will spread the beam ...

1

The wavelength of microwaves is comparatively large, if you look at the holes on a microwave oven door.

0

I assume that the laser is emitting a wavelength that the photodiode is sensitive to. If the laser has too much power, you would burn out the semiconductor. Since a laser is a collimate light source, the power lose is minimal for any distance unless there is fog. What type of photodiode do you have? Some are like transistors with low gain and others are ...

15

The type of headlight lenses shown in the images, with the rows of fringes, look like Fresnel lenses. This is essentially a lens with a series of fringes that act as prisms, each at a slightly different angle but with the same focal length. That has the effect of reflecting the non-directional light from the bulb in a particular direction. I'd think the ...

2

From your question, I guess the double mirror configuration is just an example you thought of. I suppose your question actually is about if a photon can be trapped. Then basically yes. A device able to confine electromagnetic wave or light or photon is called cavity. You should understand a photon does not necessarily means a propagating plane wave. It can ...

0

The best complex dielectric mirrors, see http://en.wikipedia.org/wiki/Perfect_mirror may reflect up to 99.999 percent of the incident energy. The loss is about 1/100,000, so after 100,000 reflections, the total intensity decreases $e=2.718$ times or so. If the distance between the mirrors is 3 meters, the light travels 3 meters times 100,000 = 300,000 ...

0

Yes & No, You can however create an perfect mirror, which does not absorb any of the photons energy however its simply not possible or even feasible at this time to create such a device without energy being conserved in the photon, but it will however it will loose its energy due to Gravitational red shift after a long-time or Red shift due to moving in ...

1

"Now, is there an increase of gravitational mass corresponding to this increase of inertial mass?" Yes, both increase by the same amount. For any object inertial = gravitational. Even antimatter will fall down at the same speed. "If the original frequency and energy of a photon “launched” from a body M were equal in magnitude to the potential energy, then ...

1

The modulating frequency shift provides the central band frequency at $f_b$ From Doppler effects, we know that if the object vibrates away from the source, the frequency $f_d$ decreases (negative), and if it vibrates toward the source, $f_d$ increases (positive). Now as mentioned, with the modulating frequency shift from which the detected frequency is ...

1

The phase-shift occurs due to the complex nature of the reflection coefficient. The so-called critical angle is given by: $sin(\theta_M) = \frac{n_2}{n_1}$ As long as $\theta<\theta_M$ we have only partial reflection and a real valued reflection coefficient $R$. As soon as the critical angle is exceeded $(\theta>\theta_M)$, we have $\mid R \mid=1$ ...

1

They are equivalent. Saying that the path length is stationary, which is what the second definition says means that it does not change with respect to small changes in the path, which is what the first definition says. This is analogous to the definition of stationary points in ordinary calculus, where a (differentiable) function $f(x)$ has a stationary ...

4

They are equivalent. The formal study of this kind of problem is called "The Calculus of Variations", and it requires that you have some level of understanding of integration and of partial derivatives. You may imagine parameterizing the path taken in any way you want, say $$\vec{f}(t;\, \alpha,\beta,\delta,\dots)$$ where the function describes the ...

0

A simple explanation of Fermat's Principle is that you can approximate the path of the light by using the path that would take the least amount of time. For example if light is passing through only one medium, its path from A to B would be a straight line. However, if light is moving through two mediums, it might take a slightly longer path in one medium if ...

0

I disagree with the other answer. I am pretty sure optical fibers (and other dielectric waveguides) do not have a cutoff wavelength. This is, in theory and within the limits of transparency of the medium, source beam quality, alignment, divergence, etc. All wavelengths can propagate! They do however have a single-mode cutoff wavelength, which is the ...

1

In the configuration you described (Infinite/focal plane conjugation) the lens acts as an angular position to linear position mapping system. Each parallel ray bundle at a given angle $\alpha$ with the optical axis converge to a position $y$ away from the optical axis in the focal plane and: $y = \alpha.f$ where $f$ is the focal length of the lens. The ...

1

Rays from the scene do not converge to a single point. Only rays coming from a single point in the "infinitely distant" scenery converge to a single point in the focal plane. Rays from a different point in the scenery converge to a different point in the focal plane. And all the different rays from all the different points in the scenery converge to their ...

4

It is not the case that all the rays will focus at the same point. Rays with different directions will focus at different points. You are probably thinking about rays parallel the the optical axis all converging on axis, at a distance equal to the focal length from the lens.

