## Change in refractive index due to embedded hologram

These are all of the old posts from the first two years of the forum. They are locked.
Updated: 2005-03-28 by HoloM (the god)
JohnFP

### Change in refractive index due to embedded hologram

Ok, lets look at this another way. The refractive index of this material changes as a function of the amount of light incident on it for a period of time. Why not measure the refractive index before you start and then measure the refractive index of the material after it has been exposed with a single beam of the laser light and time durration such that it is the same energy per square cm as a calculated constructive fringe peak. Then calculate the fringe size and spacing. Using these and the average of the "hologram", you should be able to calculate where on sin curve of the fringe an exact refractive index is.
No?

Dinesh

### Change in refractive index due to embedded hologram

Believe it or not, and Joy doesn't, I think in mathematics. If someone talks about the brightness of a hologram, I guess most people think in terms of a bright image. I think in terms of tanh squared f(delta n). Even when throwing a piece of paper into a bin abaout a few feet from me, I think' a x squared plus b x squared equals 1' (the equation of a parabola) and s equals a half a t squared.
What do I think of at night? Well right now, I'm thinking that if you could reformulate the Schrodinger equation into a non-Minkowskian metric, it may be possible to show that energy levels inside a particle may be distorted at the edge of a black hole. By using the Hawking Principle of Evaporating Black Holes, the entropy change may be a way of getting quantum gravity. Oh yeah, and can you get color control by using John's "broken mirror" effect and tilt the diffusion plate. The tilt angle will vary the amount of light onto the plate and hence vary the ratio. In DCG, this may be a way of color control.

Dinesh

### Change in refractive index due to embedded hologram

The variances are possibly due to the fringe structure. Kogelnik assumes plane beams, an infinite plane and perfectly sinusoidal fringes. In practice these are not ever achieved. One way to tell the difference between ideal Kogelnik and a practical hologram is to plot any of the parameters and compare with theoretical curves in the paper. One way I suggested was to plot the efficiency against the dephasing parameter xi (figs 12 and 13 in the paper). By noting any disparity between the curves it may be possible to correct for Kogelnik assumptions. I assume since this is in glass there is no possibility of it being a mixed grating, ie it is pure phase.
John's idea of hitting a sample of the substance with one beam may work. However, the sample size, I would think, needs to be small, about as small as the fringe width. If not, it's possible that absorption could alter the index. If, therefore a sample of thickness about a micron or so could be illuminated in such a manner as to change it's index, you could now reflect laser light from the surface, measure reflectivity in both polarisation states and use the Fresnel equations to get index. Of course, you don't need to reflect in both polarisation, it's just better perhaps to average out the difference.
One other way occurs to me. If you could expose it in such a manner as to get a "top hat" fringe profile, either by massive overexposure or by simultaneous recording of many object beams to one reference (Fourier superpostion), then you could use this as an interference filter. By placing this in some sort of spectro-photometer you'd get the reflectivity at various wavelengths (I'd expect a Gaussian peak over a narrow range). You could now use the Fresnel equations and the interference condition together to get the index difference and so, knowing the bulk index, get the fringe index.

JohnFP

### Change in refractive index due to embedded hologram

OK, let me try to explain better. Now this is a logical solution and not really a mathimatical solution.

Ok, you find out the refractive index of the material without being exposed to light.
Then you set up a holographic configuration with two beams entering from same side of the plate (transmission hologram) with a known angular separation (very close to get the fringes running nearly perpendicular to the material surface plane) and the beams are as “flat top” as possible (same intensity across the entire plate in all directions).
You measure the light intensity of each beam at a spot where the plate plane lies.
You expose that plate for a know length of time and measure its refractive index. This is your hologram.
As constructive interference is the addition of both intensities, add the “power” of each beam together (energy and time).
Now, take another plate and expose it with just a single “flat top” beam across the entire plate for the calculated power above and measure its refractive index.

Now you have 3 samples with presumably 3 different index of refractions, your base line which is the glass unexposed (homogeneous refractive index), your hologram (non homogeneous refractive index) and your top line which is the glass exposed to a single beam (homogenous refractive index).

Now calculate the fringe spacing from the wavelength of laser being used, and the angular separation. (d=wavelength/2 sin theta - where theta is the half angle between the two laser beams).

So now you have a sin wave of know spacing (fringes), with known refractive index at maximum and know refractive index at minimum. You also have a holographic sample with know refractive index. With these samples you should be able to calculate what you need.

JohnFP

### Change in refractive index due to embedded hologram

Hey Dinesh,
I hope Joseph is true and wants information and does not take advantage of our good nature in helping or trying to help others in their questions. I have seen a few times on this board and now that I am aware of your concerns I think see a different side to the story. It would be a shame if the Joseph fellow asks a question, someone on the board leads him to a path that enlightens him to a solution or he arrives at it on his own and he never comes back with a thanks, acknowledement or follow up. It does make you feel kind of used and I am becoming more and more aware of this, Nothing personal Joseph.

SO my quesion is, Joseph, has anyone on this board lead you to an idea that may help solve your problem? If so or if not so, please keep us informed of your quest and findings , you started the thread and if you solve the problem on your own, let us know how you did it. Being an inquisitive person, knowing a problem, trying to solve it, even if unsuccessful and never knowing the answer drives me crazy. I may continue to think about it for days. Put me to rest if you have an answer. I think that is what the sucess of this board is based on and it is a thin line. And that is, those that want should try to give cause the board knows there are plenty here that give with minimal taking and there are those here that take with minimal giving.If we all give honestly when we want, then the board and holography will grow.

This applies to everyone that reads this board. If you ask a question, even if it does not get answered here but some were kind enough to try and help, you should post your findings as that would be the right thing to do.

Peace to all!!