Exposure time

[lobaz]

What laser and the recording setup do you use, HoloM? 0.74 uW/cm2 sounds quite low.
Anyway, I do not work with VRP-M regularly, it is just a coincidence I worked with them and CWC2+dichromate bleach a few weeks ago. I am not sure if I remember results of the exposure test (bracketing) correctly, but I think your 100 s is not completely off. If I recall well, I would use 200 s for a transmission hologram, 4:1 ref/obj ratio.

[Din]

Why perpendicular to the object or the reference beam, Din?
Imagine e.g. a geometry of the Mach-Zehnder interferometer: the beam is split 50/50 to object/reference (at an angle 90 degrees), reflected by two mirrors, spread by the same optics, and instead of the final beam splitter, there is a plate. If the plate is facing one beam, the other illuminates it at an angle 90 degrees, i.e. in fact does not illuminate it at all. Rotate the plate, and the power from the first arm gets smaller, while power from the second arm gets higher.
If I measure the power with the probe perpendicular to the beams, I would get 1:1 ratio regardless of the plate orientation.

Petr, maybe I was being misunderstood. I meant that the plane of the detector should be perpendicular to the k vector and parallel to the wavefront.

Suppose in your Mach Zender, you were to embed the detector onto the plate, so that the plane of the detector is in the plane of the plate. Perhaps glue the detector to the back of the plate. Now, the k vector of either beam is perpendicular to the plane of the detector (and, of course, the plate) when the plate is facing either arm. When you initially face one arm, call it arm(1) of the Mach Zender, you're detector will read the maximum amount of light (not necessarily half the incoming light, that depends on polarisation, dirt on the mirror etc) from arm(1) and zero light from arm(2). As you rotate the plate towards the other arm, the detector reads R = I(1)*cos(theta) + I(2)*sin(theta), where I(1) and I(2) are the intensities of the two arms and theta is the angle between the plate/detector normal and arm(1). Of course, if the split is really 50:50, then the reading at any arbitrary angle is R = 0.5*I*[cos(theta) + sin(theta)], where I is the total light into the Mach Zender.. So, as you rotate the plate/detector, the light from arm(1), decreases, and the light from arm(2) increases. If, for simplicity, we assume equal intensities, then at theta = pi/4, the reading will be R = I/{sq rt(2)} = 0.707*I

[BobH]

I always measured my beams with the detector perpendicular to the beam being measured, because it's the amplitude of the beam I'm interested in (I think that's what Dinesh is describing). Beam ratio determines fringe visibility, and that exists in space before a recording plate is put in the same space as the fringes. Fresnel reflections modify that a little, but not significantly and I disregarded them.

Using the numbers given by measurement as described to calculate exposure time is a different issue. The intensity of the reference beam (for example) will be less at each spot on the recording plane because of the angle of its incidence, but the total power hitting the plate will be "in the ballpark" and close enough that exposure time will be proper on the second attempt (if all else is kept constant).

[holomaker]

I too use my meter when needed perpendicular as if to find the highest spot or output area, which also means I may spend a lot of time hunting too!

[holomaker]

I never evaluated density after development, thinking it cumbersome, impractical and inconsistent.

I was thinking more along the lines of an unfamiliar film, new material always has its challenges for example, is my Safelite correct? , if so, how long can I let it be exposed to the Safelite before fogging , how long does it take for my film to reach the proper density in development , and also estimated time of exposure ,this could all be done on a 1"square piece of film...

[lobaz]

Thank you for trying to explain me that, but I still do not fully understand it. Imagine following situation: the object beam (cyan) is 3x3 cm and total power is 9 mW (for simplicity, let us assume its shape is square and that the light power is uniform over its area), i.e. it has 1 mW/cm2. The plate of size 1x1 cm (black line) then gets 1 mW. The reference beam is also 3x3 cm, but it is 4x stronger than the object beam, i.e. total 36 mW, or 4 mW/cm2. However, as the angle of incidence is 60 degrees, the apparent area of the plate is just 0.5 cm2. Thus the plate gets just 2 mW. The total incoming power is then 3 mW. If the sensitivity of the material is 9 mJ/cm2, I should set the exposure time to 3 s.

beams.png (19.12 KiB) Viewed 3598 times

Is this reasoning correct? Or is it better to assume that beams are coherent, and make following calculation?

In case of coherent light, the amplitude of the object beam is proportional to 1 unit, and the amplitude of the reference beam to sqrt(4) = 2 units. Thus, the amplitude of the brightest part of the sum should be 1+2=3 units, i.e. power of the brightest part should be 3^2 = 9 mW/cm2. Thus I should set exposure time to 1 s.

