Double slit experiment -WHAt is it it showing or proving

[ZharAngel]

Anyway it looks like this thread is turning into a discussion of the photoelectric effect instead of the significance of the double slit experiment.

Lol, yeah. The double slit experiment also proved the wave-particle duality of things at the very small scale, like electrons for example. You would imagine electrons to be "billiard balls", but hey, no, they are probability waves.

 

[Andy Resnick]

<snip>
The wave theory says that since the amount of energy is determined by the wave amplitude, which is proportionate to the intensity of the light source (Poynting vector), <snip>

Small quibble in an otherwise excellent post: the Poynting vector is associated with momentum, not energy.

 

[Tw15t3r]

This is incorrect. The photoelectric effect is about electrons being liberated (emitted from metal surface) and not merely promoted from an inner shell to an outer shell. The theory which makes use of the electrons being promoted from the inner shells to the outer is the one on gas absorption of light.

Woops, thanks for correcting me.

ZharAngel has a point pertaining to the subject of this thread. Electrons, which are supposed to be particles, have proven to exhibit the same type of interference pattern when the same double slit experiment is performed, suggesting wave properties.

And regarding mashing two different frequencies of light to cause the pulsing to hopefully excite an electron (whether to eject it or not doesn't matter), one must remember that if such a method is done, only the amplitude will end up being pulsed, and not the frequency, which is what that affects rate of electron ejection.


But an interesting thing that I have not been able to explain myself is that: E=hf, but intensity is proportional to energy/(area x time). If both are energy, what are the differences? And according to the law of conservation of energy, can they be converted between each other?

 

[rcgldr]

How would physicists know if double slit experiment results aren't due to interaction with electrons in the molecules at the edges of the slits? There could be some special interaction between these electrons and either light or a free stream electrons. One property of the molecules at a sharp edge is that they are more "exposed", where both refraction and reflection could take place.

 

[Landru]

Think of it this way. Wave theory says that the energy of a wave, such as light, is given by its amplitude, not the frequency. Which means to say that given a particular wave, increasing its amplitude would also increase the amount of energy a light wave could provide to metal surface electrons. The photoelectric effect is the observation that metal surface electrons could be liberated from the surface by shining a light source of sufficient energy on it.

The wave theory says that since the amount of energy is determined by the wave amplitude, which is proportionate to the intensity of the light source (Poynting vector), if shining a low-frequency light on metal does not liberate any electrons (which are measurable as photo-current), all one needs to do is brighten that light by increasing the intensity and hence amplitude of the incident light wave. Even if we assume that the the frequency of the light affects the amount of energy associated with the wave, it was thought that one could always compensate for the low-frequency light wave by increasing the intensity of the light source.

The photoelectric effect shows that this isn't true. In other words, if one uses a low frequency light source, no matter how great the intensity (or brightness) of the light source is, there would not be any electrons emitted from the surface of the metal. In contrast, if you instead use a low-intensity but high frequency light source there will be electrons ejected from the metal.

What I don't understand and what I can't find is why a single high energy photon can pop off an electron but lots of low evergy photons won't even if the low evergy photons are collectively more energetic than the single high energy photon.

 

[Landru]

How would physicists know if double slit experiment results aren't due to interaction with electrons in the molecules at the edges of the slits? There could be some special interaction between these electrons and either light or a free stream electrons. One property of the molecules at a sharp edge is that they are more "exposed", where both refraction and reflection could take place.

Assuming the edges of the slits could and did reflect light like a perfect mirror the interferance pattern would be extremely faint in comparison to the line of site, but as it is the interference is relatively bright. The number of photons that follow the wave outline greatly outnumbers what would be accounted for due to reflection alone.

 

[rcgldr]

Assuming the edges of the slits could and did reflect light like a perfect mirror the interferance pattern would be extremely faint in comparison to the line of site, but as it is the interference is relatively bright. The number of photons that follow the wave outline greatly outnumbers what would be accounted for due to reflection alone.

My point was about refraction (the apparent bending of light), not reflection.

 

[Landru]

My point was about refraction (the apparent bending of light), not reflection.

All the same, there are more photons hitting within the interference area than are interacting with the edges of the slits.

 

[Cthugha]

What I don't understand and what I can't find is why a single high energy photon can pop off an electron but lots of low evergy photons won't even if the low evergy photons are collectively more energetic than the single high energy photon.

Well, in principle they can via two photon absorption, but this process is usually not very likely as it is a nonlinear effect. Those electronic excitation processes happen very fast, somewhere between the attosecond and femtosecond range, so two photons need to "hit" the electron in a very short time window somewhere between 10^{-15} and 10^{-18} seconds.

