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At the end of your wire (any direction) we have diffraction but I think this question can;t be solve by diffraction . Just experiment it : If after glowing your light we have complete shadow in the back of the wire , we haven't diffraction but If not , we have that .


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An aspect I learned, which is different from the other answers given here, centers around the fact that an energized lasing medium will react to photons which go through it by producing more photons along the same path, but will also spontaneously release photons traveling on random paths. Any energy the lasing medium spends doing either of those things ...


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The author confused things in section 3.1. In macroscopic EM theory of radiation (radiometry), technically irradiance at a given point of a plane is defined as time average of the normal component of the Poynting vector (normal to the plane): $$ I = \overline{\mathbf S \cdot \mathbf n} $$ As you have shown, this is function of $|\mathbf E|^2$ only in some ...


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The usual use of an image like the one you posted is to identify nonuniform regions of the material. The amount of variation in polarization at any location on the protractor is related to the amount of strain, or nonuniformity, in the material itself. As explained in this post , there are a couple underlying optical principles which cause the color ...


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In a paraxial thin-lens approximation, two otherwise identical beams from two different (but paraxial) directions produce the same image (but displaced inside the image plane). You can see this by tilting your setup such that first one and then the other source is on-axis: For a thin lense, this process only introduces changes that scale with $1 / \cos ...


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Imagine not the direction of the column but the direction of the front row. Suppose the front row of soldiers were carrying a horizontal bar, the one on the left hitting the swamp would slow down while the one on the right was still moving quickly so the bar (=wavefront) would change direction It's a slightly bad analogy. A much better one is: Imagine you ...


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General case is: $$L\equiv\frac {d^2 \Phi}{dA\,d\omega\,\cos\theta}$$ You must use this (in actual fact) difficult formula when: $\omega = f(A)$ or $A = f(\omega)$ or $\omega = f(t)$ and $A = f(t)$ (it can be temperature for example) and so on... Particular case is: $$L\equiv\frac {\partial^2 \Phi}{\partial A\,\partial\omega\,\cos\theta}$$ You can use ...


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The answer is yes. It has already been demonstrated that a lightfield display can be configured to "counter-blur" an image such that a person with a given visual refractive error (including higher-order aberrations) will see it correctly. (And no one else will.) http://graphics.berkeley.edu/papers/Huang-EFD-2014-08/


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This is one of the rare cases where the full and the partial derivative are (at least typically) the same. There is a difference if the way you tilt your projection surface influences the solid angle into which you radiate. Whoever uses the full derivative must implicitly assume that such a curious dependency cannot happen (what does the lamp care how you ...


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Project Excalibur The idea of a nuclear pumped X-ray laser was one which was investigated in detail in the Reagan "Star Wars" program of the 1980s, backed by one Edward Teller. Tests were carried out by surrounding the nuke with bundles of rods to create a one-pass laser. Apparently it was nowhere near efficient enough to be used in a military context. ...


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What exactly do you mean by "measure of absorption"? The beer-lambert law is parameterised by the absorption coefficient $\alpha$ with units $\textrm{cm}^{-1}$, $$ I(x) = I_0 \exp\left( -\alpha x \right) $$ However, you can't work out the attenuation of the light beam with the information you have supplied because you don't have a value for $I(d)$, the ...


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Yes, the refraction index will be changed, but the absorption will differ even more. Look at similar situation. (P.S. I'd like to add this as comment but I cant yet, cause of reputation)


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It doesn't quite work the way you envision it (if the refraction angle is such that you can add light, it will escape the same way), but there are optical resonators that do essentially what you want: Light incident on a mirror gets added to a light field trapped between two or more mirrors. In such setups, not quite enough light usually builds up to cause ...


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If the bottle and liquid are made of dielectric material, then the interfaces between different mediums reflect light, they don't absorb it (i.e. dissipate it as heat in the glass). This is probably a good approximation for your bottle. As a first approximation, once you have worked out your incidence angles with Snell's laww, you need to use the Fresnel ...


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I don't know if it exists, but there's no reason it couldn't. A simple diverging lens system could be used to make a virtual image of a close-in display appear at 20 feet. The more difficult problem would be making different portions of the display have their virtual images at different distances (presumably one that corresponded with the intended distance ...


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Here is a popular account of how it has already been done Simply an array of LEDs showing the scene on the other side of the car. However, the use of LEDs or any projection technology is never going to be good enough to match sunlight ie 1kW/sq m at max. Night time is different


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Thinking about it, maybe what selects the special direction is that the creases in the water that have wavefronts perpendicular to the line of sight cover each other up while the ones with wavefronts parallel to the line of sight don't. This may mean that as diffraction off small features causes the large features in the angle distribution (as it's a ...


