# Tag Info

## New answers tagged visible-light

0

Almost every digital camera in existence nowadays uses an IR filter. Just google it. Here's an example from the folks at Raspberry Pi: http://www.raspberrypi.org/whats-that-blue-thing-doing-here/

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The wire is a cylindrical reflector. The laser light that hits the top of the wire is reflected upwards; the light that hits the side is reflected sideways. This is simple reflection - no need for a diffraction explanation.

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Light has properties of waves and particles and doesn't necessarily travel in "straight lines". There is an interesting experiment (I will look up the original scientists who did it and post that info here) where they set up basically a cardboard circle in front of a wall and then shine a flashlight at it. Interestingly enough, there is a shadow in the ...

1

The colour you're seeing is from the very small fraction of light that the panels are reflecting. The vast majority of light is being absorbed to generate electricity. Why some of the panels appear slightly blue while others don't I don't know. Presumably there must be small differences in the manufacturing process. The absorptance of solar panels does fall ...

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The solar panels pictured convert light to electricity. If they were perfectly efficient then they would absorb all wavelenghts including available ultraviolet and infrared as well as visible and convert them all to electricity. In reality solar cells will be designed to absorb one set of wavelengths, though sometimes you might have sandwiches where one ...

2

This is the plot of sunlight, red at ground level. Solar irradiance spectrum above atmosphere and at surface. Extreme UV and X-rays are produced (at left of wavelength range shown) but comprise very small amounts of the Sun's total output power. As all light comes from the sun during daylight this should suffice. One can get the number of photons by ...

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The formula for the diffraction maxima involves the $\sin\theta$, where $\theta$ is the angle between the (virtual) line from the slit to the zero order maximum and the line fromt he slit to the nth order maximum. Keep in mind that $\sin\theta$ must be less than 1. This will place an absolute limit on the minimum size of the slit. The actual angular size ...

1

It depends entirely on the wavelength of the wave medium you are trying to diffract. If you are talking about light you need a monochromatic source (one wavelength) that has a wavelength in the range of 650-700 nanometers. For maximum diffraction the slit with in the diffraction mask should be of the order of this wavelength.

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I don't think you can understand how a photon is made without knowing what a photon is, and to understand what a photon is requires understanding quantum field theory - specifically quantum electrodynamics. Quantum field theory is a very odd way of looking at the world, but it works and gives predictions that agree with experiment. In quantum field theory ...

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Photons, and in fact all other elementary particles, are not assembled. They exist. We don't know how, or why, they simply do. Photons are the quanta of the electromagnetic field - in quantum field theory, we associate to the classical electromagnetic field particles called photons. For a technical account if how this (roughly) works, read my answer to "The ...

3

Whenever one makes light, one makes zillions of photons, so the way to make photons is the way to generate light: burning wood etc, striking stones, make wires incandescent, as in electric bulbs, etc, fusing nuclei as in the sun and stars and maybe more ways. It is much easier to make zillions of photons. How one photon is made takes us to the realm of ...

3

Your wording suggests a few misconceptions: It seems you are thinking of light as having a corpuscolar nature (nothing wrong with that, you are in good company). Well it turns out that things just do not work that way. Phenomena like diffraction (to name one) tell us that we cannot describe the behaviour of light thinking of it as composed by (classical) ...

1

There's a few things to disentangle in this question. Let's talk about photons first. It is not true that a "light photon" travels in a wavelike pattern. If you talk about light in terms of photons the (basic) picture you get is light as billiard balls, moving in straight lines. This is how ray tracing and its ilk work, because (in many situations) you can ...

0

It depends on what you mean by sweeping. If by "sweeping" you mean motion in cross-range (orthogonal to radial) then at least in the non-relativistic regime there should not be any Doppler shift because (the length of) the range transmitter/reflector/receiver is constant, that is the number of wave-crests per unit time does not change. To see Doppler shift ...

0

This is an effect known as time dilation. In this post, I will be taking material from the excellent book, Einstein Gravity in a Nutshell, by A. Zee. Figure 1 will be the basis for the argument. We bounce a photon around to create a clock. It is postulated that the speed of light is the same in all frames. In the rest frame, the time it takes for light ...

0

You do not have to shoot the photon at an angle in your example, this is a result of the principle of relativity. The photon appears to travel diagonally to an outside observer (an observer at rest relative to the moving mirrors). If you move along with the mirrors the photon will still appear to move perpendicular to the mirrors. You can also imagine ...

2

The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in ...

0

The window will not protect you. Due to the Maxwellian distribution of thermal energies of the weapon plasma, a significant fluence of visible light can be transported through the atmosphere, the window, and your eye prior to cataract generation. So while the window can block the infrared, it won't block the visible. Nuclear weapons generate thermal ...

