Red sensitivity of MBDCG

Dichromated Gelatin.
pluto
Posts: 94
Joined: Fri Feb 05, 2016 2:06 am

Red sensitivity of MBDCG

Post by pluto » Mon Nov 20, 2017 12:11 am

Hey all,

Put simply, I'm wondering how much power at 633nm would be needed to produce a bright 4x5 inch hologram. Not asking for anything terribly precise, perhaps to the nearest 10mW.

From searching on the forum I got the figure of 30-50mJ/cm^2. I'll use the low end figure of 30mJ/cm^2.

For a 5x4 inch hologram that would work out to 3.87 joules...

So 3.87 joules / 50 milliwatts = 77 seconds

Would it be fair to say you could achieve a bright hologram using 50mW of red light for 77 seconds?

Thanks!

Edit: So many things wrong with the math in this utterly confused post. Do not attempt to make any sense of it.
Last edited by pluto on Wed Nov 22, 2017 10:57 pm, edited 1 time in total.

Din
Posts: 224
Joined: Thu Mar 12, 2015 4:47 pm

Re: Red sensitivity of MBDCG

Post by Din » Mon Nov 20, 2017 9:37 am

You have to look at the power at the plate with some appropriate meter. Each square cm on the plate receives a certain amount of power, ie a certain amount of energy/sec. Each square cm on the plate needs 30 mJ of energy. So, you keep exposing until the sq cm on the plate gets it's 30 mJ.

For example, if the power at some point on the plate (from both beams, remember) is 0.3 mW/sq cm, then that point on the plate's getting 0.3 mJ every second. To get 30 mJ, it needs to be exposed for 100 seconds. Basically,

exposure time = sensitivity/power at plate.

Most meters, or rather the heads of most meters, have an aperture of 1 sq cm, so whatever reading your meter is giving you, that's probably the power/sq cm at the plate.

pluto
Posts: 94
Joined: Fri Feb 05, 2016 2:06 am

Re: Red sensitivity of MBDCG

Post by pluto » Wed Nov 22, 2017 10:54 pm

Ah yes, I was doing that all wrong.

This is what I found on the MBDCG wiki:
A 10x15cm plate needs about 3 minutes of exposure time at 50mW. And about 25 seconds of diffuse post exposure to harden the plate...
They need about 30mJ per square centimeter.
But it's not adding up for me...

Plate area = 150 square centimeters.
Beam power = 50 milliwatts
Emulsion sensitivity = 30mJ/cm^2

So we need 30mJ (milliwatt seconds) per square centimeter. Assume we perfectly and losslessly expand the laser beam to an area of 150cm.

Each square centimeter gets 50 milliwatts / 150cm^2 = 0.33 milliwatts per second

30mJ / 0.33mW = 90 seconds

This is about half of the 3 minute figure quoted. I have seen the sensitivity of MBDCG quoted as 30mJ/cm2 to 50mJ/cm2, so perhaps the exposure time in the wiki was using the latter figure.

Assuming my math is correct this time :)

Din
Posts: 224
Joined: Thu Mar 12, 2015 4:47 pm

Re: Red sensitivity of MBDCG

Post by Din » Thu Nov 23, 2017 9:23 am

Whoever wrote that wiki is making a lot of assumptions.

Firstly, there is confusion regarding sensitivity, power at plate and exposure. Secondly, when a figure such as 50 mW is quoted, this means 50 mW per square centimeter. The entire beam cannot be "50 mW" (unless someone has a bloody large aperture on their head!). Also, sensitivity is the energy per square centimeter, so if the sensitivity is given as "30 mJ', this means 30 mJ/sq cm, again, the entire plate does not need 30 mJ. The "square centimeter" is used as a sort of standard because the heads of most meters are about 1 sq cm. So, whatever the reading on the meter, that is the power per sq cm. The fact that the power and sensitivity is given per square centimeter is common enough that it's often left out because it's universally accepted. Thus, when you hear that a plate has a sensitivity of 30mJ, it's understood (and therefore left out) that this is 30 mJ/sq cm. However, people who do not understand this, often don't realise that the 'per sq cm' is left out because it's understood.

Consider the formula I quoted above:

exposure time = sensitivity/power at plate per square centimeter.

Now, assuming a sensitivity of 30mJ/sq cm, and an exposure of 180 seconds, this means that it is assumed that there is 30/180 = 0.16 mW/sq cm = 160 uW/sq cm. on your plate. That may, or may not be your case. It depends on the power of your laser and the optics (lenses, mirrors etc) between your plate and the laser.

