michel123456 Posted February 9, 2014 Share Posted February 9, 2014 Can someone provide a link to an accurate description of the double slit experiment. i mean, not the result, but all the details. More specifically, I'd like to know the following 1. what is the material of the shield (the blocking screen) that have the 2 slits 2. whether this shield absorbs perfectly the photons that hit or whether the photons are reflected back in the room. 3. the thickness of this shield 4. the wavelength of the emitted light. For example I have read that the original experiment by Young was made with sunligth (I suppose with visible light). Visible light has a wavelength between 380nm and 780 nm. So if the shield in which the slits are opened has a thickness of 0,1mm (that is 100000nm), that represents between 263 and 128 times the wavelenght. At a macroscopic level, it is like shooting a grain of sand of 1mm diameter across a shield 128 mm thick (13 cm thick). 1 Link to comment Share on other sites More sharing options...
swansont Posted February 9, 2014 Share Posted February 9, 2014 As long as the shield is opaque, the details don't matter all that much. Link to comment Share on other sites More sharing options...
John Cuthber Posted February 9, 2014 Share Posted February 9, 2014 The last time I tried this, I used a bit of thin copper wire and two razor blades. I imagine that Mr Young used something else. I believe that the classic school classroom version used a photographic plate, exposed + developed to make it black & opaque with the two slits cut into the gelatine with a sharp knife. the important bit is that the results don't depend strongly on the materials you use. Link to comment Share on other sites More sharing options...
michel123456 Posted February 9, 2014 Author Share Posted February 9, 2014 As long as the shield is opaque, the details don't matter all that much. well I was wondering about the details. I know that last experiments shoot photons one by one. How do you do that in practice: I don;t understand where the photon is shooted at? or is it not a ray? but something like an horizontal plane of light shooted at the vertical slits? Or if it is a ray indeed (a line) why do scientists say that they cannot know through which slit pass the photon? Or is the ray of light thicker than the 2 slits aperture? Has anyone a good description of the experiment? Thanks. Link to comment Share on other sites More sharing options...
Sensei Posted February 9, 2014 Share Posted February 9, 2014 michel123456, you are asking questions like you would never saw Young's experiment in your life... ?! Setup to Young's Experiment I bought for $7 in local physics equipment shop. Laser 100 mW green for $15. It has two frames. One frame with width of slits 0.03 mm Distance between centers of slits 0.06 mm. Second frame with width of slits 0.05 mm Distance between centers of slits 0.1 mm. 1 Link to comment Share on other sites More sharing options...
michel123456 Posted February 9, 2014 Author Share Posted February 9, 2014 michel123456, you are asking questions like you would never saw Young's experiment in your life... ?! Setup to Young's Experiment I bought for $7 in local physics equipment shop. Laser 100 mW green for $15. It has two frames. One frame with width of slits 0.03 mm Distance between centers of slits 0.06 mm. Second frame with width of slits 0.05 mm Distance between centers of slits 0.1 mm. Only over the Net, never in real life. What is this frame made of? What's its thickness? Link to comment Share on other sites More sharing options...
Sensei Posted February 9, 2014 Share Posted February 9, 2014 (edited) Only over the Net, never in real life. Find local physics equipment shop and buy stuff.. Like I said, it's very cheap. And also get diffraction grating setup as well (cost $2 for each frame here) I have 1000 slits per mm, 500 slits per mm, and 50 slits per mm. Using laser 532 nm with them, I have number of "blobs" on screen that is always prime number.. And searching for somebody with diffraction grating with different number of slits to compare results. What is this frame made of? Some kind of plastic. What's its thickness? Too small to be measured by me Fraction of mm for sure. One frame with width of slits 0.03 mm Distance between centers of slits 0.06 mm. Second frame with width of slits 0.05 mm Distance between centers of slits 0.1 mm. In other words: it's x width of slit, then x width of material, then x width of slit. 0.03mm + 0.03mm + 0.03 mm = 0.09mm Looking at it naked eye, you don't know there are two slits. Edited February 9, 2014 by Sensei 1 Link to comment Share on other sites More sharing options...
