New answers tagged

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But if you adopt the different sign convention you have just got the same result! The E-field is a vector; by making the incident and reflected waves have opposite signs on the LHS in your revised formula you have started off by assuming that the reflected E-field is in the opposite direction to the incident E-field. Then when you find that the reflection ...


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There is the question you asked, and the question you might have asked... and they are slightly different. I will answer both (I had answered the second before you clarified that you meant to ask the first... seems a shame to delete the second answer) First - the question you asked (about electronically controlling the display properties). There are (at ...


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My attempt of answer will be based on my non-specialized knowledge of programming and computers workings, because the links you provide don't seem to give out the details on the workings of the code. I'm a physicist with a non-academic conceptual and practical understanding of computers. Whatever layers your software might have, the display will react ...


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In order to see the curvature, you need a 60 degree field of view and a cloud free day. From what I've read, you need to be about 35,000 feet above the surface. Find more information in this article here: http://www.howitworksdaily.com/how-high-do-you-have-to-go-to-see-the-curvature-of-the-earth/


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As a former sailor I can assure you that you can see the curvature of the earth from the crow's nest - all it takes is a calm sea. Consider a light house that stands 50 meters above sea level. If you are in the crow's nest, say 25 meters above sea level), at what distance could you first spot the top of the light house on a calm day? The problem is ...


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If the tube was 100% reflective - like the sun tubes in some homes - then there would be no loss. In this case the tube is absorbing a portion of the light on each reflection, so if a given ray starts with intensity R, and looses a fraction x on each bounce, after one reflection the intensity would be R(1-x); after n bounces it would be R(1-x)^n. Now ...


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Here is a series of images from two point objects (stars) to illustrate the idea of resolving power. The objective lens of the telescope produces the diffraction. If the distant stars are closer together the diffraction patterns come closer to one another. Diagram (a) shows to diffraction patterns (the intermediate image in a telescope) well spaced ...


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I use a program like Remo Photo Recovery but you may want to see methods on how the UFO crash Ramey Memo from Roswell, NM was refocused.


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Halos are not exclusive to ice crystals, and rainbow colored halos are commonly reported, they usually indicate some issue with eye's focusing. One possibility is a growing cataract:"Seeing rainbows or halos around light indicates a problem with how light is filtering into the eye. Light is made up of different colours but the rays are normally focused on a ...


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Yes it would depending on how thick the smoke was. Photons intercepted by the smoke would be scattered or absorbed but the photons that still have a clear shot to the screen would contribute to the original fringe pattern. The pattern would still be the same but not as clear.


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In this first diagram the two grey rays $OXI$ and $OPI$ are the rays I used to find out where the top of the image is formed. They are called the construction ray and these are usually the only two rays that you see on a ray diagram. However in reality all rays bounded by rays $OWI$ and $OZI$ go through the lens to form the image. If I put an obstruction ...


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You are probably referring to the conventional representation of rays in geometrical optics, where typically two rays are drawn to define the image, which is exactly where they cross. You can also draw 20 of them but 2 is minimum. So this is just the property of the sharp image - all rays from the object are collected in a single point of the image. In any ...


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The answer is "Yes" but not the way you might expect. It is possible to construct a telescope mirror from rotating liquid metal.Mercury used to be used but something like Gallium is safer and better. So print a cradle for it, put in the Gallium, raise the equipment past the melting point (about 30 degC), spin gently to get a parabolic surface, and then ...


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A properly collimated laser beam is called a Gaussian Beam whose transverse magnetic and electric field amplitude profiles are given by the Gaussian function. The Gaussian beam is a transverse electromagnetic (TEM) mode. The mathematical expression for the electric field amplitude is a solution to the paraxial Helmholtz equation: The width of such laser ...


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Strictly speaking, the issue is not in different wavelengths, but in different frequencies. (The first case would still allow you, e.g., to record holograms happily from two noncollinear beams in a hypothetical anisotropic photographic emulsion.) The difference of frequencies, however, is nothing but a steadily growing phase difference between both beams. ...


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You create the original hologram by creating an interference pattern that is then recorded on the holographic film. If you use two light sources with different wavelengths they will not produce a static interference pattern but one that strobes at a frequency equal to the difference between the frequencies of the two light sources you're using. Unless you ...


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Just as Jon Custer wrote in his comment, even a perfectly collimated laser beam with a planar wavefront will diverge. The way it happens is determined by the Huygens principle, and depends on the beam profile: When the light intensity is abruptly cut by a sharp flat obstacle, the light will indeed diffract in almost all angles. A razor blade cutting a laser ...


