# Tag Info

0

Dark matter is more correctly transparent matter (aether). Background noise is aether. The "GOD" particle is aether. The Higgs Boson is aether. All matter and forces are aether; The NOUS of Anaxagoras which he described as a invisible vortex of pure energy expanding to fill the universe. Add a particle and you have aether you cannot have particles without ...

0

It's because of DLP projectors. In such projectors a 3-color wheel rotates in front of the light thus projects only one color at a time but very rapidly. Search google for DLP rainbow effect. It's common to see an be bothered by this. Source: http://en.wikipedia.org/wiki/Digital_Light_Processing#The_color_wheel_.22rainbow_effect.22

0

Summary: The best way to see the water is to try and look for the reflection of a light source (sun or light bulb) at a large angle of incidence. More Details: You see the water on your floor because it reflects light differently than the dry surface of the floor. Namely, the water forms a smooth enough surface that the light reflected off of it can be ...

0

It is on a small scale the same as a mirage in a desert. A mirage is a naturally occurring optical phenomenon in which light rays are bent to produce a displaced image of distant objects or the sky. In contrast to a hallucination, a mirage is a real optical phenomenon that can be captured on camera, since light rays are actually refracted to form ...

0

"From this source, I gather that he did argue that the light particles sped up when entering a more dense medium. However, it just doesn't make sense." It does make sense. They enter the medium at a higher speed, then slow down. If Newton had said they continue to move at a higher speed within the medium, he would have been wrong. But I don't think he said ...

2

Your friend is correct. It's not specular, because that would mean it's a mirror, and you'd only see the image if you were at the angle of reflection. Further, a mirror does not act as an image plane, so you might have difficulty ( :-( ) perceiving the image. Basically you'd need another lens to re-image the source. The diffuse surface acts as an image ...

