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#1
Nov2603, 10:49 PM

P: 17

having problems trying to understand and derive the equation for intensity in the doubleslit interference pattern. any sort of help would be welcomed. thanks



#2
Nov2703, 10:25 AM

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What kind of problems? Show us what you've tried so far and what you don't understand.



#3
Dec2203, 01:11 PM

P: 31

I was asked by an old mechanical engineer how a photon that is emitted from a point source can then give up all its energy at a discrete point many thousands of diameters away.
I gave an answer using the electromagnetic concept of radiation from a point source, thusly: Think of the electromagnetic wave as a sphere that keeps expanding and the energy distribution in the wavefront has to spread itself over a greater and greater spherical surface area. If you place an object somewhere in this expanding wavefront it will interrupt a portion of the expanding sphere (either absorbing or reflecting depending upon the characteristics of the object), the remaining wavefront keeps on going and going and expanding. The strength of an rf source dissipates with the square of the distance. Radiated energy can be starting at a point source and "spreading" three dimensionally. Take a volume of a sphere and double the radius, the volume is squared, thus the energy level at any point on the spheres surface is distributed over the spherical wavefront, and gets smaller and smaller at discrete points.  This really doesn't answer the basic question that was posed, and whether the energy of the photon is subject to the square of the distance rule. 


#4
Dec2203, 01:48 PM

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Light As A Wave



#5
Dec2303, 12:06 AM

P: n/a

It’s my understanding that a single photon or light “wave” goes out only in one direction and not as an “expanding sphere” of light. I think it could be described something like a long single (perhaps wiggling) bullet. However, when light bulbs radiate light and when radio transmitters radiate radio waves, a lot of these “bullets” are emitted out in all directions at the same time, and when all their different directions are plotted, they all go out in the form of a “sphere of radiation”. So, the sphere of radiation does not result from just one photon going out or a steady stream of them going out such as from a laser. If I am wrong about this, someone can please let me know. 


#6
Dec2303, 12:12 AM

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Wouldn’t this idea match what I just said? Light from a light bulb dims at a distance because we are shining fewer photons per sq cm on a flat white wall at a distance, because the individual photons are spreading out the further away from the source they get? 


#7
Dec2303, 01:33 AM

P: 31

In the following URL, equation (11) expresses the inverse square law for "intensity" and equation (16) expresses the inverse square law for "density".
http://prancer.physics.louisville.ed...ual/node7.html "Either a detector counting photons or one measuring energy will show an inverse square law with distance from the source." The following URL contains a "simple" explanation of the Inverse Square Law, http://www.astro.sunysb.edu/fwalter/CEN511/week2.html "All waves have the property that, once emitted, they propagate outwards from the source. As they do so the energy density in the wave decreases. This is because the total energy in the wave is fixed, while the volume of space the wave is expanding into is continuously increasing." Once a single photon is emitted from a point source in a particular direction, the photon has no connection to the source, and since it does not in itself expand, why would it lose energy? Once on its way it is self sufficient, correct? Equation (11) of the first reference implies photons lose energy. The value of P in equation (11) implies it refers to a group of photons, even though the source power emitted could be that of a single photon. 


#9
Dec2303, 11:27 AM

P: 31

Equation (11) mentioned earlier is,
[tex]F = \frac {P} {4 \pi r ^ 2}[/tex] where F is the detected energy and P is the emitted energy which is divided by what is the equation for the surface of a sphere. If I am concerned with only one photon from a point source and it is measured at a distance with a detector that is aimed at the incoming photon, there is no spreading out. Since it is not going to be measured with a detector that is spread out, the divisor is eliminated and the photon arrives at a distance with the same energy as emitted. P can be equated to the energy of one photon. Thus a single photon, whether it travels a thousand diameters or a thousand light years would arrive with the same energy that it started with. The basic question I was asked, is the energy of a single photon subject to the inverse square of the distance rule? 


