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Subject:
How a mirror reflects a photon?
Category: Science > Physics Asked by: billa999-ga List Price: $10.00 |
Posted:
24 Aug 2006 09:23 PDT
Expires: 23 Sep 2006 09:23 PDT Question ID: 759112 |
All these are related I believe: What is the physical process by which a mirror reflects a photon? Why are ~5% of photons not reflected (assuming a very high quality first surface mirror?) Why is the photon reflected at the angle of incidence? Are any properties of the photon changed by reflection? Is there any limit to the number of photons that a mirror can reflect simultaneously (assuming multiple light sources illuminating a mirror from all possible angles?) What experiment (documented) has proven the answer to these questions? | |
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Subject:
Re: How a mirror reflects a photon?
Answered By: hedgie-ga on 25 Aug 2006 00:49 PDT |
Photon has zero REST mass, but it has a momentum. Magnitude of this vector is E/c = h *ñ /c where ñ stands for Greek letter ni meaning frequency, direction is direction of travel. When photon hits a conductive surface (interface between glass and silver coating) the two components of this vector behave differently: normal component (perpendicular to the mirror surface) changes signs tangential component (parallel to the mirror surface) is unchanged Result is reflection according to the Snell law: http://acept.asu.edu/PiN/rdg/reflection/reflection.shtml Here is a video of actual experiment On a flat mirror, the angle of incidence is equal to the angle of reflection. http://www.wfu.edu/physics/demolabs/demos/avimov/bychptr/chptr9_optics.htm This mechanism is same as when reflection of a ball from a rigid surface: http://www.lhup.edu/~DSIMANEK/ideas/bounce.htm During the reflection (bounce) the momentum is transfered to the mirror (massive wall) that give rise to radiation pressure http://www.u.arizona.edu/~lilley98/ That is simple case. Things can get more complex when we consider real; materials (which are not perfectly conductive or rigid) and spin: http://www.physics.usyd.edu.au/~cross/Gripslip.pdf. Photon also has an angular momentum or spin, which is manifested as polarization of light. As you probably know, light (as other elementary particles) has particle and wave properties, and for some phenomena (photo-effect) the description by particle model is more suitable, for other (reflection is one example) the wave model is more suitable. For some effects, the complex math of QED is required. simple expose of QED http://en.wikipedia.org/wiki/Quantum_electrodynamics more complex http://hyperphysics.phy-astr.gsu.edu/hbase/forces/qed.html popular books http://www.amazon.com/gp/richpub/listmania/fullview/1SYJYH1NOKUW0/102-6271296-4091321?%5Fencoding=UTF8 For most phenomena of reflection, QED would be an overkill. When the thickness of the optical layers is comparable to the wavelength of the light, such as in antirefelction coatings of lenses http://en.wikipedia.org/wiki/Optical_coating very complex phenomena can be adequately described by (simple, matrix) calculus based on the wave picture of light. (note the picture of lambda/4 reflection in the above article). That (thin film theory) is important engineering discipline based on Maxwell equations https://www.omegafilters.com/index.php?page=tech_fildes_tft (see technical articles) which also describes non-perfect materials, which absorb certain proportion of light (in case of ordinary mirror at room temperature - about 5% photons get absorbed instead of reflected, in case of 'semi-transparent' (half-slivered) mirror, some get refracted. http://www.williamson-labs.com/optical-body.htm Let's se if I covered all of your sub-questions: 3) Are any properties of the photon changed by reflection? Yes. Momentum, and sometimes spin. 4) Is there any limit to the number of photons Well, not really, in general. However, if you shine too much light on the mirror - it will heat up (by absorbed photons) and that may change its properties. More complex is entanglement: Normally photons do not interact which each other (there are bosons, not fermions) -- Maxwell Equations are linear. But, if photons come from same coherent source (a laser beam, split in two, and brought together again, there could be interference). See SEARCH TERM : interference of photons It can get quite complex and in some cases QED has to be invoked. Here are few technical articles on this http://www.citebase.org/abstract?id=oai%3AarXiv.org%3Aquant-ph%2F0208174 ------------------------ Don't miss this paper from the Zeilinger group: " R. Kaltenbaek et al, PRL 96, 240502 (2006). In particular, read the first paragraph: Is it possible to observe fully destructive interference of photons if they all originate from separate, independent sources? Yes, according to quantum theory. The perfect interference of photons emerging from independent sources cannot be understood by the classical concept of the superposition of electromagnetic fields but only by the interference of probability amplitudes of multiparticle detection events. As stressed by Mandel ??this prediction has no classical analogue, and its confirmation would represent an interesting test of the quantum theory of the electromagnetic field?? .." --------------------- http://www.physicsforums.com/archive/index.php/t-124474.html There is number of sites, of varied depth and complexity, which illustrate these processes, as #42, #46, #48, #68 in this following list http://www.cordonline.net/laserapplets/ If something is not covered, please post an RFC. Else, rating is appreciated. Hedgie |
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Subject:
Re: How a mirror reflects a photon?
