Change in refractive index due to embedded hologram
Change in refractive index due to embedded hologram
Hello,
My name is Joseph Matchett and I work for RL Associates in Chester, PA. I have a porous glass material that has a hologram written throughout the thickness of the material. I am able to measure the bulk refractive index of the material. However, I would like to measure the change in refractive index of the material due to the hologram. If you can help me out with a way to do this, please reply via Email (joseph_matchett@yahoo.com). Thank you very much. Sincerely,
Joseph D. Matchett
My name is Joseph Matchett and I work for RL Associates in Chester, PA. I have a porous glass material that has a hologram written throughout the thickness of the material. I am able to measure the bulk refractive index of the material. However, I would like to measure the change in refractive index of the material due to the hologram. If you can help me out with a way to do this, please reply via Email (joseph_matchett@yahoo.com). Thank you very much. Sincerely,
Joseph D. Matchett
Change in refractive index due to embedded hologram
I cannot answer you question but I have a question for you. You state that the hologram is written throughout the thickness of this porous glass material. Are you sure? Is it an emulsion layer of some sort adhered to a porous glass material or is the porous glass material some sort of optical recording material of its own?
Change in refractive index due to embedded hologram
The glass is an optical recordign material. Two lasers are directed at the material, one of either side. The angle sof the two lasers are offset. I don't pretend to be an expert on holograhpic recording, but this is what I have gathered. If you'd like, take a look at http://www.lptinc.com/R&D_projects1.html for more info.
Change in refractive index due to embedded hologram
I know this sounds simple but have you tried to make a hologram in half of the material while blocking the laser light from the other half and measuring the difference in refractive index with you current method of the two halves? Or can you obtain a sample of the material without a hologram and measure its "bulk" refractive index and compare it to the holograms refractive index? Because the refractive index changes from crest to trough on a single fringe, you would have to measure the refractive index where there was constructive interference and where there was destructive interference which will be a fraction of the size of the wavelength of the laser you used.
Change in refractive index due to embedded hologram
"However, I would like to measure the change in refractive index of the material due to the hologram."
The diffraction efficency of a volume hologram is dependant on the refractive index modulation. The efficency, eta, is given by:
eta = [pi)(delta_n)(T)]/[(lambda)(cos(theta_m))]
pi = 3.141592653589796 (Joke! I pride myself on knowing pi to this many figures!)
delta n = refractive index modulation = n_bulk - n_fringes
T = Thickness of hologram
lambda = recording wavelength
theta = recording angle in the medium
If you know the bulk refractive index, you can measure the efficency and so get the change in index. The index is also depenndant on the thickness, so you need to know that.
A couple of words of caution:
You may want to check the extent to which this is a volume hologram by determining the Bragg selectivity, since the efficency calculation assumes a truly thick hologram. If the Bragg selectivity is pretty narrow, the error is not too bad. Check the loss in efficency as the hologram is turned off-Bragg on both sides. Thbis sshould be low.
You say you measured the bulk refractive index. Was this before the recording or after? The recording process also increases the bulk index. You can see this by
I = I_1 + I_2 + 2*I_1*I_2*cos(<phase difference>)
where the various I's are total intensity, intensity of beam 1 and intensity of beam 2. The first two terms lead to an overall increase in the actinic effect.
The diffraction efficency of a volume hologram is dependant on the refractive index modulation. The efficency, eta, is given by:
eta = [pi)(delta_n)(T)]/[(lambda)(cos(theta_m))]
pi = 3.141592653589796 (Joke! I pride myself on knowing pi to this many figures!)
delta n = refractive index modulation = n_bulk - n_fringes
T = Thickness of hologram
lambda = recording wavelength
theta = recording angle in the medium
If you know the bulk refractive index, you can measure the efficency and so get the change in index. The index is also depenndant on the thickness, so you need to know that.
A couple of words of caution:
You may want to check the extent to which this is a volume hologram by determining the Bragg selectivity, since the efficency calculation assumes a truly thick hologram. If the Bragg selectivity is pretty narrow, the error is not too bad. Check the loss in efficency as the hologram is turned off-Bragg on both sides. Thbis sshould be low.
You say you measured the bulk refractive index. Was this before the recording or after? The recording process also increases the bulk index. You can see this by
I = I_1 + I_2 + 2*I_1*I_2*cos(<phase difference>)
where the various I's are total intensity, intensity of beam 1 and intensity of beam 2. The first two terms lead to an overall increase in the actinic effect.
Change in refractive index due to embedded hologram
We can measure the refractive index, that's no problem. We can't make a hologram in only half of the material; it's basically all or nothing. So, I proposed the idea of making a single hologram in a sample, and measuring the refractive index from side to side so that it would intersect the hologram at some point. Then, take the refractive index of that sample. Finding the difference between the refractive index of that sample and the bulk index, we woudl knwo the change. However, since the hologram is written with a laser, there are points of bulk material within the section containing the hologram, and the holographic points are on the order of nanometers. So 1) we'd be hitting partial bulk material at the same section as the hologram and 2) we would have no idea of how many holographic points we are hitting. It's a really good idea though.
Change in refractive index due to embedded hologram
>The diffraction efficency of a volume hologram is dependant on the refractive >index modulation. The efficency, eta, is given by:
>eta = [pi)(delta_n)(T)]/[(lambda)(cos(theta_m))]
>If you know the bulk refractive index, you can measure the efficency and so >get the change in index.
