I think I may be able to go one up!
I have the same trays for the developer and the bleach and separate trays for the different types of developers and bleaches. So, If I'm using CWC2 and PBU, I have two specific trays. If I'm using CWC2 and Van Ranesse, I use the "Van Ranesse" tray for the bleach.
One of the problems with holography is that you have to be exact and precise, yet be flexible and imaginative at the same time.
Exposure time
Re: Exposure time
Sorry, guys, you are really trying hard to explain it, but I still do not get it.
The collimated beam of the area 30x30 mm2 can be thought as a pack of 6x6 "small beams" of area 5x5 mm2.
As the object beam (cyan) carries power 9 mW, one "small beam" carries 9/36 = 1/4 mW.
The plate of size 10x10 cm2 is covered by exactly four small beams, i.e. it gets power 1 mW.
The reference beam (magenta) carries 36 mW, i.e. one "small beam" carries 36/36 = 1 mW.
As the plate is covered by exactly two small beams, it gets power 2 mW.
Thus, 3 mW in sum.
Please note that no vector quantities are involved here. Every "small beam" carries a scalar value of power, and I just sum them.
Why should I assume that the magenta "small beam" carries more power? Both Bob and Din write that the energy from the reference should be 4 mW, i.e. each of my "small beams" should carry 2 mW.
(I understand that the actual recording process is very complex and the correct exposure time depends on many factors. I consider this as a theoretical exercise.)
The collimated beam of the area 30x30 mm2 can be thought as a pack of 6x6 "small beams" of area 5x5 mm2.
- beams2.png (53.45 KiB) Viewed 4048 times
The plate of size 10x10 cm2 is covered by exactly four small beams, i.e. it gets power 1 mW.
The reference beam (magenta) carries 36 mW, i.e. one "small beam" carries 36/36 = 1 mW.
As the plate is covered by exactly two small beams, it gets power 2 mW.
Thus, 3 mW in sum.
Please note that no vector quantities are involved here. Every "small beam" carries a scalar value of power, and I just sum them.
Why should I assume that the magenta "small beam" carries more power? Both Bob and Din write that the energy from the reference should be 4 mW, i.e. each of my "small beams" should carry 2 mW.
(I understand that the actual recording process is very complex and the correct exposure time depends on many factors. I consider this as a theoretical exercise.)
Re: Exposure time
Let me think on this a little.
But, I already see one problem. You state that the perpendicular beam carries 9mW of power and you state that the sensitivity is 9 mJ/cm^2. However, if the beam carries 9mw/cm^2, and the area of the beam is 3cm x 3cm, then the beam power is (9mw/cm^2)*(9 cm^2) = 81 mW. For the ref. you have 36mW, but again, if that's supposed to be 36 mW/cm^2, then the beam carries (36mW/cm^2)*(9cm^2) = 324 mW. However, if the power on the beam is 9mW for object and 36 mW for ref, each over an area of 9 cm^2, then the beam power is 9 mW/9 cm^2 = 1mW/cm^2 (object) and 36mW/9cm^2 = 4mW/cm^2 (reference). Since the sensitivity of the plate is given per sq cm, then I think it needs to be clear whether the incoming beam powers are absolute or relative to beam area.
In the former case (beam powers of 9 mW/cm^2 and 36 mW/cm^2), your "small squares" would have an area of 0.5cm x 0.5 cm = 0.25 cm^2, and so would have powers of 9 x 0.25 = 2.25 mW/small square and 36 x 0.25 = 9mW/small square.
But, I already see one problem. You state that the perpendicular beam carries 9mW of power and you state that the sensitivity is 9 mJ/cm^2. However, if the beam carries 9mw/cm^2, and the area of the beam is 3cm x 3cm, then the beam power is (9mw/cm^2)*(9 cm^2) = 81 mW. For the ref. you have 36mW, but again, if that's supposed to be 36 mW/cm^2, then the beam carries (36mW/cm^2)*(9cm^2) = 324 mW. However, if the power on the beam is 9mW for object and 36 mW for ref, each over an area of 9 cm^2, then the beam power is 9 mW/9 cm^2 = 1mW/cm^2 (object) and 36mW/9cm^2 = 4mW/cm^2 (reference). Since the sensitivity of the plate is given per sq cm, then I think it needs to be clear whether the incoming beam powers are absolute or relative to beam area.