2

Without GR, there is no reason for Mercury to not be tidally locked with the Sun, always showing the same face to the Sun (much as the Moon always shows the same face to us). There is also no reason for clocks at higher elevations to desynch with clocks at sea level. If you take these as strong evidence for general relativity, then all that the authors' ...

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Basically, absorption lines exist because absorbed photon are not re-emitted in the same direction, so dark lines can be observed. There are various reason causing this. For example, the extra energy can be dissipated as phonon in solid or strongly interacting system. Excited states can also emit multiple low frequency photon if there are meta-stable ...

2

It is not correct saying that no force is applied. A photon carries momentum see PE here so on reflection there is momentum transfer. This is the idea behind laser propulsion discussed here. Concerning the speed it is even more complicated. The fact that light gets reflected usually requires an abrupt change in the index of refraction. To get reflected, ...

2

To my knowledge, there is no discretization of the light wavelength (they form a continuous spectrum). On the other hand, there exists no infinitely narrow absorption "potential". I mean that all transitions of electrons that may correspond to a photon absorption have a finite width. Consequently, the photons have a non-zero probability to get absorbed. ...

1

As you can see the spectrum at the top of the atmosphere is continuous, with some saw tooth excesses, but still continuous. The absorption does create a saw tooth pattern, even so there is continuity. To dispel doubts here is the sun spectrum showing continuity and absorption spectra Solar spectrum with Fraunhofer lines as it appears visually. ...

1

There are two things. More photons means a brighter beam. power (Energy/sec) is proportional to the number of photons/sec. Photons with shorter wavelengths and higher frequencies have more energy. That is a bluer beam has more power. So $P = nh\nu$ where n is the number of photons/sec.

1

No. The only effect that the red color would have would be to attenuate all the colors that are not red. With white light in, you'll still get a spectrum out, but the blues and greens will be dimmer than they would be with a clear prism. (The positions of the colors might be shifted by a very small amount because absorption comes with a small change of ...