As an example, assume you have a 30 mW HeNe laser. Assume that you're making a single beam Denisyuk. If the laser initially goes through an objective and pinhole, assume that there is a 20% loss, resulting in 24 mW of laser power, and the beam expands to your 150 sq cm. Thus, no collimation, the beam simply expands and hits your plate. The 24 mW is now spread over 150 sq cm, giving you 24/150 = 0.16 mW/sq cm, actually, I just made up the numbers so it's pretty remarkable I should get the assumed power at the plate in the wiki!. However, considering I do get a power of 0.16 mJ/sq cm at the plate with these figures, then, a plate needing 30 mJ/sq cm, needs an exposure of 30/0.16 = 187.5 seconds. If you have further optics, there will be further reductions. So, if the laser beam hits a mirror, then is diverted by a mirror, assume a further 20% loss. Thus the beam exiting the spatial filter assembly is now:

Laser power = 30mJ (note, not per sq cm. This is the entire power of the beam)
20% loss after 1st mirror; laser power = 30 - 30/5 = 24mJ
20% loss due to spatial filter; laser power = 24 - 24/6 = 20 mJ
expanded to 150 sq cm; this gives 20/150 ~ 0.13 mJ per sq cm
For a sensitivity of 30 mJ/sq cm, exposure time, T = 30/0.13 ~ 230 seconds.

jeff-blyth
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Joined: Sat Feb 18, 2017 9:41 am

Re: Red sensitivity of MBDCG

Post by jeff-blyth » Sat Dec 09, 2017 7:52 am

Dinesh, I know of all people your calcs for DCG comes from years of experience and it is good that you are helping Pluto get it sorted. But I just want to play a bit of Devil's advocating here. In theory collimated returning light from a mirror as object can result in the bright fringes receiving almost 4 times the intensity of light per sq cm that would have occurred with just the single beam without any returning light and conversely in the dark fringes ---virtually no light per sq. cm. at all. Now I suppose that Pluto’s object is neither a mirror nor going to be lit by a collimated beam. Nevertheless I would like to know if you have had to take such aspects into consideration when dealing with bright refection objects in your long experience.

Din
Posts: 224
Joined: Thu Mar 12, 2015 4:47 pm

Re: Red sensitivity of MBDCG

Post by Din » Sat Dec 09, 2017 10:38 am

Jeff,not sure what you mean by "returning light from a mirror as object", but, yes, under certain circumstances, the bright part of the fringes can receive 4 times the intensity of the beam, and the dark parts receive zero. This effect is independent of collimation.

Consider two beams impinging on a plate from two different directions. There will now be a phase difference between the two beams, because they will have traveled different distances. The phase of a beam of light having traveled at any distance x, is (2π/λ)*x ). So, if beams 1 and 2 have traveled distances of x(1) and x(2) respectively, then the phase difference between beams is φ = (2π/λ)*{x(2) - x(1)}. It can be shown that the variation of intensity on the plate (the interference pattern on the plate, if you will) is

I(t) = I(1) + I(2) + 2√{I(1)*I(2)}*cos(φ)

where I(t) is the total intensity, I(1) is the intensity of beam 1, and I(2) is the intensity of beam 2. You can see the sinusoidal nature of the interference fringes.

Now, if the beam intensities were the same, that is, I(1) = I(2)= I, then the total intensity is:

I(t) = 2I + 2I*cos(φ) = 2I*{1 + cos(φ)}

The maximum and minimum values of cos(φ) are +1, and -1. In the case of cos(φ) = +1, the total intensity is

I(t) = 2I{1 + 1} = 4I

In the case of cos(φ) = -1, the total intensity is

I(t) = 2I{1 - 1} = 0.

So, for that part of the plate where the phase difference is (2π/λ)*{x(2) - x(1)} = 2π (ie the path difference at that point of the plate is exactly one wavelength), the intensity is 4 times the intensity of either beam. But, for those parts of the plate where the phase difference is (2π/λ)*{x(2) - x(1)} = π (ie the path difference at that point of the plate is λ/2), the intensity is zero. Thus, the interference pattern varies from four times the intensity of either beam, to zero.

This is a function of only path differences between the two beams, not collimation. If the wavefront is flat, ie collimated, then the path difference (of either λ or λ/2 ) is the same over the entire area of collimation, because all parts of the same beam have traveled the same distance. If it is not collimated, then those parts of the plate where the path differences are either λ or λ/2 will see this effect. But, as the beam diverges or converges, then different parts of the same beam will have traveled different distances.

Note also, this is not the case for general interference, where the intensities are not the same, and the path differences may not be either π or 2π.

Din
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Re: Red sensitivity of MBDCG

Post by Din » Sat Dec 09, 2017 10:57 am

In display holography, the beam intensities are not the same. The beam ratio is determined by factors such as the H & D curves, and the plate sensitivity. Also, different parts of the same beam, the object beam, travel different distances to the plate. Thus, any point of the plate "seeing" a specific point on the object beam, will create a fringe dependent on the path difference between the (fixed) reference path distance and the distance of the specific object point from the plate, which is, after all, the point of holography - to create a "phase map", of the varying distances of the object to the plate, with respect to the (fixed) reference beam.