michel123456 Posted February 9, 2014 Author Share Posted February 9, 2014 Find local physics equipment shop and buy stuff.. Like I said, it's very cheap. And also get diffraction grating setup as well (cost $2 for each frame here) I have 1000 slits per mm, 500 slits per mm, and 50 slits per mm. Using laser 532 nm with them, I have number of "blobs" on screen that is always prime number.. And searching for somebody with diffraction grating with different number of slits to compare results. Some kind of plastic. Too small to be measured by me Fraction of mm for sure. In other words: it's x width of slit, then x width of material, then x width of slit. 0.06mm + 0.06mm + 0.06 mm = 0.18mm Looking at it naked eye, you don't know there are two slits. No you wrote "between centers of slits" not between edges of slits. Link to comment Share on other sites More sharing options...
Sensei Posted February 9, 2014 Share Posted February 9, 2014 I reedited message, because right value is 0.03 mm, not 0.06 mm "between centers of slits" is direct translation of what's written on it small letters. Link to comment Share on other sites More sharing options...
michel123456 Posted February 9, 2014 Author Share Posted February 9, 2014 Thank you for the info So, if I made no mistake with the lot of zero's, here is a section at scale I took the smallest figures: One frame with width of slits 0.03 mmDistance between centers of slits 0.06 mm. It is supposed that the thickness is equal to the width, that is 0.03 mm The large square in the center is the gap between the 2 slits. The little square in the left slit represents a square of wavelength 532nm. 1 Link to comment Share on other sites More sharing options...
swansont Posted February 9, 2014 Share Posted February 9, 2014 well I was wondering about the details. I know that last experiments shoot photons one by one. How do you do that in practice: I don;t understand where the photon is shooted at? or is it not a ray? but something like an horizontal plane of light shooted at the vertical slits? Or if it is a ray indeed (a line) why do scientists say that they cannot know through which slit pass the photon? Or is the ray of light thicker than the 2 slits aperture? You use a low-power source, typically a laser, and attenuate it to the point that you have single photons, with a reasonable average time (microseconds) between them Link to comment Share on other sites More sharing options...
michel123456 Posted February 9, 2014 Author Share Posted February 9, 2014 You use a low-power source, typically a laser, and attenuate it to the point that you have single photons, with a reasonable average time (microseconds) between them But you still have some kind of flow I presume. I mean the ray of light has a thickness comparable to the distance between the slits, no? Link to comment Share on other sites More sharing options...
Sensei Posted February 9, 2014 Share Posted February 9, 2014 michel123456, visualize now diffraction grating frames with 50 slits/mm, 500 slits/mm and 1000 slits/mm at the same scale as in #10 post. 0.03 mm slits frame would have 16.66(6) slits/mm, if they would be continuous. So, if I made no mistake with the lot of zero's, here is a section at scale I check it in 3d application and your scale is correct. I have the same result. We can place ~56 boxes with width 532e-9 m in box with width 3e-5 m. Link to comment Share on other sites More sharing options...
michel123456 Posted February 9, 2014 Author Share Posted February 9, 2014 michel123456, visualize now diffraction grating frames with 50 slits/mm, 500 slits/mm and 1000 slits/mm at the same scale as in #10 post. 0.03 mm slits frame would have 16.66(6) slits/mm, if they would be continuous. That's all O.K. What does not change is the thickness. For better understanding, as it appear on my screen, the scale of the image in post#10 corresponds approximatively to 1 mm equal to a meter. And also a regular sheet of printer paper is 3 times thicker than what is shown on the drawing. Link to comment Share on other sites More sharing options...
