I've seen most of the emphasis put on the beam ratio's. How important is polarization for reflection and transmision holos, and what is the effect of it not being correct? (In laymans terms please.)
Just how important is polarization to making holograms??
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JohnFP
Just how important is polarization to making holograms??
For single beam holograms the correct polarization is used in conjunction with the Brewster's angle to minimize reflection off the back of the glass which causes the wood grain effect on the final hologram. The polarization would also play a very important role in a single beam if the object is shiny like a mirror or fire glazed ceramic. In a H2 copy set up, if the polarizations are not the same for object beam and reference beam when the both hit the plate, you will not get a hologram or dim at least depending on the degree of polarization.
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Dinesh
Just how important is polarization to making holograms??
Fairly important. In fact, you could say crucial. There's also a lot of confusion about polarisation in general.
Firstly, a hologram is an interfernce pattern between two beams (not necessarily, but usually) coherent. The diffraction from these interference fringes determines how bright your hologram is - the stronger the fringes the brighter the hologram (within certain constraints like sinusoidal profiles). The strength of an interference fringe is determined by mainly three factors: the phase relationship, or coherence, between the beams, the relative strengths of the beams (the beam ratio) and the polarisation state of the two beams. Assuming the beams are coherent this leaves the beam ratio and polarisation as the main factors in the strength of the fringes and so the brightness of the hologram. Ignoring beam ratio for the moment, the strength of the fringes is damped by the cos of the polarisation angle between the beams. So let's say one beam (A) is vertically polarised and the other (B) is also. Then the angle between the polarisation states is zero, cos(0) = 1 and you get the strongest possible fringe strength. Let's say A's pol is vertical and B's is horizontal. Now the angle betwee pol states is 90 degrees, cos(90) = 0 and there is no interference at all. Let's say that B's pol state is inclined at 30 degrees to the vertical. Now the angle between pol states is damped by a factor of cos(30) = 0.877, ie the fringe strength is now about 87% of what it would be for parallel polarisation and the brightness is similarly affected.
The above considers only fringe strength at the plate. There is also the polarisation effects due to reflection and transmsion. Reflections from a metallic object, for instance, maintains the polarisation states of the incoming beam so if your referencing in one plane and objecting in another - say your ref is overhead and your object beams are in a horizontal plane - the interfering beams may be polarised at right angles to each other and, as mentioned above, and you'll end up with no fringesat all. A diffuse object depolarises, so while it seems bright to the eye, you may be losing a lot of the object light due to cos term damping. The reflected light is polarised in all possible direction but the reference is polarised in one, say vertical.Now divide the circle into segments of 10 degrees. there are 36 of these segments. The polarisation states of the reflected object light from the diffuse object will exist in all 36 segments but each segment will damp the interference term by cos(10*segment number). So the light reflected off the (diffuse) object whose polarisation states are in segment 2 will dampen the fringe strength by cos(20), the pol state of the reflected light in segment 3 wiull dampen it by cos(30) etc. If I is the intensity of the light reflected off the diffuse object,the true intensity for interferometric purposes as seen by the reference beam is;
I/36 + (I/36)*cos(10 + (I/36)*cos(20) + (I/36)*cos(30) +...+(I/36)cos(360).
This is a lot less than the measured intensity of I. The number of segments, 36, was chosen because mostly the cosine function does not appreciably change over a 10 degree variation. If you wanted to be exact, you'd have to consider 360 segments at 1 degree per segment.
There are other factors as well, however a clay (diffuse and depolarising!) gecko awaits me on the table....
You might want to look at the reference beam with a polariser and orientate it to get the maximum light through it. Then look at the reflected object light with the polariser at the same angle. This is the best possible tru object light. As you turn the polariser, the light you see from the object, the true object light, is the intensity you see through the polariser damped by the cos of the angle through which the polariser is turned.
Firstly, a hologram is an interfernce pattern between two beams (not necessarily, but usually) coherent. The diffraction from these interference fringes determines how bright your hologram is - the stronger the fringes the brighter the hologram (within certain constraints like sinusoidal profiles). The strength of an interference fringe is determined by mainly three factors: the phase relationship, or coherence, between the beams, the relative strengths of the beams (the beam ratio) and the polarisation state of the two beams. Assuming the beams are coherent this leaves the beam ratio and polarisation as the main factors in the strength of the fringes and so the brightness of the hologram. Ignoring beam ratio for the moment, the strength of the fringes is damped by the cos of the polarisation angle between the beams. So let's say one beam (A) is vertically polarised and the other (B) is also. Then the angle between the polarisation states is zero, cos(0) = 1 and you get the strongest possible fringe strength. Let's say A's pol is vertical and B's is horizontal. Now the angle betwee pol states is 90 degrees, cos(90) = 0 and there is no interference at all. Let's say that B's pol state is inclined at 30 degrees to the vertical. Now the angle between pol states is damped by a factor of cos(30) = 0.877, ie the fringe strength is now about 87% of what it would be for parallel polarisation and the brightness is similarly affected.
