# Tag Info

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The general and easy rule to remember the sign convention for convex and concave mirrors and lenses or any optical component is thinking in terms of power. The power of an optical element is calculated by 1/focal length and the unit is dioptre (1/m). It is negative for components which diverge parallel incident rays of light and positive for components ...

2

The green lines are wave fronts, not normal lines. You can think of the green lines as crests of the wave. That's why they're always perpendicular to the ray. Contrast this with normal lines, which are perpendicular to the surface. You're right, incident and refracted angles are always measured between the ray and the normal to the surface. This turns out ...

2

I agree that's confusing, and that $\theta_2$ is just plain wrong. I've always seen it explained with the normal perpendicular to the surface, just like you say, and exactly as drawn in https://en.wikipedia.org/wiki/Snell%27s_law But note that this gives the same $\theta_1$ as in your drawing. But the way you've drawn your drawing, $\theta_2=\theta_1$, ...

0

umm, as long as Light/Plasma is defused by Plastic/Glass in any way/form; our eyesight will not recognize/perceive an illumination of anything. TV Shows have presented the false impression of refractive amplication for lighting and lasers being self-amplified. What SciFi Programs HAVE gotten-right is that an underground tunnel/cavern can be illuminated with ...

-1

Simply we say that when light ray hits the boundary of the 2 media at more than critical angle, it refracts so much { >90deg. } that it gets directed back into the same medium.

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The ideal perfectly smooth flat surface has translational invariance symmetry. That means that there is no mechanism for the scattering of a wave in the horizontal direction, and no mechanism for the change in the component of wave vector parallel to the boundary. That is, the horizontal component of wave vector is conserved. For light incident at angles ...

2

Imagine the speed of light to be $1$ meter per second and the speed of light in the medium with a high refractive index to be $\frac{1}{2}$ meters per second. If you have a single peak of a wave in the slower medium, that peak must move forwards at speed $\frac{1}{2}$, no matter what angle it's facing. In the faster medium, that peak must move forwards at ...

0

About 30 GHz, according to this. Check "Bandwidth-limited Pulses". Because of Fourier transform, the product of the temporal duration and spectral width is ≈ 0.44 for Gaussian-shaped pulses. See also this.

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You can apply the Huygen's construction to any distance over which the local environment is optically homogeneous. Encountering a reflective or refractive surface is encountering a non-homogeneity, so you can't draw big circles that include a mirror or a lens. You can modify the Huygen's principle so that you can use near such boundaries, but the ...

3

Your text is rather muddled, but to answer the question: the Poynting vector is normally in the direction of propagation, which is to say the E and B fields are perpendicular to the direction of prop. This is always true in a vacuum, but it turns out that in various materials, the Poynting vector can be off-axis.

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This plano concave lens has the same object and image positions as your silvered plano convex lens. Hence the same focal length. The sign convention comes from the sign convention of the u and f distances.

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As alluded to in the quotation, the time-evolution of non-relativistic matter waves is given by Schrodinger's equation, which is a linear, first-order differential equation in time. As such the future state of the system is fully specified if we specify a single boundary condition. On the other hand, photons, being massless, cannot be treated in a ...

0

Without using half shade device your eye cannot judge the exact position of extinction of light when the two nicols are placed in the crossed position.

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CO2 lasers tend to be very high power. Often they will have 100 W in the beam. They can be much higher. Some will burn through safety goggles about as fast as you can blink. Optical elements used with such lasers need to be designed for such power. When light strikes a coating, it is transmitted, reflected, or absorbed. Absorption must be kept down to a few ...

1

In radar, chirp (LFM - linear frequency modulation) is used to stretch the pulse so that the received energy is large enough for detection while having a large coherent bandwidth commensurate with the desired range resolution. There are two ways of detection, the so-called "stretch processing" that is just a correlation receiver using a homodyne mixer, the ...

0

If by constant you mean equal spacing between fringes then yes the spacing of a double slit is constant and so is any slit including diffraction grating's. Sometimes if you are not using monochromatic light the rainbows or wide spectrum will give the appearance of unequal spacing. Not counting the center fringe as far as I know the only Fringe pattern with ...

0

Conformal maps tend to be the exception rather than the rule. In general, if a transformation $T:S\to S$ on some $n$-dimensional space $S$ is of interest in physics, it will not be conformal. (Indeed, there's generally no guarantee of a useful notion of angle in that space, but even if there is such a guarantee then $T$ is still not likely to be conformal.) ...