swansont Posted February 9, 2014 Share Posted February 9, 2014 But you still have some kind of flow I presume. I mean the ray of light has a thickness comparable to the distance between the slits, no? Yes. That's not an issue with regard to getting single photons. Link to comment Share on other sites More sharing options...
michel123456 Posted February 10, 2014 Author Share Posted February 10, 2014 Yes. That's not an issue with regard to getting single photons. I don't get it. You mean that an elementary particlelike a photon is a macroscopic flow with a measurable thickness? Link to comment Share on other sites More sharing options...
swansont Posted February 10, 2014 Share Posted February 10, 2014 I don't get it. You mean that an elementary particlelike a photon is a macroscopic flow with a measurable thickness? If you count enough photons, it will look like that. Link to comment Share on other sites More sharing options...
michel123456 Posted February 10, 2014 Author Share Posted February 10, 2014 If you count enough photons, it will look like that. Sorry I ment You mean that a single elementary particle like a photon is a macroscopic flow with a measurable thickness? Link to comment Share on other sites More sharing options...
swansont Posted February 10, 2014 Share Posted February 10, 2014 Sorry I ment You mean that a single elementary particle like a photon is a macroscopic flow with a measurable thickness? No Link to comment Share on other sites More sharing options...
michel123456 Posted February 10, 2014 Author Share Posted February 10, 2014 (edited) No Ah. Then i understand that a single photon is considered as a line (a straigth path) In this case, how does it come out that the scientist does not know where the photon goes (in the left or right slit) The photon goes where the laser is pointed to. No? If the laser is pointed to the left slit, there goes the photon. If the laser is pointed to the right slit, the photon goes there. Or do I miss something? Edited February 10, 2014 by michel123456 Link to comment Share on other sites More sharing options...
Strange Posted February 10, 2014 Share Posted February 10, 2014 Ah. Then i understand that a single photon is considered as a line (a straigth path) In this case, how does it come out that the scientist does not know where the photon goes (in the left or right slit) The photon goes where the laser is pointed to. No? If the laser is pointed to the left slit, there goes the photon. If the laser is pointed to the right slit, the photon goes there. Or do I miss something? If you think about the classical version of the experiment, the light source must be able to illuminate both slits. This means that any given photon could go through either slit. The same continues to be true when there is only one photon at a time. Link to comment Share on other sites More sharing options...
Sensei Posted February 10, 2014 Share Posted February 10, 2014 Ah. Then i understand that a single photon is considered as a line (a straigth path) In this case, how does it come out that the scientist does not know where the photon goes (in the left or right slit) The photon goes where the laser is pointed to. No? If the laser is pointed to the left slit, there goes the photon. If the laser is pointed to the right slit, the photon goes there. Or do I miss something? When we have just one slit, photons entering slit in straight line, after exiting slit will be diffracted. It can be visualized in container with water even better. http://www.youtube.com/watch?v=BH0NfVUTWG4 Link to comment Share on other sites More sharing options...
michel123456 Posted February 10, 2014 Author Share Posted February 10, 2014 If you think about the classical version of the experiment, the light source must be able to illuminate both slits. This means that any given photon could go through either slit. The same continues to be true when there is only one photon at a time. But when you shoot one photon at a time? Does the light ray of a single photon illuminate both slits? Link to comment Share on other sites More sharing options...
John Cuthber Posted February 10, 2014 Share Posted February 10, 2014 But when you shoot one photon at a time? Does the light ray of a single photon illuminate both slits? A deep question. The answer is that it behaves as if it does. You still get a diffraction pattern 1 Link to comment Share on other sites More sharing options...
Strange Posted February 10, 2014 Share Posted February 10, 2014 But when you shoot one photon at a time? Does the light ray of a single photon illuminate both slits? That is the whole point of the quantum double-slit experiment: you can't know that. If you know which slit it illuminates / goes through then there is no interference pattern. If you don't know, then there is. You cannot know anything about where a photon is until you detect (and thereby destroy) it. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now