Thus, for example, if you're shooting a porcelain cat, the difference between the nose of the cat, and the ear of the cat to a specific point on the plate may be N wavelengths, but may be M wavelengths to a different part of the plate. This difference is recorded by comparing it to a fixed reference, and creating a fixed set of fringes based on either M wavelengths or N wavelengths. Different parts of the plate, therefore, will have different fringe structures, based on M (at one point of the plate) or N (on another part of the plate).

While it may be possible to "section off" different parts of the cat, determine the different distances of each part to the plate, and determine the fringe structure by the phase difference, this would be a task for the very, very obsessed type of person! Better by far to simply assume that the entire cat gives off a varying waveform, whose structure depends on the 3D profile of the cat at a particular perspective.

Having said this, in regards to Pluto's original question, the varying distances of the cat from the plate cause varying intensities, because the "cat waveform" is not collimated. So, a determination needs to be made as to what you're going to call "ratio", since it'll vary across the entire plate. The detector will (usually) only detect light at "chunks" of 1 sq cm, and the exposure will depend on the per sq cm of light at the plate on average.

jeff-blyth
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Joined: Sat Feb 18, 2017 9:41 am

Re: Red sensitivity of MBDCG

Post by jeff-blyth » Sat Dec 09, 2017 2:42 pm

Well thank you for that Dinesh, you have given me the bit of insight that when it comes to forming holographic fringes in a Denisyuk reflection holo.--- I have now grasped that the process is independent of collimation!
However I do have a big problem understanding your statements about intensity effects and collimation when it comes to the sensitivity of the recording material as in the OP. If you have a plain flat mirror as object and your beam is very divergent then after returning from the mirror the beam is much weaker than the input beam was and therefore the fringe structure and hologram will be comparatively weak because the input is only weakly modulated by the widely spread returning beam. You will have used much of the capacity of the recording material to record mostly unmodulated light. However in the case of a collimated beam returning from a flat mirror and ignoring any absorption losses, then your returning light will give almost a 1:1 beam ratio and the hologram will apparently be the brightest you can get in that recording material albeit rather broadband and not good for making a HOE. (Yes I appreciate that in a transmission hologram a 1:1 beam ratio is certainly too high and will cause distortion effects and glare due to overmodulation of the reference beam, but let’s just stick to Denisyuks in this discussion).

Din
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Joined: Thu Mar 12, 2015 4:47 pm

Re: Red sensitivity of MBDCG

Post by Din » Sun Dec 10, 2017 8:05 pm

I think the important thing to understand is that the modulation is dependent on the beam ratio and the exosure, not on the absolute intensity of either beam.

Looking at the modulation equation again, but this time, inserting the beam ratio, and changing I(1) and I(2) to I(o) and I(r), we have:

I(t) = I(o) + I(r) + 2√{I(o)*I(r)}*cos(φ)

If we now insert the beam ratio, k = I(o)/I(r), we get

I(t) = k*I(r) + I(r) + 2√{k*I²(r)}cos(φ)

The modulation amplitude is now 2√{k*I²(r), which is now dependent on √k, the square root of the ratio. This will reduce the modulation, unless k=1. Of course, other factors also affect this, such as the linear portion of the H & D curve, where you have to work in the linear part of the curve. But, a combination of ratio and exposure determines the modulation.

Now, remember that the beam is Gaussian. So, if the beam is collimated, ie a flat wavefront, then any point on the beam profile at some height r from the beam axis has an intensity given by

I = I(0)*exp(-ar²)

That is, the beam intensity drops off, as the square exponential of the height above the plate centre, or beam centre (assuming the beam is centred on the plate) . However, if you have a divergent beam, then the Gaussian spread is along the arc of the circle of the divergent wavefront. So, now, if the point of divergence of the beam from the plate centre is d,the intensity along the arc of the wavefront, at some angle θ, relative to a line drawn from the centre of divergence to the plate centre, is

I = I(0)*exp(-ad²θ²)

That is, the Gaussian is spread along an arc of the divergent waveform. Notice, the intensity is dependent on θ², so it falls off much more rapidly, as you mentioned.

Thus, what we have to consider, in terms of the ratio - which is what determines the modulation - the spread of the Gaussian linearly (collimated beam), as a ratio of the spread of the Gaussian along the arc of a circle, for a collimated/divergent recording. Alternatively, the ratio depends on the divergence angles of the two beams for a divergence/divergence recording. As the intensity of a divergent beam at any position of the plate is dependent on the square of the divergence angle, and so falls off rapidly, if you have very different divergences, then the ratio falls off much more quickly, as you approach the edge of the plate. This may lead to vignetting.

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