#10
Dec2303, 11:54 AM

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#11
Dec2303, 07:24 PM

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This was on one of your links: “Electromagnetic waves expand spherically out into space.” I think this might be a common myth which has been confounding radio and electrodynamics students and old guys like me for many years. Surely a single photon doesn’t “expanding spherically out into space”. And even if a photon is also a kind of “wave” or a small long “wave packet”, a single one wouldn’t necessarily have to “expand spherically out into space”, not any more than a single writhing snake would have to “expand spherically out into space” if he were thrown in the air. If we throw out about 5,000 snakes in all directions: up, down, North, South, East, West, back, forward, left, and right, the whole group of snakes might “expand spherically out into space”, but all the individual snakes would not do that. Each one would go in only one direction, with that direction being altered only by gravity and air resistance. 


#12
Dec2303, 08:34 PM

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So, I wonder if we could say that sound waves become weaker at a distance because we are hearing or recording or receiving a smaller section of their spherical waves. Whereas light becomes weaker at a distance because we are receiving fewer sidebyside photons, since the photons are spreading apart as they move. 


#13
Dec2303, 08:50 PM

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Doc Al,
Since you are a smart guy, let me ask you something. This might be a very stupid question, but I’ve never understood mystery of the double slit phenomenon. In this demonstration, why wouldn’t the dot of laser light hit the card the slits are on, right in the middle between the slits, and thus not go through either slit? LINK TO JAVA DEMO 


#14
Dec2303, 10:58 PM

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The slits are much smaller and closer together than in the demo you linked  microscopic even. Often, DIFFRACTION GRATINGS are used.
The size/placement of the slits is on the order of the wavelength of the light. 


#15
Dec2403, 03:44 AM

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#16
Dec2403, 08:00 AM

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Doc Al and Russ,
Ok, thanks. I have dozens of science books of all kinds that never mention that the slits are so close together a small laser dot of light can go through both slits at the same time, and every time I’ve had a chance to talk to a physicist or a physics professor, during the past 40 years, I’ve forgotten to ask them that question. Ok, now, the next question. What do they mean when they say a single electron is fired through both slits at the same time, or am I misunderstanding what they are saying about the electrons and the slits? Does the electron go through one slit or the other, or is it also split because the slits are so small and so close, and so it goes through both slits at the same time? 


#17
Dec2403, 08:42 AM

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I put the word "understand" in quotes, because if you mean how is it possible for a particle like an electron to go through both slits at once: as long as you think of a particle as a classical object (like a tiny marble) you cannot understand it! The great Feynman said: "The basic element of quantum theory is the doubleslit experiment. It is a phenomenon which is impossible, absolutely impossible to explain in any classical way and which has in it the heart of quantum mechanics. In reality it contains the only mystery ... the basic peculiarities of all quantum mechanics." 


#18
Dec2403, 09:46 AM

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Let’s say we throw a watermelon through a picket fence. The watermelon is going to split apart and go through two or more of the slits between the pickets at the same time, yet a watermelon is not a “wave”. Could a particleelectron be splitting in the same manner, with half of it going through one slit and the other half going through the other slit? Suppose I think of an electron as a 19th Century watermelon that easily splits apart when thrown through a 19th Century picket fence? Apparently 21st Century watermelons might be considered to be “waves” rather than “solid” objects. Yet another question: Could the “interference pattern” actually be a pattern caused by the splitting apart of the particleelectron, with, in some cases, a bigger part going through the right slit, while the smaller part goes through the left slit, and in other cases, because the larger part goes through he left slit while the smaller part goes through the right slit, with the two parts being deflected slightly, in slightly different directions, each time a new electron is split while going through the two slits? I suppose this question leads to this question: Would a series of several watermelons thrown through a picket fence result in a debris splash pattern on the wall behind the fence that resembles an “interferencelike pattern” that develops when a series of separate electrons are fired through two narrow slits? This may also sound like a dumb question, but I’ve seen such an “interferencelike pattern” on walls and large cards, whenever I’ve spray painted things that have slits, that are located in front of the walls and large cards. Also, before I forget, I know a guy who had a peculiar problem regarding a narrow slit through which light passed in a special type of camera he made. He got an “interferencetype pattern” with just one slit. We puzzled over this problem for months, trying to figure out all kinds of reasons for the cause of the pattern. Finally, I remembered the doubleslit experiment, and I suggested that he might be getting some type of pattern because the slit was too narrow. So, he widened the slit and the pattern disappeared. 


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