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For this solution (which was not mentioned) is is assumed that the glass is isotropic. Then according to (for example) Sutherland we have the three room dimensions $x$, $y$ and $z$ for the incoming electrical field. Now due to isotropy the 81-element-matrix only has 21 non-zero elements with $$\chi_{XXXX}=\chi_{YYYY}=\chi_{ZZZZ}$$ ...


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The middle part of the lens will just act as a rectangular glass slab. We may verify it by cutting the lens horizontally. Now we know that rectangular glass slab refract light in such a way that emergent is parallel to incident. Actually the same happens when the ray passes through optical centre. This can be observed in a thick lens. In thin lenses the ...


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I am not crazy about guys using mathematics to explain things .. so I will try to be more graphical than theirs so here you go.. 0 down vote It is easy to understand that the higher the focal length is the narrower the angle of the image from the objective lens or mirror to the eyepiece. So the narrower the angle (higher focal length that is) the deeper ...


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It is easy to understand that the higher the focal length is the narrower the angle of the image from the objective lens or mirror to the eyepiece. So the narrower the angle (higher focal length that is) the deeper the image can go into the eyepiece without losing focus. The lower the focal length is the shorter the depth of focus. It means that you have ...


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It means that fringes can be seen everywhere to the right of the double slit arrangement if the source is to the left of the double slit. They occur because coherent waves for the two slits actually overlap and hence form an interference pattern to the right of the double slits. For some arrangement used to show interference the coherent waves from ...


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The answers here is beautiful. But, i’ll give another simple example. Just take a glass square. The critical angle of glass is 42 degree. So pass a light ray through a very small slit , such that it strikes the side of the square glass slab, at 45 degrees. It will reflect and will ALWAYS strike the faces of the slab at angles greater than 45 degrees. And ...


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medium with refractive index less than unity implies medium in which particles travel faster than light speed. R.F<1 => c/v<1 =>v>c so according to special relativity nothing can travel faster than speed of light.so medium doesn't exists and the medium should be lighter than vaccum.so such a medium doesn't exist.and also if we assume the old ...


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I will answer this question in two parts: first comparing plane waves, spherical waves and cylindrical waves (which are really the same thing as I will explain). how this relates to ray optics (which is completely different). Plane waves, spherical waves and cylindrical waves are 3 different examples of doing a decomposition of a wavefield. The method is ...


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If the 2 objects are at different locations, then the green light is hitting different group of cones in your eye than the red light. You should thank to the lens in your eye which is sending the light rays to different areas on your retina, depending on the direction from which the light is coming. Without the lens, all light would hit all cones at the same ...


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Huygens's Wave Theory is what you call a first order scalar diffraction theory of light. So what does it describe and what does it fail to describe? First order means that electromagnetic effects like induced currents in surfaces etc. are ignored. These can be described by solving Maxwell's equations for the same system instead of working with the wave ...


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Neither image displays diffraction. They both illustrate shadowing. The second photo was apparently shot on a clear sunny day. All of the rays come from the sun. Some are blocked, those that aren't go straight down to the floor. The second was probably taken on a hazy day. Rays originate from the haze; they come from all directions, and make their ...


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It's no different than behaviour of rays that transit from matter to vacuum. In refraction, only the ratio of indices matters, not the absolute value of the phase speed of light. This is actually used in refractors for X rays. Properties are nothing speciall really - the phase velocity is greater than the speed of light, but that's just the speed of moving ...


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Traditional light sources (candles, light bulbs, star light) can be just as spatially coherent as the best laser. After all didn't Thomas Young perform his famous Young's double slit experiment in the early 1800s. In your case, the the SlED has a single mode waveguide that acts like a spatial filter to only propagate certain modes of light.


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The conditions for total internal reflection from an air-common glass interface are: The light has to be travelling in the glass. The angle of incidence has to be larger than the critical angle.


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For a pane of glass, the two surfaces are parallel to each other. That means that the exit angle from the 2nd surface will be the same as the entrance angle of the first. Whatever light enters (and isn't absorbed) will exit: $$n_{air}\sin\theta_{1}=n_{glass}\sin\theta_{g1}$$ and $$n_g\sin\theta_{g2}= n_{air}\sin\theta_2.$$ A quick sketch of two parallel ...


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I don't think there is a universally agreed phrase to describe this, but I think the closest is the principle of geometrical reversibility.