The above considers only fringe strength at the plate. There is also the polarisation effects due to reflection and transmsion. Reflections from a metallic object, for instance, maintains the polarisation states of the incoming beam so if your referencing in one plane and objecting in another - say your ref is overhead and your object beams are in a horizontal plane - the interfering beams may be polarised at right angles to each other and, as mentioned above, and you'll end up with no fringesat all. A diffuse object depolarises, so while it seems bright to the eye, you may be losing a lot of the object light due to cos term damping. The reflected light is polarised in all possible direction but the reference is polarised in one, say vertical.Now divide the circle into segments of 10 degrees. there are 36 of these segments. The polarisation states of the reflected object light from the diffuse object will exist in all 36 segments but each segment will damp the interference term by cos(10*segment number). So the light reflected off the (diffuse) object whose polarisation states are in segment 2 will dampen the fringe strength by cos(20), the pol state of the reflected light in segment 3 wiull dampen it by cos(30) etc. If I is the intensity of the light reflected off the diffuse object,the true intensity for interferometric purposes as seen by the reference beam is;
I/36 + (I/36)*cos(10 + (I/36)*cos(20) + (I/36)*cos(30) +...+(I/36)cos(360).
This is a lot less than the measured intensity of I. The number of segments, 36, was chosen because mostly the cosine function does not appreciably change over a 10 degree variation. If you wanted to be exact, you'd have to consider 360 segments at 1 degree per segment.
There are other factors as well, however a clay (diffuse and depolarising!) gecko awaits me on the table....
You might want to look at the reference beam with a polariser and orientate it to get the maximum light through it. Then look at the reflected object light with the polariser at the same angle. This is the best possible tru object light. As you turn the polariser, the light you see from the object, the true object light, is the intensity you see through the polariser damped by the cos of the angle through which the polariser is turned.
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Tom M
Just how important is polarization to making holograms??
"there are 36 of these segments"
Don't you get interference at both the 180 degree areas? I mean, 180 degrees out is still interfering just oposite. If constructive at a given point, then destructive at the same point with 180 degrees out. Yes, no???
Help me mister wizard, I'm sinking fast.
Don't you get interference at both the 180 degree areas? I mean, 180 degrees out is still interfering just oposite. If constructive at a given point, then destructive at the same point with 180 degrees out. Yes, no???
Help me mister wizard, I'm sinking fast.
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Ted Park
Just how important is polarization to making holograms??
My experience has been that it's possible to get good holograms with the polarization going the wrong way, but results are usually better if you can get the polarization correct. The laser diodes are polarized across the short axis of the beam, so for the easiest setup (with the wide part of the beam going across the table so it can illuminate the object and the plate in a transmission hologram) the polarization is such that you get reflections anyway.
For H1->H2 setups, I've never experimented with changing the polarization, so I can't comment on that.
For H1->H2 setups, I've never experimented with changing the polarization, so I can't comment on that.
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Dinesh
Just how important is polarization to making holograms??
"Don't you get interference at both the 180 degree areas?"
Yes you do. I was talking about the interference effects at any one point. Actually, since interference cannot be a local event, I should say that when I say 'at any given point' I mean 'over a small area'- maybe a couple of fringes.
If the reference and object are polarised in parallel everywhere, then at any one point on the plate (ie over a small area), the fringe contrast is at a maximum. However, at any given point on the plate, the object beam, ie the rays from all the different parts of the object to that point, has light from all over the object, each one with a specific polarisation state so the total effect at that point is the net effect of the (variable) polarisation state of the object rays and the (fixed) polarisation state of the reference beam. This means that the fringe contrast at any point on the plate is a diminished by the variation in the relative pol states. At differrent points on the plate, the fringe contrast varies as the pol state of the object rays vary. The total efficency of the hologram is the total diffraction from all of these fringes, which is reduced from its possible maximum by the variable pol states of the object beam.
Perhaps it's important to stress that interference and diffraction are not local effects, but are average effects over a region of space. Also, the fringe at any given point on the plate are the result of interaction of all the rays from the entire object interacting with the reference.
Yes you do. I was talking about the interference effects at any one point. Actually, since interference cannot be a local event, I should say that when I say 'at any given point' I mean 'over a small area'- maybe a couple of fringes.
If the reference and object are polarised in parallel everywhere, then at any one point on the plate (ie over a small area), the fringe contrast is at a maximum. However, at any given point on the plate, the object beam, ie the rays from all the different parts of the object to that point, has light from all over the object, each one with a specific polarisation state so the total effect at that point is the net effect of the (variable) polarisation state of the object rays and the (fixed) polarisation state of the reference beam. This means that the fringe contrast at any point on the plate is a diminished by the variation in the relative pol states. At differrent points on the plate, the fringe contrast varies as the pol state of the object rays vary. The total efficency of the hologram is the total diffraction from all of these fringes, which is reduced from its possible maximum by the variable pol states of the object beam.
Perhaps it's important to stress that interference and diffraction are not local effects, but are average effects over a region of space. Also, the fringe at any given point on the plate are the result of interaction of all the rays from the entire object interacting with the reference.