Dumb Question #214

[Tony]

We all know that if you break a hologram into tiny pieces, all the pieces still contain enough information to recreate the complete hologram: smash a glass hologram of a cup into bits and you can still see the entire cup in any of the broken pieces of glass. Providing this equation is met A/((r)^2) >= A'/((r')^2).

If you were very small, teany tiny, how small can the piece of glass get before you can no longer see anything? How many fringes do you need?

Thanks

[Holomark]

Well I am no expert, but it seems to me you don't need any glass. You need the recording material on the glass or in the film.
Now since you mentioned a formula, you obviously want a more technical response, which I do not have, but I would imagine that it would be based upon the wavelength of the light source used to illuminate your recording material. But the more of the image you eliminate the more you degrade you image.I would wonder how you would light such a tiny bit of recording material.

[Wler]

Well "all the pieces still contain enough information to recreate the complete hologram" is not quite correct, there are two issues:

1) Reduction of perspective. Think of a hologram as a window. Closing say half of the window will reduce the perspective, and thus information is lost. This may not be evident for far-away objects, but for close objects (think about one being away one inch from the window) this is quite obvious. I guess when people say this above sentence, they mean infinitely far away objects.

2) Reduction of resolution. As always in diffractive optics, the resolution grows with the number of illuminated fringes. Having just, say, two fringes illuminated will produce a very structureless blob. The more and more fringes are illuminated, the sharper the picture becomes.

There is actually a nice and instructive little experiment that can be one. You don't need to smash your precious hologram. Just illuminate a transmission hologram with an unexpanded laser beam, sampling the hologram over one millimeter-square, say. There will be an image projected off the hologram that can be captured on a white screen or wall. It will not be very sharp but recognizeable. Scanning with the beam over the hologram changes the perspective of the projected image.

The real fun thing is to show this to a layman! He/she won't find anything peculiar, because they think this is like a slide and shining a beam through a slide produces an image as usual. But then point out that the beam samples the "slide" just at one spot, and not the whole of the "slide". This is when it makes "click"

[Johnfp]

My intuition is a single fringe. I think of a fringe as a rounded (sinusoidal in both ways) prism that has has all the views that the point sees of the object.

[BobH]

I suggest doing Wler's experiment with the raw beam, but put a very long focus lens into the beam before the hologram. That way, you can move the hologram through the beam waist and see the effect of decreasing beam diameter on the image projected. Go from a couple millimeters down to maybe 20 microns or less.

[Johnfp]

You two, that would be a cool experiment. Put a small piece of a hologram in a spatial filter where the pinhole would be. Then adjust the objective in and out.

[Dinesh]

Reduction of resolution. As always in diffractive optics, the resolution grows with the number of illuminated fringes. Having just, say, two fringes illuminated will produce a very structureless blob. The more and more fringes are illuminated, the sharper the picture becomes.

ust illuminate a transmission hologram with an unexpanded laser beam, sampling the hologram over one millimeter-square, say. There will be an image projected off the hologram that can be captured on a white screen or wall. It will not be very sharp but recognizeable.

It's not so much resolution as an averaging process. By the way, this only works for transmission holograms, it won't work for reflection. This is because in a transmission hologram, the fringes are perpendicular to the plate - across the plate - and so a beam illuminates a number of fringes dependent on the fringe width and the beam diameter.

Consider a hologram made with a HeNe referenced at Brewsters. In this case, the spatial frequency is about 1370 l/mm, so the fringe width is 0.7 microns. A beam with an area of one mm square has a radius of 500 microns, so the beam will cover about 1400 fringes. The reconstruction of a hologram, especially a display hologram, relies on a range of diffraction angles over a range of spatial frequencies, giving a range of diffraction angles; this is the origin of the Fresnel zones model. So, when over 1400 fringes are illuminated, enough of the fringes are illuminated to give an average effect that will produce an image with limited depth. As the beam is widened, more and more fringes are illuminated and an average over a larger number of fringes gives more substance to the image, effectively filling it out. If the beam diameter were reduced, the number of fringes illuminated by the beam is also reduced and the fringe structure appears to be more and more uniform to the illuminating beam; as if the entire hologram consisted of only one set of fringes as far as the illuminating beam were concerned. This would effectively be a grating and the beam would produce only one diffracted direction. This is the other extreme, where there is no averaging effect over a range of fringes, they're all the same width and diffract at the same angle.

