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The Sun is white, but the reason why it is white is because your brain adjusts the white balance to make the totality of all the ambient light illuminating the objects you see, white. Now, the Sun is too bright to look directly at safely, but around dusk or dawn you can sometimes see by accident a glimpse of it, and it then looks red. So, based on the ...


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Our sun is white (or rather, it gives off a variety of wavelengths of light that, when combined, appear to be white.) During sunrise or sunset it appears yellow/orange because the air, dust, and pollution cause scattering of some of the waves, leaving the yellow-orange ones. At any other part of the day, say, high noon, the sun actually appears as its ...


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In the dark you have a single Fermi level which is spatially dependent on the net free charges $E_f(z)$. When the semiconductor is perturbed by light the Fermi level splits into two quasi fermi levels for electrons $\mu_e(z)$ holes holes $\mu_h(z)$. The quasi fermi level separation (or electrochemical potential) is, $\Delta E_f(z) = \mu_e(z) - \mu_h(z)$.


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Photons can come from a variety of sources, some of which are indeed transitions of electron levels (or nucleus levels or molecular levels), which are indeed discrete (it is also not because of wobbling). The most common way of having photons of arbitrary wavelengths is via Bremsstrahlung radiation. This is the radiation obtained when a charge is ...


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The inverse square law is "simply" a statement of the fact that a diverging cone of "energy"/particles/stuff will have a cross sectional area that increases with the square of the distance. This is a result of basic geometry and the fact that area is proportional to length squared. As the stuff-beam impacts over an increasing area as distance ttravelled ...


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SUMMARY: This is a very good question. In a lossless medium, fundamentally the answer to your question is "no, an individual ray does not lose energy in propagating" because it represents a plane wave (in photon language, a momentum eigenstate), whose intensity does not vary as it propagates. Intensity information is encoded in the flux density of rays ...


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Here is an example of a multi-slit experiment where you can see the diffraction. It would be very difficult, if not impossible, to detect whether light coming around a corner was there because of diffraction or because you were actually looking at the light source. However, if you set up a slit experiment, then shine a light through the hole, you could ...


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A study undertaken by Nutting and Nuttall at the University of Leeds found that "gold is not inherently more ductile than other face-centered cubic metals", such as copper. The authors found by experimentation that "gold is considerably less ductile in tension than silver." But when beaten foil becomes very thin, other metals tend to fragment, whereas gold ...


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One single photon is not enough for you to see objects. If there would be just one photon coming to your eye, you would first see only darkness (before the photon arrives), then you would notice a dim point for very short time (when the photon hit your retina), followed by darkness again. You can see objects because there is a constant flow of huge amount ...


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A photon is an elementary particle with zero mass, moves always with the velocity of light c, and has energy given by E=h*nu . It has spin +1 or -1 and its wavefunction also has a polarization which will build up the polarization of the emergent from many photons classical electromagnetic wave. Its energy is a property that characterizes the emitting ...


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I don't think Robotex's question on "counter" has been answered . It is more difficult to put a "counter" than to simply cover one of the slits. the counter experiment is more interesting to observe, I think. If somebody can suggest a home made electron beam, that would be super. great


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I don't think you can draw such conclusions from a simple graphic. They've probably just cut it off at somewhere around 200nm because the power output at shorter wavelengths is almost irrelevant. If you want a more accurate graph for shorter wavelengths, try this one. You can set it up for 5900K and graph values below 240nm.


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It seems to me that the person drawing the graph was a bit sloppy - the ideal black body radiation ("idealer Schwarzer Körper" - Temperatur 5900 K) does not cut off sharply at 240 nm as shown. Instead, it should look like this: when calculated from Planck's Law. I suspect some bug in the method used to calculate the values in the plot you reproduced - ...


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When you pay attention to the left or right area of your double slit experiment you will see at the end the typical intensity distribution of a single slit. So a multi-slit arrangement is nothing all as the sum of two single slits. Of course the intensity distribution depends from the distance between the two slits. Artfully created, one would see a clear ...


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The obvious solution is to aim the torch at the second hole, tho' due to the angular spread you'll still dump most of the light. Replace the torch with a laser pointer and aim that at the second hole & nearly all the light will get thru. You might consider a light pipe such as a large-diameter optical fiber, plugged into both holes so that all light ...


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I think any rods-and-cones explanation, such as the Purkinje effect, for the perceived colour of the Moon itself is nonsense, simply because the light reflected from the (full, overhead) Moon is easily bright enough to be perceived by the cones in our retina. As you say, if the contrary were true, it would also prevent us from seeing the Moon as orange when ...


