MichaelH wrote:Maybe we should start off with a thread (like this one! ) and see if the discussion merits a whole category. We could end up with an intimidating front-end if the categories proliferate too quickly.
Just to throw in my two cents, I'm working on a hologram for my father which features a P-51 airplane and because of the lighting I'm attempting to make use of some overhead scaffolding to hold a mirror that allows me to light the plane from the "overhead" as well as a side mirror that lets me light it from the front. I hadn't intended to use the scaffolding yet but needed the height in order to get the mirror up and over the model.
Next I'll probably break out my 1/2 wave plate and attempt and overhead reference setup using the scaffolding.
Remember you can fold the light path with two mirrors to accomplish a 90 degree polarization rotation without a wave plate. Kaveh's thesis pointed this out to me.
Simple bend the path up 90 deg then left (or right) 90 degrees and back to a parallel plane. If that makes any sense.
That certainly makes sense and I'll be using that method to re-orient the polarization once I mount my big laser on the wall (the laser can't be rotated and the beam will be coming down towards the table so I'll turn the beam again to get the polarization parallel with the table).
I don't want to introduce too many optics into the system because it increases the chance of error. I can do what I need (I think) with one waveplate vs two or more mirrors.
When you reflect the beam through 90 degrees to alter the polarisation, be aware of what's actually happening. The polarisation of the two beams at the plate is what's important. If the polarisations are parallel, you get maximum efficency but you risk woodgrain. Note you risk woodgrain. Whether or not you actually get woodgrain depends a lot on beam ratio. As mentioned in a reply to Colin's post in the last forum about "sheen", if the ratio is just right, there's very little of the reference left to cause woodgrain. Anyhow, if the polarisations are not parallel, the fringe density reduces by the cos of the angle between polarisation vectors. Now, if the setup is on the same plane, you can use simple descriptions like 's' or 'p' or even 'vertical' or 'horizontal'. If the setup is perpendicular (refernce at right angles to object) you can still do this, bearing in mind the 's' and 'p' does not always mean 'vertical' or 'horizontal'. The terms 's' and 'p' refer to the Plane of Incidence and so will depend on the plane formed by the last reflection. However in your P51 setup, if the object beams are coming in from angles that are not in a horizontal plane or a vertical plane, then the polarisation vectors are twisted in two different directions.
The reason that a 90 degree change in direction causes a change in polarisation is that the polarisation vector cannot be parallel to the direction of the beam. If the polarisation of the beam is vertical in a horizontal plane, then sending it 'upward', ie into a vertical direction, must cause the polarisation to shift, since the polarisation can no longer be 'vertical'. It now shifts to 'horizontal'. However, 'horizontal' can mean two different directions - asuming 'up' is the z axis, the polarisation can now go along the x- or y- axis since both of these directions are now perpendicular to the beam direction. If the polarisation vector goes completely along the x- or y- axes and then the beam itself is reflected into the x- or y- direction, the polarisation has to shift yet again (can't be along direction of beam travel). If the 'up' travelling beam polarisation shifts along both axes, x and y, then either way, you've lost half the beam power. In addition to all this, a 45 degree reflection causes a loss in reflectivity depending on the kind of mirror you have.
When you reflect the beam through 90 degrees to alter the polarisation, be aware of what's actually happening. The polarisation of the two beams at the plate is what's important. If the polarisations are parallel, you get maximum efficency but you risk woodgrain
I can only assume that by "parallel" you mean parallel to the plate. I've read at least one post or report that said efficiency is higher in that orientation but don't really understand how since a lot of light is lost through reflection off the glass.
It seems to me that the most important beam, speaking strictly of polarization, in display work is the reference beam since the object beams tend to have their polarization pretty well randomized by most subjects. If you've got a lot of specular reflection and want to maintain the hotspots, this of course wouldn't be true.
I'm at work and only able to steal the odd moment here and there to read your post and I'm afraid I didn't quite follow all of it as-is. If my response doesn't make sense, just ignore it.
I don't know how to copy and paste the way I used to and I can't see your original post, so I may be misquoting you. However, by parallel I mean the polarisation vector.
Imagine a beam of light travelling left to right. We'll call that direction the positive z axis. In this case, the polarisation - always at right angles to beam direction - is an "arrow" pointing somewhere in the x-y plane. In a right-handed system the x axis is pointing away from you and the y axis is pointing up. The polarisation is 'plane polarised' if this arrow is pointing anywhere in the xy plane - say at 45 degrees to both the x and y axis (1st quadrant) - and does not revolve. In other words, this arrow grows and shrinks but is always pointing in a direction 45 degrees to the x and y axes. If it happens to point in the y direction it's 'vertical' polarisation if along the x it's 'horizontal' polarisation. Since these axes are just by coincidence along the gravitational field we use words like 'horizontal' In space, this means nothing and only the chosen co-ordinate system matters. Now, if another beam were to intersect this beam and we assume that the two beams are mutually coherent so they can interfere, then the degree to which they interfere, the fringe contrast as it were, would depend on the polarisations of the two beams. If both beams are going along the z axis and both beams had polarisations at 45 degrees to the x and y axes, the polarisation vectors would be parallel and interference would be maximum. If one of the two beam had a polarisation direction at 135 degrees to the x axis or -45 from the negative x axis (3rd quadrant) it would be at right angles to the first beam and you'd get no interference at all. If one of the beams were at vertical polarisation, there'd be an angle of 45 degrees between the polarisation vectors of the two beams and the fringe contrast would be diminished by a factor of cos 45 = 1/(sq root)2, whatever that is. Since you program graphics, I'm assuming you're familiar with direction cosines. If so, assume one beam along (0,0,1) and one along (1,1,1). Now the polarisation angle is not only got a theta twist but a phi twist. It's twisted and turned as it were.
In a hologram, the most important polarisation is the reference, since as you say, the object beam is usually depolarised. However this assumes that the reference-object plane is either all horizontal (horizontal reference horizontal object) or at right angles (vertical reference-horizonatl object). If the reference is either vertical or horizontal but the object beam is "diagonal" then no matter what the polarisation of either is, there's bound to be loss in fringe contrast because of the azimuthal or phi twis in polarisation - the turn I mentioned earlier. This is why some holographers are paranoid about getting the beams exactly horizontal. In reality, a variation in the plane of the shoot - a turn - of about 10 degrees doesn't affect it much since cos(10) = 0.9848 or, in strict mathematical jargon, damn close to 1.
) and see if the discussion merits a whole category. We could end up with an intimidating front-end if the categories proliferate too quickly.