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This picture shows what happens to a ray passing through the center of the lens: The incoming ray hits the air-glass surface of the lens at an angle $\theta$ to the normal and is refracted. It then passes through the lens and hits the glass-air interface at the other side where it is refracted again. Because the lens is symmetrical about the horizontal ...


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The simple answer is yes, you can see it with the naked eye. It requires only that the light source is showing significant, periodic variations in intensity, and preferably with a frequency greater than the fusion frequency of the human eye (20 - 30 Hz). The greater the intensity variation, and the shorter the "on" duration, the more pronounced the effect ...


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The energy is reflected from the cavity. In general, an optical cavity acts as a variable transmissivity mirror for a light source with very narrow linewidth (a laser). As you change the length, the cavity can go between highly transmissive and highly reflective. The specifics of how tranmissive and how reflective it can be depend on the reflectivity of ...


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It is indeed possible. This was a famous experiment by Isaac Newton (published in 1672). Place a lens of focal length $f$ a distance $2f$ from the first prism. Add a second identical prism $2f$ past the lens and rotate it round until white light emerges. The lens is required to bring the rays back together. It creates an image of the exit of prism one on ...


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As you've observed in your edit, if you flip the orientation of the beamsplitter such that the thick part is now on the back side, then you still need to use the compensator plate in the other arm. You might wonder why a compensator plate is used at all. In an interferometer which uses monochromatic light (i.e. a laser) you could just shift the position of ...


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It depends on the actual materials. If you have a big need and a big budget, you can make some mirrors that have very high reflectivity. 99.9% is achievable in some cases for certain energy ranges. As energies go up, (UV and higher) it gets harder to reflect cleanly. For sunlight, most of the energy is in the optical, so you can ignore the reflectivity ...


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There is no contradiction: Your first answer, with the numerical factor 1.22, is a measure of the width of the diffraction spot from a circular aperture (e.g. from a round lense). Historically, the smallest resolvable feature has usually been defined not as this size, but as the closest distance between point sources that create an image that still has a ...


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You can look for geometric optics: Paraxial Approximation, but the case when angles aren't small (it's strange a little); In optical engineering we use it for calculations: It seems that right answer is 0.156 mm


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The previous answer is good enough but I want to add my point of view. The fact that an eye and el. devices ultimately do not register the amplitude of the radiation but its intensity (energy), the intensity we get by squaring the real and imaginary units thus they are no longer imaginary. That is, the imaginary part $k$ is an imaginary unit only in ...


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Most 1300-nm optical fiber systems use single-mode fiber. Most 850-nm optical fiber systems use multimode fiber. Therefore, typical 850-nm transmitters are designed to couple into multimode fiber. Coupling into multimode fiber only requires focussing the light down onto a spot about 50 um in diameter. Single-mode fiber, on the other hand, typically has a ...


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The event horizon is not an optical effect, so changing the refractive index of the space around it will make no difference. Adding extra mass will make a difference, but it's the mass that makes the difference not the refractive index. You couldn't make your spherical shell of glass go right up to the horizon because at the horizon no material is strong ...


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are both lines meant to be parts of a same beam of light, and these lines correspond to the edges of such beam? No, both lines are meant to be representative of different possible paths for light, which really extend through a large area, and are separated for the purpose of illustration. Supposedly the path difference can yield constructive ...


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Yes, it will, that is the whole idea. The reason is that everybody has to get the same physical answer independent of the sign convention. The problem is that you are free to choose positive radius of curvature for convex or concave surfaces, positive focal lengths for a converging or diverging lens, and so on, so, conventions are necessary standards within ...


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Don't get too precious over the term "resolution". There are many ways to define it, and indeed ultimately what you resolve with a microscope gets down to what measurement signal to noise ratio you can achieve. With a perfectly clean signal, you can deconvolve the lens's point spread function from your image and resolve features smaller than what the simple ...


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There are actually two different sign conventions in optics. Without any convention it's hard to develop any universal scientific statement (like formula) so let's make up one here: F - lens focal length, d - lens to object distance, s - lens to image distance. 1. Cartesian sign convention Small lens formula: 1/s - 1/d = 1/F ...


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Missing from these answers is that CCDs like your camera can't, in principle, see a Fourier transform. Even if you stick a lens at the focal plain you won't see the Fourier Transform! Your camera is not an interferometer! Borrowing the equation from Colin K's answer we see that the integral, can be negative, positive or even imaginary. \begin{equation} ...


