Use a Hologram of a lens as a lens?
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Dinesh
Use a Hologram of a lens as a lens?
Yes, actually. Clearly you can't use rods or squeegees, unless you had a curved one with the same curvature.holorefugee wrote:Was coating a curved surface difficult?
This was 25 odd years ago, but, as I remember, we placed the surface concave side down. Then used a small parastolic pump to pump a pre-determined amount of gel onto the top of the curved surface (the convex side). The trick was to balance the temperature and amount so that the goo slid down the curved sides uniformly and at just the right rate that it didn't leave a "bald" spot at the top and pool at the bottom edges or vice versa, ie that it didn't freeze at the top and leave almost no goo at the lower edges. This meant that you had to place the nozzle of the parastolic pump at exactly above the centre of the curvature and make sure the goo went down at exactly right angles. If the pump nozzle was twisted, you got more on one side than the other.
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walschuler
Use a Hologram of a lens as a lens?
This is a very interesting discussion, at least to me, though I wonder if the original questioner is still engaged. I would like to add a bit to this.
First, I believe no one has remarked that a hologram of a lens shows at least some lens-like behavior not mentioned above. For example, long ago I made a reflection holo of a 3 inch diameter uncoated double convex lens in front of a rustic grid of wood twigs, behind which sat a white ceramic tile. The resulting virtual image clearly shows each of these elements, and if you line up your eye correctly you see the grid magnified through the lens, while outside the lens edge the grid is unmagnified (no surprise!). Using a bright source to view with you also see internal and external lens surface reflections of the source in the lens, as you would in real life. At least one of these is a real image, in that it appears to the eye to clearly hang in space between the hologram and the eye, and a white card placed at that location in front of the hologram will reveal a focused image of the light source. You could split hairs and say that because it is a reflection hologram this is mirror-like behavior, but since this the property of the original lens, I would call it lens-like.
If you made a transmission image (I don't normally make them) of a convex lens, then looking at the virtual image using a laser you should see internal and external reflections, and using a white source you should see colorful real images of the viewing source. Also, if set up for the real image, the main focus should be visible. It should give a dispersed color image, as would the actual lens, but red would be more dispersed (diffracted) than blue, the opposite of the behavior of the refractive effect.
So, Bob and Dinesh (and?), what say you?
First, I believe no one has remarked that a hologram of a lens shows at least some lens-like behavior not mentioned above. For example, long ago I made a reflection holo of a 3 inch diameter uncoated double convex lens in front of a rustic grid of wood twigs, behind which sat a white ceramic tile. The resulting virtual image clearly shows each of these elements, and if you line up your eye correctly you see the grid magnified through the lens, while outside the lens edge the grid is unmagnified (no surprise!). Using a bright source to view with you also see internal and external lens surface reflections of the source in the lens, as you would in real life. At least one of these is a real image, in that it appears to the eye to clearly hang in space between the hologram and the eye, and a white card placed at that location in front of the hologram will reveal a focused image of the light source. You could split hairs and say that because it is a reflection hologram this is mirror-like behavior, but since this the property of the original lens, I would call it lens-like.
If you made a transmission image (I don't normally make them) of a convex lens, then looking at the virtual image using a laser you should see internal and external reflections, and using a white source you should see colorful real images of the viewing source. Also, if set up for the real image, the main focus should be visible. It should give a dispersed color image, as would the actual lens, but red would be more dispersed (diffracted) than blue, the opposite of the behavior of the refractive effect.
So, Bob and Dinesh (and?), what say you?
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walschuler
Use a Hologram of a lens as a lens?
Also, one further note. The curved holographic grating Dinesh described is a new age ingeniously executed solution to the problem of making a curved grating. This problem was also solved about 100 years ago (need to check that), by people who made conventional gratings, including RW Wood, whom Dinesh mentioned above. Back then a conventional concave mirror was made, then metal coated to make a mirror surface, and this was then scribed by fine needle driven off a very fine thread screw to make the ruled grating. This element then typically became the combination of disperser and camera mirror, saving path length and optical elements in spectrographs.
