by Dinesh » Fri May 23, 2014 6:31 pm
marwaj wrote:This laser has two controllers and if I want I can turn on only one diode , then I get half of power 35mW , I'm curious if then the light will be still cross-polarized..
With only one beam, assumed coherent (or semi-coherent), you cannot have two independent polarisation perpendicular states. The polarisation must be either linear or random, ie unpolarised. If you have two beams from two sources inline with each other, I suppose it's possible (but odd) that the two beams are linearly polarised perpendicular to each other. If the two wavelengths are the same, then this will result in elliptical polarisation. If the two wavelengths are slightly different, but ve-e-e-ry slightly different, then you'll again have elliptical polarisation, but the ellipse will rotate, since the frequencies of the two E vectors are slightly different; it depends on the degree of coherence and the phase difference between the two E vectors. If they're in-phase, the ellipse is described by E(1)cos(w(1)t) + E(2)cos(w(2)t) with w(1) - w(2) >~ 0. If the frequencies of the lasers, or their wavelengths are very different, then the ellipse will orientate in random directions very fast and you'll again get "random" polarisation.
marwaj wrote:You suggest that I can try to shot hologram without make separation of polarization?
Well, I'm not sure what you're trying to do. Let's say you're trying to make a single beam Denisyuk. Let's also say that the object is shiny (eg coins) and the object does not de-polarise. In this situation, let's call one polarisation state A, and the other one perpendicular to A, we'll call B. So, A and B are directions perpendicular to each other - they may be "up and down" and "left and right", for example. In this case, both A and B will hit the object. A reflects A back, interferes with A but does not interfere with B, since B is perpendicular to A. Now you have "A fringes". At the same time, B reflects back and interferes with B, but does not interfere with A, for the same reason. Now you have "B fringes". What you'll record is a set of A fringes and B fringes. But, if the wavelengths are slightly different, the A fringes will have a very small fringe width different with the B fringes, ie there will be a small difference in spatial frequency. This small difference in spatial frequency will create a beat frequency, which shows itself up as a set of whirly lines across the image which will look like moire fringes.
[quote="marwaj"]This laser has two controllers and if I want I can turn on only one diode , then I get half of power 35mW , I'm curious if then the light will be still cross-polarized.. [/quote]
With only one beam, assumed coherent (or semi-coherent), you cannot have two independent polarisation perpendicular states. The polarisation must be either linear or random, ie unpolarised. If you have two beams from two sources inline with each other, I suppose it's possible (but odd) that the two beams are linearly polarised perpendicular to each other. If the two wavelengths are the same, then this will result in elliptical polarisation. If the two wavelengths are slightly different, but ve-e-e-ry slightly different, then you'll again have elliptical polarisation, but the ellipse will rotate, since the frequencies of the two E vectors are slightly different; it depends on the degree of coherence and the phase difference between the two E vectors. If they're in-phase, the ellipse is described by E(1)cos(w(1)t) + E(2)cos(w(2)t) with w(1) - w(2) >~ 0. If the frequencies of the lasers, or their wavelengths are very different, then the ellipse will orientate in random directions very fast and you'll again get "random" polarisation.
[quote="marwaj"]You suggest that I can try to shot hologram without make separation of polarization?[/quote]
Well, I'm not sure what you're trying to do. Let's say you're trying to make a single beam Denisyuk. Let's also say that the object is shiny (eg coins) and the object does not de-polarise. In this situation, let's call one polarisation state A, and the other one perpendicular to A, we'll call B. So, A and B are directions perpendicular to each other - they may be "up and down" and "left and right", for example. In this case, both A and B will hit the object. A reflects A back, interferes with A but does not interfere with B, since B is perpendicular to A. Now you have "A fringes". At the same time, B reflects back and interferes with B, but does not interfere with A, for the same reason. Now you have "B fringes". What you'll record is a set of A fringes and B fringes. But, if the wavelengths are slightly different, the A fringes will have a very small fringe width different with the B fringes, ie there will be a small difference in spatial frequency. This small difference in spatial frequency will create a beat frequency, which shows itself up as a set of whirly lines across the image which will look like moire fringes.