4

Like all good physicists Newton proposed a hypothesis for why light refracts when it crosses a boundary between different refractive indices. And his hypothesis makes a certain amount of sense: If the initial velocity of the light is $v$, then the velocity parallel to the boundary is: $$v_p = v\sin i$$ and the velocity normal to the boundary is: $$... 1 I'm afraid that it is not easily possible to take the luminous flux and obtain the insolation (as radiant flux). Here's why: The luminous flux F is calculated from the radiant spectral power distribution J(\lambda) by weighting each wavelength with a luminosity function y(\lambda) as per$$ F = c \int J(\lambda)y(\lambda)\mathrm{d}\lambda$$where ... 8 You don't have a fluorescent light but an LED light that time-multiplexes different colors to achieve white. The fan blades act as a stroboscope and make the switching frequency of these LEDs visible. Or... I am wrong! So back to the drawing board: I did the experiment. Most fluorescent lights in my house do not show this effect, at all, but one does, ... 2 There exists a good answer to what is color which stresses that the question is mainly about physiology of the eye. I will address the last part: Basically, what properties of the paint material makes it absorb the red wavelength the least? Light has a specific wavelength . If that wavelength impinges on the eye receptors, it gives the colors we have ... 1 The energy of a wave is h*f. The energy in an electromagnetic wave is where E is the electric field of the wave as it moves with velocity c. How i imagine this is each cycle consisting of 1 quanta. There are no quanta in the sense you are imagining it. The classical wave above emerges from a zillion of quanta, called photons. For a ... 1 The light that you can see in the air is actually light that has been reflected from particles in the air (dust, water vapour, etc.) towards your eyes. So when the beam is pointing away from you, some of that light will still bounce off those particles and come to you; this is what allows you to see the beam. However, when you look at the water, it acts more ... 0 If the beam light or what you say as cone is above the water what you will see is the reflection of light from the lighthouse and not the beam, as you can see it produces a cone shaped light with the help of a parabolic shaped reflector it have or either it uses other type, the point is that the light is focused and under tha law of reflection for specular ... 1 One has to distinguish the two frameworks: the classical, light; the quantum, photons. The classical electromagnetic wave, of which visible light is a part of the frequency spectrum, emerges out of zillions of photons, the quantum of light. This happens because the photon has an energy E=h*nu, where h is the Planck constant and nu the frequency of the ... 0 There's quite a lot to say about this, 18490. The tl;dr for Mythbusters fans is at the end. Most materials are somewhat transparent. This is why you often need two or three coats of paint on a wall. The light enters the material, travelling many microns before it is eventually absorbed or scattered back. The result is usually that most of the light is ... 0 At around the 1\mum particle size you're in the Mie scattering regime, and this makes life hard because there isn't a simple analytic formula for the cross section due to Mie scattering. However if you're prepared to consider particle sizes significantly smaller than the wavelength of light, say 0.1\mum and smaller, then it's easy to show there is an ... 0 Good question. The key is to realize that a mirror-like surface send rays of light into your eyes directly from the source; they just bounce simply off of the mirror. What you see is the source, and the color of the source. Not so for a diffuse object. In this case light from the source hits the mirror and does not bounce directly into your eye. It ... 6 Physics first: light and radio waves are the same thing, just at vastly different frequencies. Radio works between 1\text{ Hz} and approx. (10^{11}-10^{12})\text{ Hz} (at the moment). The frequency of visible light is around 10^{15}\text{ Hz}. The range between these frequencies is usually called infrared radiation and is of little interest for ... 6 Frequency modulation in VLC is not done by changing the frequency of the light itself. The light intensity is pulsed at a high frequency to create a carrier wave. This carrier wave is then frequency modulated to transmit the information, typically using frequency shift keying. That's why it works with white light, where the frequency of the light isn't well ... 3 Any waves that appear on one side of an interface must immediately leave on the other side. There's no room to queue them up. This means that in a frame where the interface is at rest, the frequency on both sides is always the same. Once you have that, then it becomes obvious that the wavelength in air will always be the same as well. So from the ... 1 Yes, the term "shadow" can refer also to something or (dare I say) someone that is dark, shady, inconspicuous, etc. One can also use it as a verb; to shadow someone is to follow them closely. Like "I'm having the new guy shadow me for a while until he learns how to do everything". 5 Yes! Any beam that is blocked by an object will basically make a shadow. For example, the IceCube detector can see the moon's cosmic ray shadow. 1 Yes, for example a Crookes tube shows an electron shadow. The area I live (Chester, UK) is in a rain shadow. 2 For example, acoustic shadow (http://en.wikipedia.org/wiki/Acoustic_shadow ). 2 You can think of super massive object like black holes which can bend light. Near the event Horizon you could get a 180 degree turn for light and thus see the earth back in time. But I do not think this is practically possible as earth is small and dark (compared to stars) and this layer would get compressed really thin as some small deviation in the ... 4 The term dispersion refers to the speed of light in a material having a dependence on frequency (or equivalently wavelength). The refraction angle's dependence on frequency is caused by the material dispersion, not the other way around. In all materials the refractive index will have dispersion but it's often the case that certain materials in certain ... 4 The whole point of Snell's law is about taking into account wavelength! Remember that the fundamental property of the light is its frequency. Wavelength, on the other hand, is not a fundamental property of a beam of light since the wavelength changes all the time as it passes through various media. The index of refraction of a medium is a dimensionless ... 7 Snell's law is given by n_1 \sin i =n_2\sin r  where n_1 is the refractive index of the material the light is initially in and n_2 the refractive index of the material the light is going into. These are not constants for a given material and change (although very slightly) with the wavelength of light. Thus for different wavelengths of light the ratio ... 1 The "slowing" of light in a medium can be entirely explained using a classical wave-based approach. An incoming EM wave wiggles the electron clouds around the atoms in the material. These electrons clouds re-emit a much weaker EM wave having a very small amplitude. This re-emitted wave is 90-degree phase shifted from the original wave but superposes with ... 2 This is a nontrivial problem in materials science. People have done a lot of work both with different materials and different surface structures to create "ultra-black" absorbing structures for use in optical systems. A big part of the problem is the statistical nature of photon absorption. There's only a certain percent (<100) chance of absorbing a ... 1 It happens in Special Relativity because of time dilation. The time between crests when wave packets are emitted as observed by the emitter, is different to the time between crests when the wave packets arrive and are observed by the receiver. So the observed frequencies are different if there is some time dilation effect between the two observers. In GR, ... 7 Your hypothesis is basically correct, in theory, even in a vacuum. Light consists of electromagnetic radiation. According to classical electrodynamics, electromagnetic fields in a vacuum are linear, which means that one light beam will pass right through another, completely unaffected. But according to quantum electrodynamics, at electric field ... -1 Yes and no. Photons don't interact in free space. So a beam of light can't block another beam of light in vacuum. Photons can interact due to the nonlinearity of the medium. So it's plausible to block another beam of light if you have the right mediators. It's however not the light itself becoming a solid. See, for example, electromagnetically induced ... 0 The process of light propagation is described by the Maxwell equations.$$ \nabla\cdot{\bf D} = \rho  \nabla\cdot{\bf B} = 0  \nabla\times{\bf E} = - {{\partial{\bf B}}\over{\partial t}}  \nabla\times{\bf H} = {\bf J} + {{\partial{\bf D}}\over{\partial t}} $$These equations say (in simple terms) that: change in the electric field is causing ... 1 This is primarily a biological question. We (humans in particular, mammals specifically) can't see radio waves because our bodies do not have the sensors to detect them. We can detect light in the visible spectrum because the rods and cones of our retina that constitute our light sensors absorb photons in that spectrum, ultimately activating neurons in our ... 16 No. There is no void left by the lack of an aether. The very notion of aether should serve as a warning as to how catastrophically analogical reasoning can fail. "Water waves are in water, sound waves are in air, therefore there must be something in which light propagates." This is flawed logic, and decades of physics were arguably hindered by adhering to ... 0 There are many possible cases of total transmission at an interface (assuming that the media are lossless), and below is a list of the underlying physics: if the media are impedance matched any the given incidence angle. Brewster's angle can be seen as a special case. Usually, at normal incidence, this condition would require the materials to have the same ... 6 As most people know, "let there be light" is a famous biblical quote, from Genesis. Now, on to the teacher's shirt. Those equations on his back are Maxwell's equations. "Let there be light" is a joke, because Maxwell's equations describe electromagnetic fields, and light is a form of electromagnetic radiation, so the equations can be used to describe ... 2 Briefly, the formula E=mc^2 applies only particles at rest in an inertial frame of reference. Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula E=mc^2 to a photon. In more detail, the four-momentum of a particle has components (\frac{E}{c}, \vec p) The four-momentum for ... 1 The correct general equation is E^2=m^2c^4+p^2c^2, since m=0, E=pc. A photon's momentum is p=\hbar k, where k is the wavenumber (k=\frac{2\pi}{\lambda}). This would be consistent with E=h\nu where \nu is the frequency. 4 A "sharp" tip typically has a finite curvature; there will be a very small part of the "tip" that is therefore angled at such a way that light will be reflected off it. The sharper the tip, the smaller the radius of curvature, and the smaller the "twinkle" or glint. The second effect is diffraction: Light that passes an object will be diffracted. For ... 2 You are asking about the number of photons fired from a device such as a laser at a unit time. We cannot say what the precise number of photons would be at any interval due to a version of the Heisenberg uncertainty principle:$$\Delta E \Delta t \geq \frac{\hbar}{2} It basically states we cannot suppress our uncertainty about energy fluctuations in time, ...