2

To start with the double slit experiment gives interference even when the beam is composed by one photon at a time. The spot on the screen a photon/particle the statistical accumulation the interference seen as expected classically too. The joint comes because the photon as a quantum mechanical entity has a wavefunction that is the solutions of Maxwell's ...

2

It's simply the diameter of the fiber core. In a single-mode fiber, only the lowest-order mode fits physically into the fiber.

2

Any optical fibre - and any optical waveguide in general - has a cutoff wavelength (and therefore a cutoff frequency or a cutoff energy) because wether light is confined in the film region (guided modes) or escapes to the substrate (substrate-radiation modes) depends on the propagation constant, $\beta$, which is related to frequency trough the dispersion ...

4

The diffusion approximation is one solution to the radiative transfer equation. In general, the choice of applying this particular solution depends on the optical limit, as you say. For an optically thin medium, radiation will travel and may interact along the way. This is not characterized as a diffusive process, because the beam can interact with the ...

3

An optically thick medium is one for which the mean free path of a photon is low. This means that a photon won't be able to travel very far before it interacts with the matter than makes up the medium. The measure of optical thickness, optical depth, does depend on the volume of material in the medium. For example, for a material with a fixed density, ...

0

You can assume either electric or magnetic field for simplicity, they both are present in light as it's electromagnetic wave. Now, mode is a sustainable pattern in the fiber optic cable. Imagine waves on a string, only the integral multiples of half wavelengths effectively form a mode. Let there be a complex output pattern at the output of a string. It ...

2

It's not possible. There is a field of study called non-imaging optics dedicated to this kind of problem. See this Wikipedia page. FWIW, I don't agree that this question belongs somewhere else. I think it could live here, or in Engineering. Not sure where you would get better responses in this case.

0

A ball lens will introduce spherical aberration to your image. Spherical abberation enlarges the focus point to a small focus area. The sharp part of the focus point is refracted by the center of the ball lens. The outer rays will introduce more an more spherical aberration an blur your image. Find the optimal compromise between sharpness an image ...

0

I'm not sure what goes into a film emulsion. As to small particles, it's basically the black-body principle. No material is perfectly reflective, so the more reflections a light ray undergoes, the more attenuation occurs. A pile of tiny particles essentially acts as a "light trap." One of the better absorbing materials (up until the development of ...

0

No there isn't. As your drawing shows, even for monochromatic light, the exit angle is a function of both entrance angle and location (which side of each "tooth" the light enters). Your term "distortion" is not well-defined here. For one example, a perfectly flat plate placed in a converging cone of light will cause the final spot to be astigmatic. If ...

4

Your observation is linked to the "Optical window in biological tissue". Like you already suspected, the absorption of blue light in tissue is higher than the absorption for red light. Best read the related wikipedia article, where all relevant effects are nicely illustrated. http://en.wikipedia.org/wiki/Optical_window_in_biological_tissue

1

It depends on what you are really trying to do. There might be differing schemes, or it may not be possible at all. You can possibly use a fiber. You focus light into the fiber with a microscope objective. In order to make this work, you need a way to secure the fiber, and a way to move the fiber distances on the order of microns reliably in three ...

1

Answer: you can ignore the coating (assuming monotonic index of refraction); light does not exit if incident ray is beyond critical angle Reasoning: Snell's Law states $$n_1 \sin \theta_1 = n_2 \sin \theta_2,$$ where $n_1$ and $n_2$ are indexes of reflection within media $M_1$ and $M_2$, and $\theta_1$ and $\theta_2$ are the angle between the normal and ...

0

The laser may be lasing on multiple modes (multiple longitudinal modes, multiple transverse modes, or both). Each of the individual modes may have a very long coherence length, while the many modes taken together would have a much shorter coherence length. Suppose that one of the modes has, say, 40% of the power, and that four other modes have 15% of the ...

0

How about the book "Nonlinear Fiber Optics" from Govind Agraval? It starts on page 5 with a section "Fiber Losses" and there is a whole chapter about "Pulse propagation in fibers". Might be that you'll have to use numerical methods if facing a complicated model, though.

0

Use an ellipsometer. It's a common tool in semi-conductor processing and thin layer coating specialists for optics. It uses polarized light and measures the relative change in polarization angle with depth. From this you can get index, thickness of a layer stack.

1

I have recently published a paper on Zn3N2 nanocrystals (http://pubs.rsc.org/en/content/articlelanding/2014/tc/c4tc00403e#!divAbstract). The material appears to have a direct band gap around 1eV and makes nice nanophosphors so if you can work out how to p and n dope it I am sure you could make an LED.