[BobH]

My thought would be your first case, but to be honest, I always split the difference. I'd measure the beams as you describe, giving 1mW/cm^2 for the object beam and 4mW/cm^2 for the reference. 1+4=5mW total power and 9mJ/cm^2 gives 1.8s exposure time. Even if making multiple exposure pixels on the hologram (when the exposure is critical), a couple of test plates (probably needed to troubleshoot something else, like uniformity or stray light problems) will allow the exposure to be optimized. The inconsistency between power meters, their use in different labs, with or without fresnel reflection compensation .... all that stuff makes comparisons a bit useless really.

[Din]

It's actually a tricky point.

Consider the fact that the actinic reaction is energy dependent. It takes a certain amount of energy to transform the state of the emulsion to whatever physical change records the hologram (eg Ag -> Ag+ for silver and change of density for dcg). However, energy is a scalar quantity, not a vector quantity. If the beams are collimated, there is no variation of energy across the wavefront as a function of distance, the inverse square law is not valid for a collimated beam. Thus along the wavefront, at all points on the wavefront, the wavefront transports the same energy - 9mW and 36 mW in your case. Now, if the plate were 1cm x 1cm, then the normally incident beam would dump 1 mW on the plate, or 1mJ/sec. But, the inclined beam also dumps 4 mW on the plate, only, the bottom of the plate "sees" the 4 mW at a later time - there is a phase lag. It's this phase lag that causes the holographic data to be recorded, but the total energy on the plate at all points is 5 mW or 5mJ/sec. If the sensitivity were 9 mJ, then the exposure would be 9/5 seconds.

Consider the following situation. I have two rods that are 3cm long. On these rods, at appropriate intervals, I place candles. Let's say I place a candle every 0.2 cm. The candles on one of these rods burns 4 times more strongly than the other, say I've coated the wick with Pot Nitrate or something. At some point I have a tray of alcohol. I now move one rod perpendicular to the tray and one rod at an angle to the tray, say at 60 deg. I assume (or measure) that the "weak" candles transfer 9mW of energy and the "strong" candles transfer 36 mW of energy. Both these rods "meet" at the tray of alcohol. Now I ask, how long will it take for the alcohol to evaporate? If the latent heat of vapourisation of alcohol is L_alc and the specific heat of alcohol is c_alc, and I assume that the boiling point of alcohol is 80 C and that the alcohol is at room temp (~20 C), it takes an amount of energy given by mL_alc + mc_alc*(80-20) J of energy.However, the transfer of energy by the rods to the tray is not directionally dependent since energy is a scalar quantity. The lower part of the tray receives the energy from the angled rod - the heat from the candle at the end of the inclined rod - at a later time, but it still receives the same heat from the candle, ie, there is a "phase delay" for the energy from the candle on the angled rod.

Thus, the k vector determines the direction of the energy transfer, a vector quantity, but the quantity of energy transferred is a scalar.

[Din]

My thought would be your first case, but to be honest, I always split the difference. I'd measure the beams as you describe, giving 1mW/cm^2 for the object beam and 4mW/cm^2 for the reference. 1+4=5mW total power and 9mJ/cm^2 gives 1.8s exposure time. Even if making multiple exposure pixels on the hologram (when the exposure is critical), a couple of test plates (probably needed to troubleshoot something else, like uniformity or stray light problems) will allow the exposure to be optimized. The inconsistency between power meters, their use in different labs, with or without fresnel reflection compensation .... all that stuff makes comparisons a bit useless really.

In a practical situation, you're quite right. The number from the meter - and the sensitivity of the plate - are merely numbers and should be treated as a guide rather than an absolute. In a display hologram, the object light is usually so variable that it's impossible to say "4:1" or some such ratio. I remember it bothered me when I first started as to which part of the object beam were you supposed to take as your 4:1 standard. I remember Dave Traynor from Richmond saying that you take an average. But, unless the object beam were perfectly uniform and fell off uniformly, the average had to be a weighted average. It was Peter Miller who finally told me to forget all that 4:1 stuff and simply wiggle your fingers over the plate. This works as a guide to "tune in" the actual final exposure.

Of course, over and above exposure, there's also the variability of development which has to be taken into consideration. There's more of a black art to development than there is to shooting.

[BobH]

There are soooo many variables that affect the final image in a hologram, and the only way I found to keep them all under control is to keep as many as possible constant. That's why I always mix chemicals the same way, develop in a given bath for a determined time and don't evaluate the density during development, and measure with constant technique. Then, exposure time can be adjusted to "walk it in", and beam ratio too in a controlled way.