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Think about how electromagnetic radiation is generated. Electrons and protons emit photons. The wavelengths of this photons depends from the surrounding temperature, for the protons it depends from the chemical elements (H, He, ...., C). Furthermore it depends from the gravitational potential and the velocity of gas streams. Perhaps this are not all ...


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First of all, a celestial body could appear blue for a simple reason: it may be blue, indeed. This is the case of Neptune. Its being blue simply means that the probability that a photon is reflected off Neptune is higher if the frequency (color) of the photon is basically blue than when it is green, yellow, or red. However, the Moon isn't actually blue. The ...


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The quick answer to your question is this : TV remotes typically operate in the infra red (IR) and our eyes are not sensitive to IR but camera sensors are. I guess you are intrigued by the fact that when you operate a TV remote facing a camera, you can typically see a red light on the IR diode of the remote in the camera screen, whereas with naked eyes you ...


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A common wavelength for an IR emitter for a remote control is 980 nm. The human eye can only detect wavelengths up to about 725 nm efficiently, so we don't see the beam from the IR remote. (In the case of an 850 nm IR signal you may see some residual emission within the human sensitivity range). Cameras, on the other hand, typically are made with silicon ...


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I may not be understanding the source of your difficulty. There are three facts here. First, the speed, frequency and wavelength are related as $v = \lambda \nu$. Second, the frequency of light remains the same when crossing the interfaces between media. This is a consequence of ensuring that the continuity conditions implied by Maxwell's equations are ...


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When there is a resonance in electron response, both refractive index and absorption change rapidly: the refractive index has a "jiggle" in the vicinity of the resonance, like this sketch (adapted from this earlier answer by John Rennie - but I disagree with the "n=1" label so I cut it off...: As you can see there is higher refractive index at the low ...


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Although there are already some excellent answers, I believe they are a little complex. Please allow me to offer a simplistic answer. Let me start with the analogy of sound waves and the ear. The sound enters the ear and causes certain cilia to vibrate in response to the frequency and amplitude of the sound wave. Similarly a photon (as a wave), enters the ...


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The hydrogen discharge tubes typically used in student labs are not designed for long-term use. After a couple of years, the tubes leak and air gets mixed with the hydrogen. This causes them to get dim and the weaker lines are almost impossible to see. It has nothing to do with the power supply and everything to do with how new the tube it and how many ...


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Photons are energy. When a photon hits your retina, that energy is absorbed and converted to electrical energy in your optic nerve.


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Light from all over the place hits your eyeball fairly randomly. The lens forces light from a specific angle to hit a specific part of the retina. This HowStuffWorks article shows how the mechanics of that work. The only major differences between camera lenses and eyeball lenses is that we can dynamically alter the shape of the lens to focus on different ...


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Photons can be created and destroyed freely, since they don't have charge or mass. Turn on a light, and you create many photons. Any body (made of atoms) not at absolute zero temperature will spontaneously emit photons. They are consumed just as easily. Most any bit of bulk matter will absorb a photon in the electrons on the surface, transforming the energy ...


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Imagine a spring-loaded trap with a hole that's sized such that only a particular size of object can enter the hole and trigger the trap. The molecules involved in vision are like that trap, with a bond having an electron energy gap tuned to the visible frequencies of light, encapsulated in a specialized protein that transforms the absorbed energy into a ...


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Shortly, the energy of the photon goes over to the electron. But energy is a vague concept. In material sense, could the photon, or better, the electron's electric field and the electron's magnetic field be quantized? I developed a model with two different quanta. Photons, electrons, positron's, protons, neutrons, ... are made from this quanta. Photons are ...


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From the wiki article on color vision as an illustration of how photons are absorbed: Perception of color begins with specialized retinal cells containing pigments with different spectral sensitivities, known as cone cells. In humans, there are three types of cones sensitive to three different spectra, resulting in trichromatic color vision. Each ...


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You cannot increase the amount of energy that the lamp has already emitted. Not unless you add more energy in some other way. For typical household lamps the easiest way to double the apparent illuminance of a specific area (a newspaper for example) is to note that some light is being emitted in directions that do not directly illuminate your desired area. ...


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You need to increase the power into the light. How easy that is depends on the light. If you have a 60W incandescent, the easiest way is to buy a 100W bulb. Depending on the fixture, the extra heat may be a problem. If you have a CF or LED bulb, you may again be able to find a higher wattage equivalent. For a standard incandescent, without replacing ...


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The answer to your question is the obverse of it: we assign a color to an object based on the wavelengths which are reflected to our eyes (or in the case of filters, transmited to our eyes). That means other wavelengths are absorbed. The absorption of wavelengths is based, primarily, on the chemistry of the object. Red dye applied to cotton cloth is a ...


