where do the photons produced by a lightbulb come from??
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I have no physics degree but I think this might help you.  The answer will vary slightly depending on how contemporary the physics you use to answer it is but here is an inexact answer which should give you the general idea.\n\nSo the first thing you need to know is that all matter is not in existence already.  Matter is energy in a special form where it is sort of tied up in a wiggly knot.  You can see the relation between the two in einstein's\n\nm=E/c^2   (just an alternate form of E=mc^2)\n\nwhere\nm is mass(the property that quantifies matter)\nE is energy\nc is the speed of light\n\nSo from that there is an equivalence between matter and energy.  Matter can be created by knotting up energy, or destroyed by liberating the energy it was made with.\n\nSo, without going into really advanced physics, all mass is not conserved (meaning it can be created or destroyed) but energy IS conserved, there are all kinds of rules involved, but you can interchange the two of them.\n\nWhen you release energy as photons from matter you are converting some of that matter (tied up energy) into energy.  If you could weigh the matter that radiated the photons well enough you would be able to see that it weighs less!\n\nIn "pair production" a photon can create matter; an electron and positron.\n\nAnother example of energy being released from matter is nuclear fission.  Where the matter going in weighs more than the matter after the process.  The difference of mass is the energy freed from the uranium fuel.\n\nIt's like the photon is some of the stuff that the electron is made of... (this is a loose analogy) like some of the waggling rope that the electron knot was made of.  The electron knot can lose some excitement (energy) by shedding a section of the length of the rope(photon) which goes zipping away as pure electromagnetic energy.\n\nAll the abstractions like neutron, proton, electron, photon aren't really such absolute and fundamental things that are seperate and different from each other.  They are more like energy tied up in different ways.
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Energy is measured in watts, so I assume you mean a voltage source of 200v.\n\n(1)\nThe voltage source of 200v applied to the light bulb (laminar) will be 200v.  Since the light bulb is designed for 130v, the supply voltage will likely blow the bulb.\n\n(2)\nHere you have two possibilities, depending on your circuit.\n(a) If you have a parallel circuit, then each light bulb will have 200v across it, and again the bulbs will likely blow because the voltage is to high.\n(b) If you have a series circuit, the voltage across the light bulbs will be subject to their resistance.  Assuming the resistance is the same, then the voltage will be the same, that is, 100v each.\nThe voltage across each light bulb will be equally proportional to it's resistance ratio.
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You mean fireflies?  They produce light through a process called bioluminescence.  Check the web page cited below for the exact answer to your question.
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I'd say option B. Each photon contains one fragment of the image. But actually, there are no such things as photon particles. The "waves" that "photon particles" create are actually part of the photon themselves.
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http://en.wikipedia.org/wiki/Light_bulb#History_of_the_light_bulb
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Photons are produced by atoms when a bound electron moves from one orbital to another orbital with less energy. Photons can also be emitted by an unstable nucleus when it undergoes some types of nuclear decay. Furthermore, photons are produced whenever charged particles are accelerated.\n\nIn physics, the photon (from Greek φως, "phōs", meaning light) is the quantum of the electromagnetic field, for instance light. The term photon was coined by Gilbert Lewis in 1926.\n\n \nThe photon can be perceived as a wave or a particle, depending on how it is measuredThe photon is one of the elementary particles. Its interactions with electrons and atomic nuclei account for a great many of the features of matter, such as the existence and stability of atoms, molecules, and solids. These interactions are studied in quantum electrodynamics (QED), which is the oldest part of the Standard Model of particle physics.\n\nIn some respects a photon acts as a particle, for instance when registered by the light sensitive device in a camera. In other respects, a photon acts like a wave, as when passing through the optics in a camera. According to the so-called wave-particle duality in quantum physics, it is natural for the photon to display either aspect of its nature, according to the circumstances. Normally, light is formed from a large number of photons, with the intensity related to the number of them. At low intensity, it requires very sensitive instruments, used in astronomy or spectroscopy, for instance, to detect the individual photons.\n\nPhotons are commonly associated with visible light, but this is actually only a very limited part of the electromagnetic spectrum. All electromagnetic radiation is quantized as photons: that is, the smallest amount of electromagnetic radiation that can exist is one photon, whatever its wavelength, frequency, energy, or momentum. Photons are fundamental particles. They can be created and destroyed when interacting with other particles, but are not known to decay on their own.\n\nPhotons have zero mass and zero electric charge, but they do carry energy, momentum and angular momentum. Photons are always moving, and photons in a vacuum always move at the vacuum speed of light with respect to all observers, even when those observers are themselves moving. The energy and momentum carried by a photon is proportional to its frequency (or inversely proportional to its wavelength). This momentum can be transferred when a photon collides with matter. The force due to a large number of photons falling on a surface is known as radiation pressure, which may be used for propulsion with a solar sail.\n\nThe actual upper limit of the photon rest mass compatible with observations depends on additional assumptions. A conservative consensus limit of the Particle Data Group is 6×10-17 eV based on changes in magnetohydrodynamics which would contradict solar wind observations.\n\nPhotons are deflected by a gravitational field twice as much as Newtonian mechanics predicts for a mass traveling at the speed of light with the same momentum as the photon. This observation is commonly cited as evidence supporting Einstein's theory of gravitation, general relativity. In general relativity, photons always travel in a "straight" line, after taking into account the curvature of spacetime. (In curved space, such lines are called geodesics).