0

Before the photon is absorbed, i.e. whilst the photon quantum field is in a pure, one photon state, you can define the following probability amplitudes that uniquely specify the one-photon state: $$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\psi\right>\right.\\ ... 0 You are right that the light travels from a place in the far past. However, the light itself does not experience the passing of time. In relativity, there is an important distinction to be made between massive objects and massless objects. For massive objects you can always find a frame of reference where they are at rest. In this frame time passes at "a ... 0 The way to measure time dilation would be to compare the time the moving object measures (its proper time) to the proper time of an observer. The problem with light is that proper time is defined as:$$d\tau=\sqrt{-g_{\mu\nu}dx^\mu dx^\nu} $$In flat spacetime g=\eta=\text{diag}[-1,1,1,1]. Now, with light, the line element g_{\mu\nu}dx^\mu dx^\nu=0 so ... 0 In special relativity, time dilation is determined by the Lorentz factor$$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$Say the photon travels some time T in our frame of reference. It is obviously travelling at the speed of light. We see that the Lorentz factor diverges as we go to the speed of light. Thus, in order for the photon's time in our frame to remain ... 3 The laser goes through the bagel hole each time until it hits the ground (assuming the mirrors are set up nicely orthogonal to your uniform gravitational field). To see why, the equivalence principle is all we need. You can imagine thrusters powering an elevator without gravitational fields, set up your mirrors and do the experiment, and that's what you ... 2 … photons undergo twice the deflection from gravitational fields as do physical objects. I don't think this is a correct assessment of the situation. Rather, there is a Newtonian way to predict the deflection of light due to gravity: assume that the light is made of corpuscles with effective mass m=E/c^2 which enter the gravitational well with speed ... 0 Maybe the explanations you got until now were sufficient, but I would just want to add something simple. Let's represent a light-wave traveling in the direction x as Ae^{i\phi} where the phase of the wave is (\text i) \ \phi = kx - \omega t = 2\pi\left(\frac {x}{\lambda} - \nu t\right). Consider a wave-front (a surface on which the phase is ... 0 There are a lot of intertwined ideas here. Let me try to tackle just part of it. When a photon interacts with a medium, it causes local polarization - that is, electrons are displaced by the E/M field of the photon. This interaction leads to a slowing down of the wave - and, as you pointed out, a shortening of the wavelength. However, at this point the wave ... 2 I'm not sure if this is worth an answer but anyway- darkness is not a physical entity in itself. It is simply where light isn't. As it is not an entity, it doed not have to go anywhere when you turn on a light, it just seizes to exist as light enters the place where beforehand it wasn't. 2 Darkness is nothing, it is not a physical entity. Darkness is what you have when there are no/very less photons, and thus you can't see anything. When you light a lamp, the photons from the lamp bounce all over the room, and hence the darkness "disappears", so to speak. Absence of light doesn't exist when there is presence of light. 0 I'm not sure if I'm missing something here, but is the answer not c, the speed of light, 186000mph? A photon is a particle of light, but as it has no mass it can travel at the speed of light. As for the second part, yes it can be influenced by gravity. As an example, gravitational lensing (the bending of light) around galaxies, making objects behind appear ... 1 The Poynting vector \vec{N} is the power per unit area of your beam. If the beam is perfectly absorbed, then the force is given by$$ F = \frac{1}{c} \int \vec{N} \cdot d\vec{A}$$So, providing you have the beam incident normally upon something, the force on it will just be the power of the laser divided by the speed of light. Of course, if the light is ... 0 In principle, yes, but for clear air the scattering is very very weak, and the scattered light would probably be drowned out by other background sources, especially the blue sky. It would be easier at night. If the air is not clear, but instead is carrying dust or water droplets or smoke, the beam would be easily visible and recordable, again much more ... 0 Yes The image below shows the Keck telescope's laser guide star. It is designed to excite sodium atoms in the mesosphere. These excited sodium atoms flouresce and act as an artificial star which allows the telescope to correct for optical aberrations caused by Earth's atmosphere. 2 The technology is described in this Wikipedia article. The bottle is acting as a light pipe. It doesn't create any light itself, it just transfers sunlight outside the building into the building. Here in the UK light pipes aren't much used, probably because electric light is cheap and natural light tends to be rather variable. And of course light pipes are ... 3 It would be better to comment this but I only just joined, so no reputation...you'll also have to forgive me if this is a silly idea, I'm not a physicist :), it's just my 2c. For the sake of argument, assume that (from the perspective of an observer on the "ground") the light deflection really is twice as large as the bagel's rate of fall - or at least ... 5 Terrific photo - good that you were able to get it and don't apologize. This diagram might explain it: The sun is "below the horizon" as demonstrated by the green dashed line. A mirage can be formed by rays following the blue line (exaggerated scale, showing a layer where light can be reflected because of a sufficiently large change in density). If the ... 5 Someone else with knowledge of the mathematical details may be able to say more, but I think the statement that the deflection of light is twice the Newtonian value is just looking the initial angle that the light approaches the gravitating body from far away, at comparing with the final angle once long after it has passed the body and is far away again--in ... 1 Light does indeed bear momentum. If you have a quantity E joules of light in a plane wave beam, it has linear momentum \frac{E}{c} in the direction of propagation. So if it bounces off a mirror, the impulse transferred to the mirror is 2\frac{E}{c}, in the direction of propagation, at least at first. Most certainly, the light bounces back. If the ... 1 The radiative transfer equation is a simplified model for describing light transfer. Of course it is possible to derive the radiative transfer equation by the Boltzmann equation for a photon density function f(x,t):$$ \partial_t f(x,t) + v_x \partial_x f(x,t) = (\partial_t f(x,t))_{coll}.  Here, the term $(\partial_t f(x,t))_{coll}$ is the gain and ...

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The light falling on the surface also determines the colour. If a piece of cloth which appears green in sunlight is viewed under red light it will appear black, because the pigments on the surface of that cloth have the property of reflecting only green light absorbing all other wavelengths falling on it. As red light has only wavelength corresponding to red ...

3

In the book "Physics of the Plasma Universe" Dr. Anthony Peratt puts candle flames near the bottom of "energy in electronvolts" portion of the 'plasma spectrum'. If you look at the chart below, you'll see candles flames about midway (ok, cosmologically) between the ends: solar bodies and laser radiation terrestrial flames interstellar charged gases ...

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The red, orange, yellow, and white parts of a candle flame results from glowing soot. The color in this part of the flame is indicative of the temperature. The spectrum in this part of the flame is fairly close to that of a black body. The blue part of the candle flame at the bottom of the flame results from chemiluminescence. Chemiluminescence is not black ...

3

The optical isolator component is active. It consumes energy and so is no different (thermodynamically speaking) from the heat-pump in a refrigerator. If you are talking about a passive component that draws no power then, a surface that allowed light pass in one direction only would violate the second law of thermodynamics. To see why, imagine two rooms, ...

3

If "wet" causes the material to become less scattering, and thus darker, because of a smaller change in refractive index between the fibers and the liquid, then the experiment to do would be to change the refractive index of the liquid. If you can see a change in "darkness" as a function of refractive index (making sure to correct for surface reflection ...

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