0

If atoms have well defined energy levels and those differences correspond to the frequencies of light that can be absorbed, how is it that opaque objects absorb all or most visible light frequencies get absorbed Photons in almost all frequencies hitting an object are absorbed in different ways (absorbed, reflected, refracted, scattered, ...

2

If atoms have well define energy levels and the differences correspond to the frequencies of light that can be absorbed, This is correct how is it that opaque objects absorb all or most visible light frequencies get absorbed and you basically don't have any visible light coming out on the other end? The crux of the matter is the word "objects". ...

0

Tadeas Bilkas answer let me think about the sence of all and all time citing the quantum mechanics. I write his answer in terms of common mechanics and get the same result: You have an emitter of balls which radiates just one single ball but in a spherical area. You place a lot of baskets some meters apart (with same distance) from the emitter. Mathematics ...

2

In case you "run out of photons", you must switch to probabilistic description of quantum mechanics. Let's consider an extreme case: You have an emitter of spherical waves which radiates just one single photon. You place a lot of detectors some meters apart (with same distance) from the emitter. QM says that the photon propagates as a probabilistic wave to ...

2

The classical electromagnetic field given mathematically by Maxwell's equations can be proven to emerge from a confluence of individual photons, which photons are described by the Quantum Mechanical form of Maxwell's equations. Thus the classical wave is made up by zillions of photons with energy $h\nu$, where $\nu$ is the frequency of the classical wave. ...

-1

At very low energy, the difference between photons and E&M waves becomes clearer: After every spin-flip in a hydrogen atom, "something" with E=6 µeV and spin=h/(2*pi) is released. Is that "something" a photon? Nobody can detect a photon with such a low energy. But if you have a radio receiver at 1420 MHz, you get a clear signal. The electric component of ...

1

"Running out" of photons simply means that your wavefront is absorbed or scattered in a different direction or something like that. Either way, the original wave is "consumed", so you loose intensity or photons, depending on which picture you like better. For the case of a single photon source: One photon can only interact with one electron. However, there ...

1

I think you have invented the zone plate, a kind of specialized flat circular diffraction grating that acts as a lens. It consists of a set of concentric transparent zones of decreasing widths. The width you have derived is the diameter of the first zone. Subsequent zones interfere constructively by allowing paths differing by integer multiples of the ...

Top 50 recent answers are included