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Dinesh
Use a Hologram of a lens as a lens?
Bill, assume the the setup you describe (lens, grid, screen) without a holographic plate and a reference beam. Then, assuming that the lens is sitting directly on the grid and that you don't have a super-fast (amazingly fast!) lens, the lens would create a virtual, magnified image of the grid behind the actual grid. The lens would act like a magnifying glass. Using the standard lens equation, the image distance, v, is:
v = u*f/(f-u)
where u is the object distance and f the focal length. If u>f, which it would be if your grid was right behind the lens, then the image distance is negative, creating a virtual image.
So, what's being recorded? Well, the light from the "laser" (the quotes are because I'm assuming there is a collimating mirror or lens which generates the reference) will hit the grid outside the lens diameter and would of course scatter back into the plate - the standard Denisyuk arrangement. The light that hits the lens itself would converge onto the grid because of the converging property of the lens, scatter off the grid, hit the back face of the lens and diverge (thus creating the virtual image that you'd see if you were actually looking directly into the lens). So, you have the forward scatter off the grid outside the circumference of the lens and the diverging light from inside the lens circumference that when traced back, would have created the virtual image of the grid. The hologram would record this particular wavefunction and recreate the exact same wavefunction. So, you've captured the light from some specific geometry of lens and object from some specific perspective.
I would suggest that capturing a wavefunction of a lens from a specific geometry at a specific vantage point is not a lens, per se, but the reconstruction of a specific wavefunction which happens to be coming from that particular lens. This "lens" would not obey the lens equation in the general form and it would not alter light coming from a random direction at some random divergence, as a real lens would do. In a real sense though, all holograms are re-creating a "lens" because the process whereby the recreating source is diffracted and creates an image is similar to a large number of lenses within the emulsion, each one taking light from some specific direction with some specific divergence and "focusing" it. You may remember that this was an early model of a hologram. Back then (pre-Kogelnik) they used an MTF model to determine the efficiency of a hologram and compared it to the standard MTF's of a lens.
By the way, this is the answer to Jeff Blythe's "impossible hologram". He was capturing a specific form oflight from a corner cube (the reflection of his face, as I recall) at some specific direction and so it acted as a corner cube from only that direction. It would not have acted like a general corner cube.
v = u*f/(f-u)
where u is the object distance and f the focal length. If u>f, which it would be if your grid was right behind the lens, then the image distance is negative, creating a virtual image.
So, what's being recorded? Well, the light from the "laser" (the quotes are because I'm assuming there is a collimating mirror or lens which generates the reference) will hit the grid outside the lens diameter and would of course scatter back into the plate - the standard Denisyuk arrangement. The light that hits the lens itself would converge onto the grid because of the converging property of the lens, scatter off the grid, hit the back face of the lens and diverge (thus creating the virtual image that you'd see if you were actually looking directly into the lens). So, you have the forward scatter off the grid outside the circumference of the lens and the diverging light from inside the lens circumference that when traced back, would have created the virtual image of the grid. The hologram would record this particular wavefunction and recreate the exact same wavefunction. So, you've captured the light from some specific geometry of lens and object from some specific perspective.
I would suggest that capturing a wavefunction of a lens from a specific geometry at a specific vantage point is not a lens, per se, but the reconstruction of a specific wavefunction which happens to be coming from that particular lens. This "lens" would not obey the lens equation in the general form and it would not alter light coming from a random direction at some random divergence, as a real lens would do. In a real sense though, all holograms are re-creating a "lens" because the process whereby the recreating source is diffracted and creates an image is similar to a large number of lenses within the emulsion, each one taking light from some specific direction with some specific divergence and "focusing" it. You may remember that this was an early model of a hologram. Back then (pre-Kogelnik) they used an MTF model to determine the efficiency of a hologram and compared it to the standard MTF's of a lens.
By the way, this is the answer to Jeff Blythe's "impossible hologram". He was capturing a specific form oflight from a corner cube (the reflection of his face, as I recall) at some specific direction and so it acted as a corner cube from only that direction. It would not have acted like a general corner cube.