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NO.  When the lightbulb burns out the filament breaks.  The filament is what the electricity flows through to create light.  When this breaks there is no path for current to flow, therfore no electricity is used.
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What is a bundle of light energy that acts like a particle called?  Photon\n\n How does a flourescent light work?  An electical potential creates a plasma of a gas.  These excited gases emit light.\n\nWhat do we call light that is produced from heat?  infrared\n\nWhat are some uses of ultraviolet light?  Sterilization. Inducing chemical reactions.\n\nWhat electromagnetic waves can we see?  This is known as the visible spectrum.\n\nWhat controls the amout of infrared radiation given off of an object? Temperature\n\nWhat are infrared waves? If you book says heat, it is wrong.  Infrared is no different than any other EM wave.  It is the part of the EM spectrum that can excite vibrations in molecules.  It just so happens that the black body radiation given off from the temperatures of objects seen on earth is mostly in the IR.  \n\n How can a microwave heat your food and not your plate?  Microwave oven excite rotational levels in water.  The plate does not have water and is not heated.
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Plants use a very small portion of the light generated by a standard incandescent bulb.  They sell grow lamps which generate a wider spectrum of light that plants can use.
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Light pushes electrons outs of the atoms in the cell making electricity
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By electrickery of course! Nah basically in layman's terms the electrical energy passing through the gas inside the tube causes the gas to give off photons of light. There is no wire element getting hot and glowing, just gas. Different gas equals different colour light think of "neon lights". Others stuff is also put in along side the gas such as metal elements and often sodium. Street lamps are usually "sodium" bulbs. If you want to find out more use the term "gas discharge lamp" or similar and you'll be more successful in your searches.\n\nthis is worth a look http://en.wikipedia.org/wiki/Gas_discharge
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See "How Light Bulbs Work":\nhttp://science.howstuffworks.com/light-bulb.htm\n\nThe full answer is:\n\nLight bulbs have a very simple structure. At the base, they have two metal contacts, which connect to the ends of an electrical circuit. The metal contacts are attached to two stiff wires, which are attached to a thin metal filament. The filament sits in the middle of the bulb, held up by a glass mount. The wires and the filament are housed in a glass bulb, which is filled with an inert gas, such as argon.\n\nWhen the bulb is hooked up to a power supply, an electric current flows from one contact to the other, through the wires and the filament. Electric current in a solid conductor is the mass movement of free electrons (electrons that are not tightly bound to an atom) from a negatively charged area to a positively charged area.\n\nAs the electrons zip along through the filament, they are constantly bumping into the atoms that make up the filament. The energy of each impact vibrates an atom -- in other words, the current heats the atoms up. A thinner conductor heats up more easily than a thicker conductor because it is more resistant to the movement of electrons.\n\nBound electrons in the vibrating atoms may be boosted temporarily to a higher energy level. When they fall back to their normal levels, the electrons release the extra energy in the form of photons. Metal atoms release mostly infrared light photons, which are invisible to the human eye. But if they are heated to a high enough level -- around 4,000 degrees Fahrenheit (2,200 degrees C) in the case of a light bulb -- they will emit a good deal of visible light.\n\nThe filament in a light bulb is made of a long, incredibly thin length of tungsten metal. In a typical 60-watt bulb, the tungsten filament is about 6.5 feet (2 meters) long but only one-hundredth of an inch thick. The tungsten is arranged in a double coil in order to fit it all in a small space. That is, the filament is wound up to make one coil, and then this coil is wound to make a larger coil. In a 60-watt bulb, the coil is less than an inch long.