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If you're calculating something like "Where is the image in this lens setup?", there shouldn't be any "convention" in the final answer. "The image is right here, where I'm pointing" $\leftarrow$ that statement is objectively true or false, it cannot depend on a convention. Therefore any conventions you use in the process of calculating this thing should ...


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I realize this is an old question and you have probably moved on... Flicking through the paper, the thing that strikes me is that the wave is normally incident. Unless that is the geometry you are using, the equations are not going to work for you. While an arbitrary wave front can be thought of as being composed of a smarties of plane waves, those wave ...


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The problem is tackled and answered here.


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Fluorescence is mostly red-shifted with respect to the excitation wavelength, as part of the energy goes to excite molecular vibrations. However, the reverse process also happens: if a molecule was vibrationally excited before electronic excitation, it can contribute this energy to fluorescence, which in this case is blue-shifted (so-called hot bands). The ...


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Trying to break my bad habit of answering in comments, putting my previous stuff down here now. As I said, I'm largely drawing on a blog post I did a couple years ago, working out the color of the sky. I had a spectrum, and I wanted a color. As I mentioned above, you can't just take a spectrum and output an equivalent wavelength--only some colors are so ...


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Given an unsharp image, you can use deconvolution to attempt to reconstruct the original image, but the resolution you obtain will be limited by the noise in the unsharp image. The theoretical limit for the case of astronomical images where the objective is to resolve double stars, is investigated in this paper.


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No, because you have no information on how far objects are, and that affects the spread of the light. If you were taking the photograph of a 2D plane (a painting for example), you probably could retrieve more information, but it would still have a considerable quality loss, since in the borders you don't have the entire spread, some would be out of the ...


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The normal nonresonant Raman scattering happens when a photon interacts with a molecule; the molecule absorbs the photon momentarily and re-emits it with slightly less energy. In an energy diagram, that looks like this. The frequency of the incoming photon is $\omega_i$, and the frequency of the scattered photon is $\omega_s$. The thick lined level is ...


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Losses are usually taken into account by considering the complex relative permittivity $\epsilon_r$ where the real part determines the polarization of the medium, while the imaginary part is related to to the losses in the material. This description is equivalent to the classical optical description in term of complex refractive index. As in non magnetic ...


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You're overthinking the theory. Think practically and the answer is obvious. The standard eyeball has a crystalline lense whose focal length is varied by the ciliary muscles. The maximum and minimum accommodation of focal length is what determines far point and near point respectively. The near point and far point have to be regarded as properties of the ...


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There is a perfectly valid physical explanation: use Maxwell's equations to find how the wave propagates beyond the aperture. Use as boundary conditions the idea that the wave is completely killed off beyond the aperture. Sound reasonable? This results in an integral. Turns out that the integral can be interpreted, at least to a very good approximation, as ...


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It's not quite as fictitious as you think! Imagine a circular wave pool. To make perfect circular waves of wavelength 5m, you would need a cylinder of radius 5m to rapidly go in and out of the water, displacing a lot of water for a brief period of time, periodically. Now imagine ocean waves of wavelength 5m going through a slit 5m wide. Inside that 5m wide ...


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Apply a parity transformation to your solution (but first please consider the region with non vanishing potential to be $[-L/2, +L/2]$ otherwise it will be notationally not convenient to shift by $L/2$ the final solutions, but it is up to you) $$ y(x) = c_1 e^{-jkx} + c_2 e^{+jkx} + \frac{jk}{2}\int_{-L/2}^{L/2} dt e^{jk|x-t|}\eta(t)y(t)$$ gives $$ ...


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The response time of the human eye is something like 60 Hz. This means that you need to draw the full line in 1/60th of a second. Based on your drawing (which isn't particularly clear) you are trying to draw a line using roughly 90° of mirror rotation. Put this all together and you need the mirror to be rotating at an angular speed of 900 rotations ...


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There are several ways to approach this problem. If we can estimate the power density achieved in $W/m^2$, then the temperature that can be reached follows from the Stefan-Boltzmann law. First method: 1) Take the total power collected, and see the size it got focused down to. You state the area of the mirror array is 0.6 m$^2$ (roughly), and with power ...


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Simple geometry tells you that the size of the image $s$, for a given distance from mirror to screen $d$, is a function of angle according to $$\tan\frac{\theta}{2} = \frac{s}{2d}$$ For example, an image of 1 meter at a distance of 10 meters needs the mirror to sweep through an angle of $2 \cdot \tan^{-1}(0.05)\approx5.7°$


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When a camera takes a picture, the shutter is open for some small amount of time, during which the sensor takes in light. If an object is moving, it is reflecting light towards the sensor from a range of locations during the time the shutter is open, so by the time the shutter closes, the image shows the object appearing to exist in a range of positions. ...



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