# Tag Info

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Perhaps I'm not clear on what you mean, but my answer would be "no". You make no statement about the focal length of the mirror, which means this should apply to one with any focal length; if we choose an infinite focal length, i.e. a flat mirror, then if the rod is moving away from the mirror with velocity $V$, from the rod's viewpoint its reflection will ...

1

The wave is not an object travelling through space and time. At any fixed time the field has values at each location. If you want to find the phase difference between two times and two places you have a fine formula. But if you are taking an imaginary snap shot and looking how the field is different at different places then the $\Delta t$ will be zero. ...

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It's quiet simple: The photon is absorbed by an electron, the direction of the electron's spin flips and sends a neuron to the brain. A lightbulb works the other way around: Electrons are forced to flip their spin and therefore they are emitting photons.

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This paradox is an artifact of describing the mirror classically. Quantum mechanics is itself free from paradoxes, and it is easy to see that this paradox vanishes the moment you describe the mirror as a quantum mechanical object. A simple way to see what goes wrong is to apply the uncertainty relation. Suppose that we have a freely floating mirror and we ...

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I am not totally sure of this answer, which is why I asked the question. However, I think the answer is that only relatively short wavelengths can pass through an annular aperture. Specifically, I think that if the outer radius of the annulus is R and the width is W, where W << R, the maximum wavelength that passes through is approximately 2R ...

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I'd like to add another take on Cape Code's answer. Holography works because, given reasonable physical assumptions, solutions to the Helmholtz wave equation are uniquely defined by the values of the solutions on one plane. So if we can light a phase / amplitude mask encoding a particular wave equation solution on a plane with a plane wave from a laser, the ...

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I assume you ask if there is also reflection when refraction occurs. The answer is yes. It happen when light (or any kind of wave) passes from a medium 1 to another medium 2, those medium being characterized with different index (i.e. wave velocity). At the interface between the two medium, in the case of light wave, the electric field has to obey certain ...

1

Everything is about electromagnetic waves. At the border of medium one of the components of electric field have to be continuus and other component of electric displacement field. Such calculations leads to Fresnel equations

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The underlying principle is to use interferometry and the Doppler effect to remotely measure the velocity of a reflecting surface. When a moving object is illuminated with coherent light it reflects it with a wavelength shift proportional to its velocity. This is the well-known Doppler effect. The frequency shift relates to the source's velocity as ...

2

Because of diffraction. When the photographic plate is exposed, it blackens and change its refractive index in a spatially varying manner. When illuminated again by the reference beam it can be considered as an amplitude transmittance. In 2D you would define it as the complex function: $$t(x,y)= T(x,y)e^{i\theta(x,y)}$$ Let's say your reference beam can ...

2

Use a sodium lamp positioned above the prism and slightly to the side to create a reflection at the interface. Look for interference patterns at the prism/prism plate interface. If the spacing between dark fringes is small (and you have many fringes), you have a problem. If you have very few fringes and the spacing is large, the interface is close to being ...

0

I once worked with a laser technician who was a specialist with the high voltage supplies and circuitry for pockels cells, he told me klystrons were used to produce the very short high voltage pulses to drive the cells. Might not be easy to get. Maybe a rotating mirror to sweep a beam quickly past a thin slot?

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Imagine a typical path. It starts at event $A$=$(ct_A,x_A,y_A)$ and ends at event $B$=$(ct_B,x_B,y_B).$ (Note I'm using $w=ct$ as the fourth dimension so we can measure everything in meters.) In the frame of the fluid this typical path has length $l_0.$ In both frames the separation is null. We can actually choose the origin of the fluid frame and your ...

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A sum-frequency system with a "hot" mirror could act something like an optical switch: Unlike a switch, the output frequency will be different from either of the inputs. Edit: For an example of sum-frequency generation crystals see: Thorlabs Introduction To Periodically Poled Lithium Niobate (PPLN) (PDF) Thorlabs also sells hot mirrors.

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Picture yourself looking into a large mirror on the wall. Now picture the mirror is made up of smaller, tiled mirrors. You will still see your reflection. If you begin to remove the tiles, so that there are only a few left, you can still use them to reconstruct the image of your face that was given by the original mirror. This is what is happening with ...

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I think the TiOPc on printer drums may be TiOPc nanoparticles embedded in some kind of organic binder, rather than solid TiOPc. See this: http://patents.justia.com/patent/20140054510 for example for the challenges of dissolving TiOPc in anything. The binder on the other hand should be easy to dissolve; try acetone first, if that doesn't work, perhaps ...

1

Greatly rewritten based on feedback in comments In order to understand this issue, it is worth considering what a telescope (or any optical / radio imaging system) really does. Taking a simple parabolic mirror, the shape is chosen such that the total path length for all rays "from infinity" to the focal point is the same. By making the path lengths the ...