3

$r=1.5r_s$ for the Schwarzschild solution corresponds to the unstable maximum of the effective potential for a photon, therefore you won't be able to see much in practice, since practically every photon on this orbit will either fall in the black hole or escape to infinity.

3

I think you have some problems with the sign convention. At first, ask yourself the question "why did I need the sign convention?" For the moment, forget the sign convention and let's derive the formula related to lenses from the scratch using only geometry. Let us consider: p= distance of the object from the optical centre q= distance of the object ...

8

The distance where light has a circular orbit is actually $1.5r_s$ not the event horizon. This distance is known as the photon sphere. In principle a shell observer hovering at this distance could indeed see their own back. The proper distance is indeed just $2\pi r$, however the object would look bigger than expected because the curvature of spacetime has ...

1

There are many different forms of the mirror and lens formulas, each varying slightly in appearance, and each using a different set of rules as to what distances, focal lengths, and radii are positive or negative! The most important part of any formula is what I call the "wheres". Mixing one formula with another set of "wheres" leads to total chaos... In ...

0

Something may be wrong with the statement of the problem. A diverging lens cannot form a real image (see this SE answer). If the problem meant to specify a virtual image, try this: turn the system around so that the virtual image is on the left. This presents a less confusing situation w.r.t. the sign convention. In any event, I'm not sure which ...

4

As described in the link you provided, Raman scattering is any scattering that changes the frequency/wavlength/energy of the light by transfer of energy to or from the matter that scatters it. If the matter absorbs energy it is called Stokes Raman scattering (sometimes shortened to just Stokes scattering). If the matter loses energy it is called anti-Stokes ...

0

If you had a neutral plasma (which can be the free charges in a metal) and you pulled the negative and positive charges apart and let them go, they would oscillate due to the electrostatic potential. This is an excitation known as a plasma oscillation. A Plasmon is the quasiparticle associated with plasma oscillations (analogous to phonons being the ...

1

Ray lines can always be drawn using Snell's law. But If you want to find the image only by basic properties(lines parallel to principle axis and passing through the optical center) of lenses then find the image for the first convex lens. Now if this image happens to be at the left of the second concave lens, you can take it again as an object for the ...

0

So, you use mirrors to focus sunlight onto a small area of water to increase the evaporation rate, and making the air at that point more humid. But the mirrors are deflecting the sunlight away from other areas of water. This lowers the temperature there, reduces the evaporation rate, and decreases the humidity of that air. So you have low and high ...

2

In the absence of noise they can both work the same (assuming you know the exact amount of defocusing, and you over-sample the Airy disk) In the presence of realistic noise, you are better off focusing the object due to the details of the noise. For intensity images, you are (likely) dealing with Riciean distributed data, and you are better off using a ...

3

If the Airy disk is smaller than a pixel (rather common), then you want to defocus. Star trackers on satellites do this in order to get sub-pixel pointing accuracy. If the Airy disk is much larger than a pixel, then you probably don't want to defocus. In the latter case the situation is complicated by aberrations and the problem of modeling the shape of ...

2

There's no difference between plasmon and plasmon polariton. Both of them indicate the resonant excitations involving electromagnetic wave and collective electronic motions simultaneously. "surface" stresses that the excitation in many cases occurs at the interface of a metal and a dielectric. However, there exist bulk plasmons as well. So "surface ...

0

I don't know why they say a thermodynamics argument is needed. Here is a tutorial on basic optical design that shows it from Snell's law and simple geometry.

2

While rob is correct about the quantum mechanical picture I think that this case is at least as easy to understand in the classical description. Classically circular polarization can be described in terms of a time-varying linear polarization, so lets just look at two points on a wave. I'm going to chose a beam in the $+z$ direction at examine two points ...

2

From a quantum-mechanical perspective, circularly-polarized light is made of photons with their spins parallel to their momentum. The mirror reverses the photons' momentum but does not affect their spins, so the dot product $\sigma\cdot p$ changes sign. Both the quantum and classical approaches are examined in Beth's 1936 measurement of the angular ...

1

Yes, this is possible. The device that makes this possible is called a polarizing beam splitter, which will transmit or reflect light according to its polarization. Thus, it will split diagonal or circular light into its horizontal and vertical components, and when used in reverse it will undo the process (it has to). Note, however, that you will in general ...

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