If a reflection hologram, one with Bragg selectivity, were illuminated with a narrow beam, then that beam would have to pass through the emulsion, hitting succesive Bragg planes (at the Bragg angle else there is no diffraction) and diffract from them. However, the efficiency of a Bragg hologram is determined by the area of the Bragg plane illuminated and a narrow beam would only illuminate a tiny part of the Bragg plane, so the efficiency would be very low. Eventually, scattering and losses would decay the beam before it got very far into the emulsion.

The resolution of a hologram is primarily dependent of on the spatial coherence of the reconstruction source, ie the width of the source. This is because each geometric point on the source reconstructs the image from a specific direction. The larger the source size - the greater its spatial coherence - the larger the number of reconstruction sources and the larger the number of images all superimposing on each other and causing a fuzz in the image

I suggest doing Wler's experiment with the raw beam, but put a very long focus lens into the beam before the hologram. That way, you can move the hologram through the beam waist and see the effect of decreasing beam diameter on the image projected. Go from a couple millimeters down to maybe 20 microns or less.

Airy radius, q, is given by

q = 1.22 f*(Lambda)/D

For the above beam, D=2mm = 2000u, q=20u and (assume) lambda=0.6u giving

f = D*q/(1.2*lambda) ~ (40,000/0.7)u = 57.15mm ~ 22.5in

[Dinesh]

If you were very small, teany tiny, how small can the piece of glass get before you can no longer see anything? How many fringes do you need?

At the limit, very roughly about lambda. So, if you're illuminating with a HeNe, the piece of glass should be about 0.6 microns. This follows from the grating equation:

sin(theta) = lambda/d

Since sin(theta) cannot be greater than 1, the diffracting aperture would have to be greater than lambda. This would give you diffraction, but would not give you any discernible image. To "see" an image, you'd have to average over a number of fringes (see my above comment) and so it would depend on the complexity of the image. That is, if the "image" were simply a point of light, then the diffractive aperture must be lambda or greater. If the "image" were two points of light, you'd need at least two fringes, each one with a slightly different spatial width. The ability to differentiate two spots of light would depend on the distance from hologram to the screen, since the divergence of the two spots would begin at the plate and get further apart as the screen were moved away.

[Johnfp]

That is, if the "image" were simply a point of light, then the diffractive aperture must be lambda or greater. If the "image" were two points of light, you'd need at least two fringes, each one with a slightly different spatial width

So what am I missing here. If image were 1000 points in a line, then we would need 1000 fringes?

Couldn't one fringe actually refract the beam like a cylindrical lense and form all 1000 points? The instesity may be weak, but...

How many points are in a 8 foot sqare room filled with objects that can be used as the object beam (of course you would need a nice coherence length) to create a transmission hologram only 2" square. Are the number of fringes in that 2" hologram equal to the number of points in the room?

I believe a single fringe (if larger then the illuminating wavelength) could reproduce that entire room of billions of points. It would be a fringe that looks like the topigraphical outline of a city with many different shaped and buildings with differnt heights.

In DCG this would be in the gelatin
In Silver this would be the arrangement of silver through the depth of the fringe.
To me a fringe is not a line or even a sinusoidal line but a 3d prism with many peaks and valleys contained within it.

[Johnfp]

Actually in thinking of the experiment with two slits in which thye shot a single photon at a time at them and watched what happened on an LCD screen, I change my thought to 2 fringes. I will try to find the exeriment if I can and post it.