In an hypothetical example you now have a monochromatic light source of 10 mW power (this is already more than one of the HeNe lasers in my lab has), which emits photons of 1 eV energy (somewhere in the infrared), but you need exactly 2 eV to liberate the electrons. A quick calculation shows, that there are about 6,2 * 10^{16} photons emitted per second, so the number of photons arriving within a femtosecond is about 60 photons. Now you also have to consider, that the time window is smaller than a femtosecond, the beam will be very large compared to the cross section of the electron, conservation rules apply (spin, for example) and by far not every photon will interact with the electron. So even with a VERY optimistic estimate, you will have a few thousand two photon absorptions per second and a few liberated electrons.

So the reason, why the high energy photons work better, is that you have to compare the number of photons arriving within a certain time window (high energy case) to the number of photon pairs arriving within a certain time window (low energy case). So unless you go to very, very high intensities, liberating electrons with low energy photons won't work.

 

[Beto Pimentel]

I guess we have been going along the wrong way in this discussion since the beginning, although I am sure all the messages have added for the comprehension of the subject. I suspect the "double slit experiment" first referred was not exactly Young's original 1801 experiment, but rather low energy (single photons) double slit exposure. Imagine a source of light very, very dim, placed in front of a double slit which, in turn, stands in front of a very sensitive photographic plate. This is all in a completely dark chamber, and supposely one can replace the photographic plate at will. Now assume the experimenter turns the source on and expose the photographic plate for a very short time, then replace it and exposes the second plate for a little bit longer, and so on, until the last plate, which is exposed for quite a long time. When the plates are developed, a funny thing shows up: in the first plate, the experimenter observes a few bright dots, as if particles of light had been shot through the slits and hit the plate (awkwardly, some might even hit the plate in regions in the geometrical shadow of the slits, but eventually the experimenter could think of the particles hitting the very edge of a slit and going astray). For the second plate the experimenter will only notice that there will be proportionally more bright dots, which makes sense, for longer exposure time means more time for the source to shoot more particles. However, as the experimenter develops more and more plates, it starts to become clear that there are regions of the plate where it is more probable to find a bright spot, and other regions where it is less probable to find a bright spot. Apparently, no matter how long the exposure time is, there are equally spaced areas where NO bright point appears. This pattern is PRECISELY equivalent to the pattern formed by the interference of a wave crossing the two slits.
THIS kind of experiment indeed shows the wave-particle duality (note that the pattern does not appear fainter when the source is dimmer), and usually it is to this kind of experiment people refer when discussing the fundamental mystery of Quantum Mechanics.

 

[Landru]

Well, in principle they can via two photon absorption, but this process is usually not very likely as it is a nonlinear effect. Those electronic excitation processes happen very fast, somewhere between the attosecond and femtosecond range, so two photons need to "hit" the electron in a very short time window somewhere between 10^{-15} and 10^{-18} seconds.

In an hypothetical example you now have a monochromatic light source of 10 mW power (this is already more than one of the HeNe lasers in my lab has), which emits photons of 1 eV energy (somewhere in the infrared), but you need exactly 2 eV to liberate the electrons. A quick calculation shows, that there are about 6,2 * 10^{16} photons emitted per second, so the number of photons arriving within a femtosecond is about 60 photons. Now you also have to consider, that the time window is smaller than a femtosecond, the beam will be very large compared to the cross section of the electron, conservation rules apply (spin, for example) and by far not every photon will interact with the electron. So even with a VERY optimistic estimate, you will have a few thousand two photon absorptions per second and a few liberated electrons.

So the reason, why the high energy photons work better, is that you have to compare the number of photons arriving within a certain time window (high energy case) to the number of photon pairs arriving within a certain time window (low energy case). So unless you go to very, very high intensities, liberating electrons with low energy photons won't work.

Thanks, that's a perfect and clear answer. So the photons must actualy come into contact with the electron and that it's very unlikely that two photons would strikes an electron at once, or ever close enough together. Therefore a single photon must contain all the energy needed to get the job done.

So, puting this together, the photo electric effect proves the existence of photons because it shows that it only matters how powerful any given photon is and not how many of them you focus into the beam. However if light had been waves, then it would never miss those electrons, in which case both intensity and frequency would be a factor in the photo-electric effect, and not freqeuency alone.

 

[Landru]

Someone said in another thread that they tried the double slit experiment at home and then, maybe joking, wanted to add a photon detector to see the wave function collapse first hand. Is there a practical way to do this at home or does it require fancy, expensive equipment?