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x-ray diffraction is not caused by atoms absorbing radiation. x-ray diffraction and diffraction by gratings do have an underlying mechanism in common. In both cases, when two incoming rays of waves (x-rays or light waves, in the case of optical diffraction grating) both rays bounce of the crystal or grating, they have travelled a different distance, say ...

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A crystalline solid is more than a bunch of atoms arranged nicely, because the atoms are "feeling" each other. I don't see scattering at crystals as the interference of absorption/emission of each atom separately (and I don't think that's the usual approach either), but rather as reflection at planes determined by the crystal structure as a whole. So, Yes, ...

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Although there are already some excellent answers, I believe they are a little complex. Please allow me to offer a simplistic answer. Let me start with the analogy of sound waves and the ear. The sound enters the ear and causes certain cilia to vibrate in response to the frequency and amplitude of the sound wave. Similarly a photon (as a wave), enters the ...

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I assume you're talking about an image that could be presented on a printed page, or on a standard LCD display, where the light from any one point is emitted in all directions. The answer is (unfortunately) no. When your eye is properly focused on an image, all the light which ends up at a specific point on your retina comes from a specific point on the ...

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As user1631 explains, deconvolution will yield the required image, but this will contain negative intensities. However, if you just raise the background gray value then it should be possible to deal with this issue. So, if we imagine a picture of the night sky and me watching it with my glasses removed. If the stars are deconvolved with the correct point ...

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Photons are energy. When a photon hits your retina, that energy is absorbed and converted to electrical energy in your optic nerve.

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Light from all over the place hits your eyeball fairly randomly. The lens forces light from a specific angle to hit a specific part of the retina. This HowStuffWorks article shows how the mechanics of that work. The only major differences between camera lenses and eyeball lenses is that we can dynamically alter the shape of the lens to focus on different ...

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Photons can be created and destroyed freely, since they don't have charge or mass. Turn on a light, and you create many photons. Any body (made of atoms) not at absolute zero temperature will spontaneously emit photons. They are consumed just as easily. Most any bit of bulk matter will absorb a photon in the electrons on the surface, transforming the energy ...

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Imagine a spring-loaded trap with a hole that's sized such that only a particular size of object can enter the hole and trigger the trap. The molecules involved in vision are like that trap, with a bond having an electron energy gap tuned to the visible frequencies of light, encapsulated in a specialized protein that transforms the absorbed energy into a ...

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Shortly, the energy of the photon goes over to the electron. But energy is a vague concept. In material sense, could the photon, or better, the electron's electric field and the electron's magnetic field be quantized? I developed a model with two different quanta. Photons, electrons, positron's, protons, neutrons, ... are made from this quanta. Photons are ...

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From the wiki article on color vision as an illustration of how photons are absorbed: Perception of color begins with specialized retinal cells containing pigments with different spectral sensitivities, known as cone cells. In humans, there are three types of cones sensitive to three different spectra, resulting in trichromatic color vision. Each ...

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Keep in mind that your eyes perceive a fixed field of view. That is, you see a certain angle in front of you. They do not see a fixed distance across your line of sight. This means that the farther away for you you look, the more distance there is from one side of your FOV to the other. When you look at parallel railroad tracks that maintain a certain ...

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The distance between the two is assumed to be constant. However, when you look at an object that is far away, it "seems smaller". What that means mathematically is that the angle from one end to the other, as seen at your eye, is smaller. Now you can't tell the difference between something that is "small and close", and something that is "big and far". So ...

3

The diagram you want to use looks something like this: Depending on how much attenuation there is in the membrane, you need to consider the potential of multiple reflections (or not). I actually analyzed this problem in some depth - considering not only the intensity of reflections on the different surfaces, but also multiple reflections and even the ...

1

Typically the order of magnitude of angular spread of a fringe in a pattern can be given by lambda/Width. If lambda is much smaller than width the diffraction pattern will be difficult to see. In the other case of lambda much greater than the width the entire screen would be covered by just the first fringe.

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There are several options to increase the signal size: 1) increase the length of the path (look at the slab from side to side instead of from the front - you may have seen a "green edge" on a glass coffee table - that's because you see light coming through a much thicker slab of glass) 2) Use multiple sheets of glass on top of each other. Be careful about ...

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If the slit is much smaller than the wavelength, there will not be significant path-length difference between beams passing through different points inside the slit. If the slit is much larger than the wavelength, most of the beam passing through the slit will not be affected by the edges of the slit. I also guess that Born had in mind Fraunhofer diffraction ...

