If one were to try to store two images on a plate they would have a couple of options that I have tried. One is to tape off half of the plate and shoot one scene then tape off the other half of the plate and shoot the other scene.
A second method would be to change the reference angle by 30+ degrees between exposures.
Both of these methods allow one to view one image at a time. But if I wanted to see both images at the same time, would it work to rotate the polarization by 90 degrees between exposures? Since the replay light has random polarization, wouldn't both images be viewable simultaneously? Would this be better for efficiency than simply making a double exposure with the same polarization?
"But if I wanted to see both images at the same time, would it work to rotate the polarization by 90 degrees between exposures?"
The fringes themselves are polarisation independant, they're just variations is density/refractive index. In other words they're scalar variations. So, without going into too much detail (wait for the video of the PCG meeting for that!)the efficency of a hologram depends on the density variation, which in turn depends on the angle between the polarisation vectors of the object and reference.
Let's assume you have two objects that both depolarise in such a way that the distribution of polarisation states of the reflected light is random. In this case, no matter what the polarisation of the reference, the proportion of light intensity in the object beam with polarisation state at some angular seperation, d_theta would be d_theta/2pi. In other words, if the polarisation of the reference beam were vertical, you might set the light intensity of the object that contributes to efficency to be that part of the object light that has a polarisation state of +-10 degrees from vertical. In this case, the proportion of object light that contributes to effeciency (purely from a polarisation standpoint)is 20/360 of the actual reflected light from the object. If the polarisation of the ref were horizontal and the object depolarised randomly, the proportion of the object light that contributes would still be 20/360, under a simil;ar argument. So the efficency would be the same in both cases, whether you shoot simultaneously or not.
Let's say the object(s) did not depolarise randomly, but maintained the polarisation state of the reference, ie the polarisation states of the reflected light from the object(s) were roughly the same as that of the reference, in both objects 1 and 2. Then vertical polarisation ref would have a strong efficency with object 1 and none with object 2, whereas the horizontally polarised ref would have a strong eff with object 2. Once more, the efficency for both objects would be the same. However, it would now make a difference whether you shot simultaneously or not.
You can pretty much follow this line of reasoning with several other scenarios. If you're up on you're statistics, you should be able to calculate the number of variation formed by two refererence states, two object states and two objects that can take on either state. In my schooldays, we had endless problems like: If I have 2 red books and 5 green books how many ways are there of arranging the books so that no two red books are together. I hated those problems! Unfortunately theoretical physics is full of 'em and I had to get used to them.
By the way there is a field of Polarisation Holography where you bring in the beams of opposite polarisations from opposite directions on the same object. This is a way of mapping the polarisation states of an object. HOEs are a completely different sort of animal. There are also ways of using polarisation states to manufacture certain kinds of HOEs.
Couldn't you just shoot one image then replace the object and shoot the second object with the reference beam at the same angle? Then both would be visible at the same time in the replay.
I could not responed directly to your message. I guess my goal is to think of the efficiency of storing two holograms. If I make a hologram of an object, then change the reference beam angle I have to make the exposure longer on the second exposure in order for the images to be of the same brightness, right? If I were to encode the reference beams with two polorization states would the extra exposure time be eliminated? Or is it simply a function of not have much un-exposed silver to work with on the second exposure?
So you're making multiple exposures and, yes you're right that the exposures have to be increased. The question is: Would the exposure times be constant if the polarisations are crossed? In a word, I don't think so. OK, that's 4 words!
Basically, the problem is one of reciprocity. The reason that you have to increase the exposures is that the silver atoms formed from the first exposure on the individual grain form a nucleus for subsequent silver formation on the same grain (you have the essay I wrote on Silver Halide). This increases noise if you expose a second time immediately. If you wait a few seconds, then reciprocity failure begins to set in and you get a lower efficency for subsequent exposure, the actual amount of efficency loss depending on the time delay between exposures among other things (this is assuming you haven't altered the sensitivity of the plate with TEA for pseudo-color). In order to maintain the same efficency, you need to expose for longer, again the exact amount depending on your initial sensitivity and the time delay between exposures. As far as I know, the reciprocity effect, ie the binding, or not, of silver ions to wandering electrons and their stability, is not affected by the polarisation of the beam but the energy transferred to the plate. This energy transfer is a scalar effect and not a vector one. However, to the best of my knowledge there is no paper dealing with some kind of "vector" theory of reciprocity.