From: sorwin-ga on 24 Aug 2006 15:28 PDT |
I understand that photons have mass, hence are influenced by gravity. When a photon is reflected by a mirror, I wonder how much force is exerted on the mirror due to the change in momentum of the photon ? |
Subject:
Re: How a mirror reflects a photon?
From: qed100-ga on 24 Aug 2006 16:39 PDT |
Hello sorwin, Photons have no mass. But they fall in gravitational fields nonetheless. Remember the findings of Galileo: objects in a given gravitational field fall with identical accelerations independent of their masses. This result can be computed directly from Newtonian principles. Newton's force law: F = ma (force is proportional to the product of mass & its acceleration) Newton's gravitational law: F = GMm/r^2 (force is proportional to the gravitational constant, G, the masses of two interacting objects, M & m, and inversely proportional to the square of their distance, r^2) So, GMm/r^2 = ma. Dividing both sides by m yields GM/r^2 = a. This is consistent with gravity as determined by general relativity. In GR a region with momentum-energy content greater than its surroundings will induce differentiation in the geometric properties of space. External objects immersed in this distorted space will accelerate according to the properties of space right there where they are. Gravity is then due to local conditions, rather than to the distant Earth, etc. In this theory it's absolutely required that light, without mass, will fall in a gravity field just like any other object. But even though the photon is massless, it still carries momentum, p = h/lambda, with h = Planck's constant, and lambda = the wavelength. So a light interactant material will be subject to the momentum of light incident upon it. |
Subject:
Re: How a mirror reflects a photon?
From: qed100-ga on 24 Aug 2006 16:57 PDT |
billa999, I suggest you read Feynman's book, QED: The Strange Theory of Light & Matter. Keep in mind that light reflection is a classical approximation. For large numbers of photons (intense light), the angle of incidence will tend to equal the angle of reflection. But on the atomic scale, a single photon will be absorbed by a single orbital electron bound to a single atom, raising the electron to a higher energy orbital. A very short time later the electron will re-emit a new photon, surrendering energy and falling to a lower orbital. The direction taken by the photon is entirely random; it could scatter anywhere. But large numbers will interfere with one another. Their frequency phases will superpose both destructively & constructively. Depending upon the overall configuration of the system, the sum total of all these interfering phases will average out to the gross behavior of light with which we are familiar on the large scale. But for more in-depth insight to this whole scheme of things, read the book. |
Subject:
Re: How a mirror reflects a photon?
From: qed100-ga on 25 Aug 2006 05:35 PDT |
It should be noted that, classically, the angle of incidence ought to be expected to equal the angle of reflection, due to conservation of momentum. Momentum-carrying light, incident upon a reflective surface, has a component of motion parallel to the surface, and that component has its own component of momentum. It's not surprising then that it's observed (on the classical scale) to scatter off the surface at an angle which carries that same magnitude of momentum. In fact, the classical conservation laws, including for energy, are generally consequential to the quantum mechanical nature of interactions. |
Subject:
Re: How a mirror reflects a photon?
From: billa999-ga on 25 Aug 2006 11:18 PDT |
I think mirrors do just fine one photon at a time. For example the deep space Hubble image. Taken over many days were only seeing the occasional photon but the image was most certainly formed. |
Subject:
Re: How a mirror reflects a photon?
From: qed100-ga on 25 Aug 2006 17:00 PDT |
Hi billa, Let's say that an astronomical object is so weak as its image reaches Earth that not more than one photon interacts with the Hubble mirror at any one instant. Quantum mechanically, each photon can scatter in any specific direction at random. But not all directions will have equal probability. There will be a spectrum of possible scattering angles, with a narrow interval containing those with the greatest likelihoods. A single photon may scatter in any direction. But for many scatterings of many photons, the average scattering angle -the angle of reflection- will tend toward the classical limit. |
Subject:
Re: How a mirror reflects a photon?