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Lenses can exhibit chromatic aberration due to a color (wavelength) dependent index of refraction in the lens material. Chromatic aberration can cause the three colors from the red-green-blue (RGB) display to separate when viewed off the lens axis: Left: on axis. Right: off axis. In less pronounced cases, this might lead to a ghosting of images. ...

2

To add to Rob Jeffries's answer: the absorption data for glass are separate from the refractive index and are measured by measuring the attenuation of light through a known thickness of glass, after taking account for the reflected amounts as described in Rob's answer. Theoretically, the refractive index and the absoption data are united in a complex ...

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Possibly some semantic confusion here. Glass, with a simple refractive index, does not "absorb" light, it is transparent. Therefore the amount of light that emerges on the other side, for a given angle of incidence, is independent of the thickness of the glass. Instead, some of the incident light is reflected from the first boundary as the light enters the ...

2

I am hypothesizing based on the information you gave in the question and clarifying comments. The fact that the distance between the "shadow" (or ghost) and the number increases with distance tells me that the shadow is produced by your eye+glasses, rather than the screen itself. Most likely, you are seeing reflections from the front/back surface of your ...

1

Any formula in physics comes with a set of definitions of what each variable in the equation represents, and how to interpret positive or negative values. This is particularly rue in the case of lens and mirror formulae. In each case, a different form of the equation, with a different set of definitions, will give the same correct result. In this case, ...

1

You can make beams of light that have orbital as well spin angular momentum and they can go through an annular aperture. So circular versus linear isn't enough. Linear has the phase advance orthogonal to the advancement with the polarization at a fixed angle. Circular has the polarization rotate but with an annular filter you can give the wavefront a twist ...

1

The origin of this equation is reasonably well explained in Abramowicz (1991). If you take a relativistically expanding enevelope and only consider Thomson scattering, then as the electron scattering cross-section in the co-moving frame $\sigma_T$ is independent of frequency, then the mean free path of a photon in the co-moving frame is independent of the ...

1

Laser Mirrors are used for beam-steering in demanding laser applications. Laser Mirrors are Optical Mirrors that have been designed for specific laser types or wavelengths.Optics’ Laser Mirrors feature dielectric coatings that have been optimized for high reflectance at specific laser wavelengths. Edmund Optics’ dielectric coatings typically feature greater ...

1

I'm going to focus on the information contained in the light field itself. This excludes from the discussion many if not all "superresolution" techniques, which directly or indirectly make use of information further to that in the imaging light field[footnote 1]. It is true that you can do a deconvolution to get somewhat below the traditional diffraction ...

1

If I can expand a little bit on Sofia's answer the polarization of the medium opposes time variations in the electric field thus slowing down the phase velocity of the wave. This can be seen from Ampere's circuit law (the 4th Maxwell equation) which is central as you stated in arriving at the wave equation describing light. It can be written in vacuum as ...

2

The main effect here is lack of field flatness, or, cited in more geometric terms, the deviation between your imaging system's focal surface and the surface which you're imaging. An imaging system generally images a plane onto the surface of an ellipsoid (approximately). i.e. light from a point surface on a plane will be converge to its tightest focus on ...

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Maybe this is a bit too general an answer but I hope this helps someone in the future. Here is the lenses makers equation write explicitly with the sign conversion terms,  \frac{1}{f} = \left(n-1\right)\left( \frac{d\left(n-1\right)}{\beta_1\beta_2n\bigl|R_1\bigr|\bigl|R_2\bigr|} + \frac{\beta_1}{\bigl|R_1\bigr|} - \frac{\beta_2}{\bigl|R_2\bigr|} \right) ...

0

There are different sign convention in optics. Once you accept one sign convention you should make sure that the answer is in the same convention. Application of the signs twice guarantee the same. How this magic happening? At the time of derivation you applaying the signs for U,V,F thar guarantee that U,V,F are in our convention. At the time of solving ...

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There is a fallacy in your question which might be the source of your confusion. It is not all light that gets focused to a point, it is only rays parallel to the lens axis! For the case of two (point source) stars, if the rays of one star are parallel to the lens axis, the rays of the other star will not be (not co-linear), therefore its image will appear ...

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You could try using contrast. The contrast in the image is highest when the image is in best focus. You can analyze the contrast information and move the object, lens, or sensor into a position that gives the maximum contrast value. The downside to this is that you will need to typically adjust your opto-mechanics through the full range of motion to find ...

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In many cases people do seem to say "Fourier frequency" when they mean "frequency". However, when dealing with data defined only on discrete time points the phrase "Fourier frequency" has important meaning. Consider a sequence of $N$ values $\{ x_n \}$ where $n \in \{1, 2, \ldots N \}$. This situation comes up all the time if we have a physical signal ...

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