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Lenses can exhibit chromatic aberration due to a color (wavelength) dependent index of refraction in the lens material. Chromatic aberration can cause the three colors from the red-green-blue (RGB) display to separate when viewed off the lens axis: Left: on axis. Right: off axis. In less pronounced cases, this might lead to a ghosting of images. ...


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Confusion arises when one has not learned that color perception and frequency associated color are two different things: the first is connected with biology and the second with physics. Perception of color begins with specialized retinal cells containing pigments with different spectral sensitivities, known as cone cells. In humans, there are three ...


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There are (at least) a couple problems with your proposals: In experiment A, you start with "a laser ... which can be controlled to emit individual photons." However, such a laser is, as far as I know, impossible to make, because the stimulated emission process that generates the output in the laser is a random process, emitting photons at random times. ...


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One has explored radio waves experimentally. It was found, that outside the emitter (antenna rod) exist both a swelling magnetic field and a swelling electric field. The electric field was parallel to the rod, the magnetic field perpendicular to the rod. This radio waves are electromagnetic radiation, but they are a special case of EM radiation. The most ...


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You can't tell directly. But you can look at a bunch of it and notice that it is at the same temperature here there and everywhere in all directions in every single place where your view isn't blocked by some moon planet star or galaxy. And that temperature is quite cold it is hard to get something that cold. And either there are many things with the same ...


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Distance is equivalent to time. The time at which the cosmic microwave background was emittied was the time when the universe made a phase transition from being a plasma to being atomic matter. During this time, the universe finally became transparent. Before this time, the universe was so hot that all matter was opaque. The CMB is the wall that ...


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I am hypothesizing based on the information you gave in the question and clarifying comments. The fact that the distance between the "shadow" (or ghost) and the number increases with distance tells me that the shadow is produced by your eye+glasses, rather than the screen itself. Most likely, you are seeing reflections from the front/back surface of your ...


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One way to think of this is to imagine a high-speed camera which takes 1000 frames per second (which is 40x as many as a regular film camera). Then imagine putting the frames into groups of 40, and averaging each group, so that you now have a film with a normal framerate, where each frame is the average of 40 high speed frames. The resulting film would be ...


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The problem is that images aren't formed by one ray of light, they're formed by countless rays of light! A "bundle" of light from the sun ("bundle" meaning a ray localized in space but containing many many photons) will hit the surface of a blade of grass, and all the photons will disperse in a diffuse manner in every direction. So a bundle of light hits a ...


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You have picked a very striking picture to illustrate your question, although the northern lights do not always have such a sharp lower border to them. However, let me advance a plausible explanation. The green northern lights are formed high up ($\geq 100$ km) in the Earth's atmosphere, largely by photons at 557.7 nm emitted from excited oxygen atoms. This ...


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The answer is that your view or sight is different from the bare images made by the imaging optics of your eye on your retina. A "view" also includes signal processing from the brain that tracks what you fix your gaze on. Light can pass between the blades and form an image of the retina for at least some time. It's true that there are also blades blocking ...


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Because human eyes and brains are slow, they cannot resolve the motion of the blades, but only see the average of the moving blades and the image in the background (this is actually primarily really due to the slow reaction time of the cones, which is slow, as is demonstrated by the fact that a 24 frames per second video does not appear as single images but ...


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The eyes are measuring the number of photons of each color that are hitting a given point of the retina – that are coming from some direction. This is a function of time, $f(t)$, for each point. However, when this function is changing too quickly, the eye can't see the changes. Effectively, the eye may also see the average of $f(t)$ in each period of time ...


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A slightly different take on your question with more emphasis on the thermodynamics of light. If the body emitting light has a well defined temperature and optionally the electronic charges in the material have a well defined chemical potential. The emitted light can be represented as a photon gas which has related thermodynamic properties just like an ...


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Thermal radiation is emitted by any surface having a temperature higher than absolute zero. So the short answer to your question is yes. Light (electromagnetic radiation) of any frequency will heat surfaces that absorb it. In case of Fluorescence, the emitted light has a longer wavelength (lower frequency), and therefore lower energy, so that's why you feel ...


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There is a sort of duplicate answer here Heat into light and perhaps by reading it, you might get a better picture of how the spread of electromagnetic frequencies process works. I am not sure that it answers your question directly however, so hopefully this answer, or a better one that comes later, will fill in any blanks. When visible light is ...


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You've got your work cut out for you, so this answer can only point you in a direction. This calculation can be difficult because there is a profusion of pieces of terminology with subtly different meanings. Find a book/other source (but for the straight dope I suggest a book) which discusses radiometry and the differences between radiant flux, radiance, ...



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