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Light Emitting Diodes and How They Work.\n\nhttp://electronics.howstuffworks.com/led.htm
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All natural light sources generate a "spectrum" of frequencies that approaches the "black body" radiation curve. The peak of that curve denotes the temperature of the emitting object. Certain frequencies can be affected in transit via interaction with materials as in emission and reflection nebula (ie. various places within the Orion constellation). The sun emits UV radiation commensurate with its black body ideal.\n\nUNLESS someone can show that the moon absorbes ALL UV frequencies on reflection then by definition:\n\nMOONLIGHT DOES INDEED CONTAIN UV FREQUENCIES\n\n===============================================\nSome of you people amaze me, you answer questions that you know NOTHING about. Answers like "No because it is reflected light" are a case in point. Is your blue shirt invisible in the mirror because the light coming from your shirt is lost as it reflects off of the mirror. I THINK NOT! Reflected light IS LIGHT. Some of it is absorbed, reflected, refracted but it is light. YES; the UV portion in moonlight is very weak, hey guess what -ALL THE LIGHT frequencies in moonlight are weak; THATS WHY MOONLIGHT IS DIM.\n==============================\nUV is NOT produced only from objects above 24000 degrees. It is produced at ALL temperatures, but the lower the temperature the lower the amount of UV that is produced. PLEASE Google "black body radiation"
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Interesting thought.\n\nThe first answer is right for leaving a point source, I would think. Most flashlights have a source and a parabolic mirror. Light from the focus of a parabola reflected off the surface will come out in parallel beams.\n\nLet's change your question to a flashlight giving out a nearly instantaneous pulse of light. I think we would see 2 pulses coming out of the light. The first would be a spherical shell from the light coming out directly, without reflecting off the parabola. Following that would be another pulse from the reflected light. I believe this would be slightly concave, since the path followed by the light towards the middle would have a slightly longer path to follow. \n\nThis is making a few simplifying assumptions, such as a point source for light and a perfect parabola. I would imagine in reality the distributed nature of the lightbulb would make for an irregular shape.
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It work off of Einstein's photoelectric effect. Light is an electromagetic wave propagating through space, but due to Quantum Mechanics it can also be viewed as particles. The energy of these particles, called photons, is dependent upon the frequency of the light. The higher the frequency the higher the energy. If the frequency and thus the energy of a photon is high enough it can knock an electron out of its atomic orbital. \n  Now the problem is that one can not predict in which direction the electron is going to travel, however there are special transitionary elements between the metals and non metals called semiconductors. These semiconductors are divided into two classes p-type and n-type. Now if one places a film of n-type semiconductor against a film of p-type semiconductor they produce a diode which allows electric current to only flow in one direction from the n-type side to the p-type side. So only the electrons that are knocked from the n-type side to the p-type side are compelled to stay put, but as more and more electrons are knocked over to the p-type side a a voltage potential starts to develope.
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A photon is a quantum of the vibrational energy of the space grid (as in the image above). At any one point on the wave, photons are produced out of the zero-point and then oppositely, photon holes are produced. Hence, it must be the case that these oscillations up and down come in discrete steps--or quanta.\n\n \n\nWhen a particle from the quantum vacuum acquires energy from a charged particle of matter, it may expand from the addition of this excitation energy to become an ordinary photon. Energy, therefore, expands the size of a virtual photon in ground state.\n\n \n\nA virtual photon, thus, is a "collapsed" short-range quantum that becomes an ordinary photon when given energy from an excited charged particle. Hence, the primary reason that we do not observe the photons that exist in the quantum vacuum, is because the quanta of space-time that compose the vacuum have a very minute amount of energy within them and are, therefore, too small to be observed directly. However, such vacuum quanta may become observable and measurable as ordinary photons after emission.\n\n \n\nWhen actualization takes place, we may measure photons and photonic energy hitting a measuring device.  But, while in-flight, we may not speak about photons at all, because technically speaking they are not photons.  So, when Planck's equation says that energy comes in lumps E = hf, this only refers to a state of an EM wave after an actualization or measurement has happened.  Prior to this we have only Kant's noumenal "wave" condition.  \n\nphotons must have higher frequencies, by exactly one hertz, than the expected difference of the energy levels of the electrons from which they received energy. Additionally, photon wavelengths must include a very long 3 x 10^8 meters to be added to all emitted wave fronts