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On rainbow formation: https://en.wikipedia.org/wiki/Rainbow#Supernumerary_rainbow . See a bit more about supernumeraries here http://www.atoptics.co.uk/rainbows/supers.htm, and http://www.atoptics.co.uk/rainbows/supers.htm . Classical rainbow is explained by geometric optics, which applies on big enough droplets. When these approach the scale of light ...


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The concave mirror will form a real image, so by convention it's focal length is positive. The diverging lens does not form a real image, but a virtual one, and thus is negative. Sometimes a negative lens is used for a beam expander for a laser and no focus is formed


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any angle over theta 90 will mean that the diffraction will be going behind the diffraction gratings which is impossible. so 90 is the maximum that you can get this is why you have to round down the decimal answer you will get.


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NOT A PHYSICIST, BUT: I was just passing through and saw your question. I've been interested in "light transfer through rapped fiber" for many years. A consistent original source would be open to computation, but the Sun is actually not consistent. There would be high's and low's of the Sun's spectrum, thus you couldn't regularly optimize the direct ...


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the Earth curves at the rate of 157mrad per km travelled. The refractive index of dry air at sea level is 1.00029, but at a height of 1km, the air pressure is 12% less and (neglecting temperature density and humidity), the refractive index would be 1 + 0.00029 * 88%. The difference, 0.000035 means that light at 1km altitude travels 35mm further for every km, ...


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with brigh field illumination all the different planes in z are illuminated (or excited for fluorescence). Out of focus objects will increase the background noise on the object in focus, decreasing the signal to noise ratio. So effectively the maximum resolution of the microscope cannot be attained.


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The further your object is from your optic's focal plain, the more blurry it gets. There's not much more to that statement as far as I can deduce.


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The answer depends on how close you are to the focus. Consider the (mathematically) simplest case, a "collimated" Gaussian laser beam with wavelength $\lambda$. If the ($1/e^2$) radius of the beam is $a$ and the focal length is $f$, we can approximate the convergence angle of the light after the lens to be $\theta_0 = \arctan (a/f) \approx a/f \textrm{ ...


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Yes, it's necessary for example on some telescopes to keep the image the same way up on a camera as the telescope tracks across the sky. VLT naysmyth focus There are a couple of optical designs, using either rotating prisms or a rotating set of mirrors. Look up field rotator. eg http://www.ing.iac.es/~eng/optics/documents/OPT-WHT-001.pdf


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If, indeed, the light is collimated (note, there are limits to how well collimated a beam of light can be because of diffraction) and the lens that focuses it is so-called "diffraction limited" (ie. the lens doesn't abberate the light) then the spot created at the focus does obey the inverse square law. However, keep in mind that it is not truly focused to ...


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Yes you can. Moreover, you don't need any lenses and such things, you only need a medium with Refractive index $n(\vec r)$. Using Fermat's principle (https://en.wikipedia.org/wiki/Fermat%27s_principle) you can calculate the path of the light travelling through a medium with $n(\vec r)$. You just have to choose a proper $n(\vec r)$ function.


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Two points from a very practical point of view based on laser systems. Imagine there is a small dust particle on an optical element in your path. Diffraction from that little particle will create a pattern that expands with the propagation along the path. If you insert a $4f$ system, you "cut out the propagation" on a distance of $4f$: the plane at a ...


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The answer lies in the band structure of the two materials. The band structure describes how the electrons in a solid are bound, and what other energy states are available to them. Very simply, the band gap for transparent diamonds is very wide (see this link): Normally, diamond is not a conductor: all the electrons live in the "valence band", and you ...


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Diamonds are unstable compared to coal (or more exactly, graphite) so high temperature and pressure are required for diamonds to form from graphite. The reason that coal (graphite) is black and diamonds are clear has to do with how the carbon atoms are connected together in the two different forms of carbon. In diamond each carbon atom is bonded to ...


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You could also think about the perception of this in terms of spheres of space. Lets say you are in a very small room which is hemispherical in shape. There is a 6 foot high door in the wall which seems to be very big - and relatively, it is - it takes up about 20% of overall area of the wall. Now imagine a much bigger hemispherical room that you are ...


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The size of an object while seeing it through eyes depends on the visual angle and not on the real size of the object. If the object is near to the eyes the visual angle will be higher and when it is far away the angle will be smaller. According to that, the size will differ. The size is proportional to the visual angle.


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I found this paper: http://users.aims.ac.za/~jweiner/AJPIAS_64_8_986_1.pdf The authors mention that Weisskopf says that a layer of thickness λ/2 at the surface of an object is responsible for the reflection of light upon it. They then take the microscopic perspective in which the reflection and transmission result from the scattering of light by the ...



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