15

You say: For a non-transparent object like a brick, when the light is absorbed by an electron it will eventually be re-emitted. but this isn't true. In a solid the excited state can decay by transferring energy to lattice vibrations instead of emitting a photon. This means the energy of the incident photon is converted to heat and the photon is lost ...

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For an object to be transparent, the light must be emitted in the same direction with the same wavelength as initially. When light strikes a brick, some is reflected in other directions, and the rest is re-emitted in longer, non-visible wavelengths. That is why a brick is opaque to visible light. Some materials we consider transparent, like glass, are ...

2

A real image is one through for which the rays forming the image pass through the point where the image is seen. A virtual image is one where they do not, but projections of the rays do. Real image Virtual image In the latter case the rays (the solid red lines) do not actually pass through the image (the larger of the two arrows) only its projections ...

1

After reading some fundamental mathematics and physics or better to say becoming a sophomore you can start reading these books but absolutely some topics need more than fundamental mathematics and physics. Fundamentals of Photonics (Bahaa Saleh, Malvin Teich): This book provides an introduction to the fundamentals of photonics. Fundamentals of Photonics ...

0

Seems like they've found a way to do it! http://phys.org/news/2015-11-device-theoretically-bit-infinite-amount.html The problem with my initial idea about using an ordinary container is described succinctly in the above article: When light is put inside a cavity, it basically interacts with the matter that surrounds it (e.g., glass or metal), and this ...

0

Your question includes both the conversion (since you speak of processing) and light propagation. Conversion involves electronics, as @Nasha mentions, and thus is directly impacted by the slew rate. Light propagation speed is reduced (with respect to that in vacuum) by the refractive index of the material. The physics causing the finite slew rate is also ...

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In case someone else have this question, I finally found that Goodman (Introduction to Fourier Optics, ISBN 9780974707723) explicitly states that the Fresnel approximation is indeed valid in the far-field.

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Your laser cavity is a Fabry-Pérot interferometer. The free spectral range tells you how close two neighboring laser modes can get: $\Delta \nu=\frac{c}{2nl}$ (for a linear resonator, length l, refractive index n). The resolution of your spectrometer needs to be smaller than this free spectral range. You can increase the free spectral range by either ...

0

The sentence can be proved by Fermat's principle and it is not relevant to wave nature of light. Waves have some properties like velocity (and could be different in different media), phase and the important property called interaction. When we say light is a wave, we must check the above properties in it. And the first experiment that showed that, light is ...

1

Each laser is made up of three main part: pump, amplifying medium and cavity. White laser is semiconductor laser; which means the amplifying medium is a semiconductor. In semiconductor lasers, pumping is electrical; which means the amplifying medium requires electrical power to be stimulated. The semiconductor of the white laser is an alloy made of zinc, ...

1

Suppose the object (the LED) is distance o from the lens and we want the image to be distance i from the lens: To get that to work, the focal length f of the lens must satisfy: $$\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$$ In your case, you want $o=20$ and $i=150-o=130 \text{ mm}$. To achieve that, you need a lens with focal length:  f = ...

4

The semiconductor lasers that are used to produce white light are powered by electricity, so that solves your problem. As lasers emit monochromatic light (1 single frequency or colour), they cannot be white. To get white laser light, you start with a blue laser. This blue light is directed at some phosphorescent material (a "phosphor") which, when activated ...

1

In a sense, the approximations that you can make in the paraxial approximation are easier to handle, since you don't have to consider many of the aberrations that come about when dealing with lens systems. But ease of use is not the main reason why it is used. In the paraxial case, spherical aberrations of the lenses can be ignored, because the focal plane ...

0

I suspect that it's a quantum mechanical question. The location of the point where the mirrors meet can be represented by a mathematical expression and various values in that expression can be chosen to make it as small as we like. Similarly, the light wave is expressed with other values which can make it wider than the mirror's point. Then we have a ...

0

Because the derivative of the mirror angle is discontinuous, the mirror does not have a specific angle at x = 0, so the reflection angle is undefined at x = 0. On the other hand, you might consider the effect of shining a bundle of light rays with a finite diameter (centered on x=0) on the origin. Obviously, half of the light will go to the left and half to ...