Perhaps I am missing somethign, and I don't intend for that to sound rude. But I don't know the change in index, so I can;t compute eta from that calculation. Is there another way of calculating eta so that I can then go back and solve for delta_n?
>eta = [pi)(delta_n)(T)]/[(lambda)(cos(theta_m))]
>If you know the bulk refractive index, you can measure the efficency and so >get the change in index.
Perhaps I am missing somethign, and I don't intend for that to sound rude. But I don't know the change in index, so I can;t compute eta from that calculation. Is there another way of calculating eta so that I can then go back and solve for delta_n?
Change in refractive index due to embedded hologram
"I don't intend for that to sound rude."
Not at all.
"Is there another way of calculating eta so that I can then go back and solve for delta_n?"
You don't need tocalculate eta, you measure it. Eta is the ratio of the power incident on the hologram to the power of first diffracted order. If the hologram is truly thick, there should only be one order so 'first order' should be superfluous. be careful to measure the diffracted power as a ratio of the incident reconstruction beam, as opposed to the zero order beam. I've seen wildly fantastic claims of incredibly high efficencies because it was measured relative to the zero ordeer.
By the way, I wrote this too fast, the actual eta is the tanh^2 of the argument given above.
So here's a possible procedure:
Measure the power of the expanded (if necessary) laser beam before the hologram is in place. Call this I_0. Put the hologram in place. Hit the hologram with this laser beam and measure the power of the first (hopefully only) diffracted order. Note the angle of the diffracted order wrt the reconstructed beam. Call this diffracted power I_1. (Note, by the way that I'm using 'I' which is, strictly speaking, energy and not power - this doesn't matter since the time term will cancel anyway.). Call the bulk index, after exposure n_b and the fringe index n_f So, now:
delta_n = |n_b-n_f| This should be positive-definite, but I never remember which way it goes, hence the modulus!
eta = I-1/I_0 = tanh^2([(pi)(delta_n)(T)]/[(lambda)(cos(theta_m))]
From which,
n_f = [(arc tanh (sqr root(I_1/I_0)))(cos(theta))(lambda)]/[(pi)(T)] + n_b
I suspect the 'T' is going to be difficult to measure. theta should be the angle of the diffracted order, assuuming normal incidence. If the beams are counter-propagating, forget the cos(theta) term.
Now revolve the hologram slightly, say 2 degrees, re-measure eta. Repeat for about a 10 degree variation on either side of the efficiency peak. Plot efficiency against angle. Consult Kogelnik as to whether or not you truly have a volume hologram. If not, calculate
d(eta)/dT and fit accordingly. If so, Bob's your uncle!
Not at all.
"Is there another way of calculating eta so that I can then go back and solve for delta_n?"
You don't need tocalculate eta, you measure it. Eta is the ratio of the power incident on the hologram to the power of first diffracted order. If the hologram is truly thick, there should only be one order so 'first order' should be superfluous. be careful to measure the diffracted power as a ratio of the incident reconstruction beam, as opposed to the zero order beam. I've seen wildly fantastic claims of incredibly high efficencies because it was measured relative to the zero ordeer.
By the way, I wrote this too fast, the actual eta is the tanh^2 of the argument given above.
So here's a possible procedure:
Measure the power of the expanded (if necessary) laser beam before the hologram is in place. Call this I_0. Put the hologram in place. Hit the hologram with this laser beam and measure the power of the first (hopefully only) diffracted order. Note the angle of the diffracted order wrt the reconstructed beam. Call this diffracted power I_1. (Note, by the way that I'm using 'I' which is, strictly speaking, energy and not power - this doesn't matter since the time term will cancel anyway.). Call the bulk index, after exposure n_b and the fringe index n_f So, now:
delta_n = |n_b-n_f| This should be positive-definite, but I never remember which way it goes, hence the modulus!
eta = I-1/I_0 = tanh^2([(pi)(delta_n)(T)]/[(lambda)(cos(theta_m))]
From which,
n_f = [(arc tanh (sqr root(I_1/I_0)))(cos(theta))(lambda)]/[(pi)(T)] + n_b
I suspect the 'T' is going to be difficult to measure. theta should be the angle of the diffracted order, assuuming normal incidence. If the beams are counter-propagating, forget the cos(theta) term.
Now revolve the hologram slightly, say 2 degrees, re-measure eta. Repeat for about a 10 degree variation on either side of the efficiency peak. Plot efficiency against angle. Consult Kogelnik as to whether or not you truly have a volume hologram. If not, calculate
d(eta)/dT and fit accordingly. If so, Bob's your uncle!
Change in refractive index due to embedded hologram
Dude, is all that stuff stuck in your brain? What do you do, eat formulas for breakfast? I dream at night about different holograms, I can only imagine what you dream about!
hahaha!
Peace!!!
hahaha!
Peace!!!
Change in refractive index due to embedded hologram
I believe you're talking about the Kogelnik theory. I have been using that up to this point to give a model of what should be my delta_n. However, even after using it, there are a number of variances in each sample I have, even if they are all supposed to be the same, that makes a model insufficient. Is there a way to physically measure the change?