Itoero Posted January 10, 2017 Posted January 10, 2017 In the double slit experiment you can detect through which slit a photon goes by placing a detector. But the detector prevents interference so the wave behavior is no longer visible. If you place the detector at the beginning of the laser beam, can you then create light that behaves only as a particle? In other words, does the detector changes the behavior of a photon or only how we observe the behavior? 1
Delta1212 Posted January 10, 2017 Posted January 10, 2017 Roughly, photons interact as particles, but travel between interactions as waves. That's overly simplified, but I think it works for the sake of this example. The double-slit experiment allows us to observe the wave behavior of individual photons by allowing them to interfere with themselves on their way to being detected as if they were a wave passing through both of the slits. This affects where the final interaction takes place on the screen at the end. By placing a detector, you are creating an interaction at one of the slits. It's no longer a wave with a double slit between it and the screen it will eventually hit and interact with. It's a wave traveling to the slits where it interacts with the detector and then sends out a new wave from that point which has a clear path from the slit it passed through to the screen, so no interference pattern. If you place a detector at the beginning of the experiment, the wave will travel to the detector, interact with the detector and then propagate again as a wave with the detector as a starting point. If there is a double slit after the detector, then you'll get the interference pattern exactly as normal. If you don't have the double slits in between, then you won't get the interference pattern. Obviously. 1
Klaynos Posted January 10, 2017 Posted January 10, 2017 The thing to try and understand is that photons are not waves nor particles but something else that has some properties of both.
Sensei Posted January 11, 2017 Posted January 11, 2017 (edited) Can you make light that behaves as particle? Yes, I can. In f.e. photoelectric effect. https://en.wikipedia.org/wiki/Photoelectric_effect Edited January 11, 2017 by Sensei
Itoero Posted January 11, 2017 Author Posted January 11, 2017 Thanks for the very clear explanation Delta1212 The thing to try and understand is that photons are not waves nor particles but something else that has some properties of both.So the wave particle duality is a way of describing the probabilistic behavior of particles? Yes, I can. In f.e. photoelectric effect. https://en.wikipedia.org/wiki/Photoelectric_effect So electrons can leave a material (gain kinetic energy)when they reach a higher energy level. Is this photo electric effect what enables a solar cell? Back when I 'studied', there stood a solar car in the middle of our building.
Sriman Dutta Posted January 11, 2017 Posted January 11, 2017 A photon shows the properties of both waves and particles. This is the wave-particle duality.
Sensei Posted January 11, 2017 Posted January 11, 2017 (edited) So electrons can leave a material (gain kinetic energy)when they reach a higher energy level. Suppose so you have metal from I,II group, and point white light source at it, there will be visible electrons, leaving small traces. Length of trace tells its kinetic energy. And you see traces with small length and traces with long length. Their kinetic energy is energy of photon (white light- full spectrum), minus energy needed to liberate them. Photon is gone from system after being absorbed, and photoelectron is emitted instead. Then the next step is using various monochromatic light sources (or split white on prism, rotated accordingly that only one color +- little tolerance, is reaching our metal) Say you use red, and see absolute nothing, no photoelectrons emitted. Then you use yellow (true yellow, not mixture of RG like in TV), and some start appearing, but they have very little energy. Then you use green, and you see their kinetic energy is higher than for yellow one. Then you use blue, and you see their kinetic energy is even higher than for green.. etc. Edited January 11, 2017 by Sensei
Strange Posted January 12, 2017 Posted January 12, 2017 Say you use red, and see absolute nothing, no photoelectrons emitted. Then you use yellow (true yellow, not mixture of RG like in TV), and some start appearing, but they have very little energy. Then you use green, and you see their kinetic energy is higher than for yellow one. Then you use blue, and you see their kinetic energy is even higher than for green.. etc. And it is important to note that the intensity (energy or brightness) of the light sources makes no difference. A single blue photon can cause an electron to be ejected but 1 billion red photons will have no effect.
swansont Posted January 12, 2017 Posted January 12, 2017 And it is important to note that the intensity (energy or brightness) of the light sources makes no difference. A single blue photon can cause an electron to be ejected but 1 billion red photons will have no effect. Almost. Multi-photon ionization is possible, but this is a nonlinear effect — you need high intensities to see this (and it becomes increasingly rare as you require more photons. It varies as IN where N is the number of photons being absorbed, until you get to really high intensity) 1
Strange Posted January 12, 2017 Posted January 12, 2017 Almost. Multi-photon ionization is possible, but this is a nonlinear effect — you need high intensities to see this (and it becomes increasingly rare as you require more photons. It varies as IN where N is the number of photons being absorbed, until you get to really high intensity) Thanks. I did wonder if that could happen.
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