0

Regardless of whatever physics is assumed, if the cone has perfect symmetry, then, by symmetry, the ray is directly reflected back along the direction it came from.

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This isn't a physics phenomenon, but a biological one - look up persistence of vision. Basically our eyes have a limited response time so if we flash two different images quickly we will see an average of both images. TV uses this to appear to give us smooth motion where in reality it is flashing images every frame. If we assume we need a 25Hz frame rate ...

1

Positive dispersion means that the lower frequencies are ahead of the higher frequencies in time, and this is caused by the fact that the crystal's group index (derivative of refractive index) is smaller for lower frequencies, giving them faster group velocity going through the crystal. To compensate, the pair of prisms gets the higher frequencies to catch ...

2

There is, as far I know, no reason that you can't achieve to produce a polarized laser if you control the polarization of the individual lasers and use an appropriate grating. One of the major advantage of incoherent combining is that it doesn't require phase locking or polarition locking of the invidual lasers. So, if you consider controlling the ...

0

It is very unlikely that such a system would be viable as an astronomical telescope since it it would be essentially unsteerable. The detection of very faint objects typically requires long exposure times, but I doubt if the solar array would be capable of getting an exposure of more than a few seconds before the target moved out of the field of view. In ...

2

You can indeed build a telescope on these principles and indeed thinking about this kind of idea leads to an active branch of astronomy: very large radioastronomy arrays. At optical frequencies the idea is hard to make practicable on the ground (but see David Hammen's answer for space telescopes): the mirrors would need to be aligned to produce a ...

5

Regarding the question raised in the title, Is it possible build a telescope on a field of mirrors? The answer to this question is a resounding "yes". A number of existing and planned telescopes use arrays of mirrors, depicted below. Noteworthy amongst them include the James Webb Space Telescope, which will use an array of eighteen hexagonal mirrors, ...

14

To augment Rennie's answer with a graphical representation let me post this diagram: Imagine, if you will, not a single beam of light but a series of wavefronts. When part of the wavefront slows due to a different density, the wavelength also compresses, thus introducing the characteristic bend.

0

Your comment explains better what you want By coherently combine, I mean to combine to different laser sources to produce a beam of greater intensity even if I wouldn't use the word "coherently". You can combine, or superpose, as many beams as you want with any polarization just using beam splitters (better non-polarizing). Contrary to polarizers, they ...

23

When you say light bends I assume you are talking about refraction i.e. the change in the angle of the light given by Snell's law. You ask: If I'm running straight, and I get slowed down, shouldn't I still be running straight? but suppose one foot get slowed down while the other one didn't. In that case you would turn in the direction of the foot that ...

0

Here is the full sentence from that Wikipedia article: The reflections from the low-index layers have exactly half a wavelength in path length difference, but there is a 180-degree difference in phase shift at a low-to-high index boundary, compared to a high-to-low index boundary, which means that these reflections are also in phase. Observe ...

0

I'll rephrase the question as I interpret it: Electromagnetic waves are drawn like this: source: https://commons.wikimedia.org/wiki/File:Onde_electromagnetique.svg Suppose this wave comes up to a vertical slit (a slit in the z-direction). What if the red arrows are longer than the slit? Then the wave won't fit through. But if the arrows ...

2

remember that the amplitude is not an amplitude in space, it is an amplitude in the sense of the intensity of the electromagnetic field. The spatial amplitude is given by the wavelenght

0

Refraction of Light is not a Thermodynamic process. If you study the QED basis of refraction, you notice that the difference happens in time. The speed of light is constant, and a photon which is refracted, doesn't actually travel any slower, it just travels a longer path, and needs thus more time. If it hit's somewhere, then it's not refracted. It's gone, ...

1

Indices of refraction are really complicated functions and depend quite a lot on all sorts of difficult-to-model molecular properties. What you really need is to determine over which wavelength range you need your approximation to be valid, and then look for experimental data that you can fit to, or models from people who have already done the fitting. ...

1

If the bounces draw a finite star, then by symmetry the ray goes out (it's where the star get closed). This case is for angle $a$ between 2 bounces ( $a = \pi-2\theta$ ) is such that $ka=0[2\pi]$, i.e. when there exist $k$ and $n$ such that $a=\frac{n2\pi}{k}$, i.e. when $\frac{a}{2\pi}$ is a fraction ( or equivalently, $\frac{\theta}{\pi}$ ). If not, an ...

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