From: billa999-ga on 26 Aug 2006 05:02 PDT |
Hmm, that is something to consider. Suppose we did something like this: Have a small half sphere with the inner surface coated with a photo sensitive film. Place this on top of a first surface mirror. Put a light source at the top of the sphere so photons, one at a time would enter, hit the mirror, and be reflected directly back to the source. Now some claim that some photons will be randomly reflected in some other direction. These photons should hit the photosensitive material. Assuming we continue the experiment until we get quantifiable results when the photo sensitive film is developed, what will the results be? A uniform number of the random photons hitting all areas? More hitting near the light source? What will we see? |
Subject:
Re: How a mirror reflects a photon?
From: qed100-ga on 26 Aug 2006 11:43 PDT |
Hi again, This sort of experiment has been done. Not necessarily with the exact apparatus you describe. But experiments have been done in which low intensity light -one photon at a time- is allowed to generate an image on a detector. The most common variant of this is diffraction of light through either a tiny hole or thin slit. Intense light is, as one expects, diffracted, generating a diffraction pattern on a detector (for example, a photographic film, or a charge-coupled device). The pattern for the small hole is an "Airy" pattern; concentric rings of alternating light & dark. For the slit it's parallel bands of light & dark. In experiments such that the source is so weak as to emit only one photon at a time, a photon may pass through the opening on its way to the detector. The photon then registers at some location on the detector. Let's say that we only expose the detector 'til exactly one photon registers. Let's also say that we repeat this many times, one photon each. We'll end up with a large number of images of single photon exposures, distributed randomly. If we then superimpose all the single images to display them in one, cumulative image, the distribution becomes obvious: the single photons choose their paths randomly, but the overall distribution for many will clearly be that of the classical diffraction pattern. In fact, if you allow a detector to be exposed to an increasing number of single photon events, it'll accumulate the same sort of distribution. As the number of photon exposures increases, the distribution tends toward the classical limit. Now I don't know off hand if the precise sort of arrangement you've proposed has been done. But it has been done with diffraction as I've described, and yours basically starts out with diffraction (at the entry hole). So I'll say with confidence that your arrangement will tend to generate a classical distribution in the same way. |
Subject:
Re: How a mirror reflects a photon?
From: willh-ga on 05 Sep 2006 13:07 PDT |
". . . Interestingly, all Hubble images are created with black-and-white cameras. Ones and zeros are sent to Earth. Color is dropped in later with the popular Photoshop program. http://www.space.com/scienceastronomy/astronomy/color_universe_020625-1.html |
Subject:
Re: How a mirror reflects a photon?
From: qed100-ga on 05 Sep 2006 14:43 PDT |
I'm not sure what the color separations in an astronomical photo have to do with the original question, but whatever. When that article mentions that the colors are "dropped in later with Photoshop", it doesn't mean that the colors are applied randomly or arbitrarily. The image sensors on the Hubble telescope are monochrome, but each image is photographed three separate times, each through its own primary color filter. What the CCD then registers is a field of intensities in a single elemental color. With three such images, each filtered for one primary color, a faithful full color image can be created. The data are downloaded to Earth, and the original color field registered by each filtered image can be reproduced by simply instructing the image editing software to color all pixels the color which was admitted by the filter on the telescope. Three color separations are produced in this way, which can then be composited to make a full color image. This is basically the same thing that higher quality digital video cameras do. They pass the image through a beam splitter, which then distributes the original image to three separate CCDs, each registering its own primary color. This maximizes the control which can be exerted over the color balance in the video. |
Subject:
Re: How a mirror reflects a photon?
From: rawbmobile-ga on 13 Oct 2006 10:25 PDT |
qed100 and billa999 Just a couple of corrections: qed you almost got it, where you went slightly wrong is the statement of "a single electron in an orbital ..." to paraphrase you. Metals are characterized by overlapping energy states in the outer orbitals which smear into bands because of quantum mechanics. Because of this a given electron is not associated with a given atom, a more accurate model is that these electrons smear out and form a conductive plane. the electrons are confined by the work function of the surface but are available to interact with photons. If the photon has sufficent energy, it will liberate an electron (einstein won a nobel for this work), if it doesn't the energy is then kicked back out in the form of an photon. Energy is conserved. Some energy is lost due to scattering and electron interaction whilst it is excited. This is more of a quantum mechanical view. A more classical view is that the metal has a complex (as in using imaginary numbers) index of refraction which is causes reflection. As to mass and momentum and gravity... This is kinda hard to get ones head around. Photons are massless but have momentum, it is ultimately tied up with why mass has energy. But to keep on topic, photons are strictly not affected by gravity. However, gravity (through Einsteins general theory of relativity) does affect and can warp spacetime, which photons notice. This is the effect of gravitational lensing that you no doubt have heard about as a proof. Good question billa999, a seemingly simple question can reveal some subtle aspects. Hope this helps |
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