In the former case (beam powers of 9 mW/cm^2 and 36 mW/cm^2), your "small squares" would have an area of 0.5cm x 0.5 cm = 0.25 cm^2, and so would have powers of 9 x 0.25 = 2.25 mW/small square and 36 x 0.25 = 9mW/small square.
Re: Exposure time
A quick question that occurs to me. What does "sensitivity of the plate" mean? Is it the energy required for the minimum possible change to record a latent image, or the maximum energy required. So, for example, in silver halide, the mechanism of recording is the ionisation of silver into a silver ion - the ionisation energy. Is the "sensitivity of the plate" the ionisation energy to ionise one mole of silver, or the ionisation energy to ionise just one silver atom. Generally, of course, every grain already has a speck on it. So, to distinguish exposed grains from unexposed ones, the exposure must create new specks. Every speck on the grain is the result of one silver atom being ionised. So, is the sensitivity of the plate, the energy required to create some number n of new grains on every speck, below which it becomes impossible to distinguish between exposed and unexposed grains? Or, just one speck/grain? I'm, of course, not including reciprocity effects here.
Re: Exposure time
I mean the second option:
As for sensitivity of the plate: I assume the same meaning as in the plate specifications, see e.g. Slavich plates specs at http://materials.geola.lt/index.php/hol ... -sensitive. For example, the specs for PFG-01 material state that exposure 100 uJ/cm2 leads to density approx. 1.5 and diffraction efficiency approx. 50%. I am not exactly sure what does it mean in terms of latent image formation, etc. Let us just assume that for PFG-01, a good rule of thumb says "expose to 100 uJ/cm2".
My question is thus: if the "sensitivity for the best result" of my hypothetical material is 9 mJ/cm^2, what exposure time shall I use? My guess would be 3 s, as the plate has area 1 cm^2 and it is illuminated by four small object beams (= 4x0.25 = 1 mW) and two small reference beams (=2x1=2 mW), in total 3 mW.
However, both Bob and Din guess just 1.8 s. I am confused.
Then, the "small object beam" of area 5x5 mm^2 carries power 0.25 mW, the "small reference beam" carries 1 mW.Din wrote:However, if the power on the beam is 9mW for object and 36 mW for ref, each over an area of 9 cm^2, then the beam power is 9 mW/9 cm^2 = 1mW/cm^2 (object) and 36mW/9cm^2 = 4mW/cm^2 (reference). Since the sensitivity of the plate is given per sq cm, then I think it needs to be clear whether the incoming beam powers are absolute or relative to beam area.
As for sensitivity of the plate: I assume the same meaning as in the plate specifications, see e.g. Slavich plates specs at http://materials.geola.lt/index.php/hol ... -sensitive. For example, the specs for PFG-01 material state that exposure 100 uJ/cm2 leads to density approx. 1.5 and diffraction efficiency approx. 50%. I am not exactly sure what does it mean in terms of latent image formation, etc. Let us just assume that for PFG-01, a good rule of thumb says "expose to 100 uJ/cm2".
My question is thus: if the "sensitivity for the best result" of my hypothetical material is 9 mJ/cm^2, what exposure time shall I use? My guess would be 3 s, as the plate has area 1 cm^2 and it is illuminated by four small object beams (= 4x0.25 = 1 mW) and two small reference beams (=2x1=2 mW), in total 3 mW.
However, both Bob and Din guess just 1.8 s. I am confused.
Re: Exposure time
Ok, let's look at it like this. Assume both beams are contra propagating, that is, the beams are traveling in opposite directions. Let the plate be between the beams. This is the classic Denisyuk geometry with zero angles between the beams. Let one beam, beam(A), have power P(A) in units of W/cm^2, and the other beam (B) have power P(B), in the same units. Then, the number of photons/second for a beam with power P, is given by
N = (P*lambda)/(h*c).
So, for beam A, the number of photons/sec is
N(A) = {P(A)*lambda}/(h*c)
and, similarly, the number of photons/sec in B is
N(B) = {P(B)*lambda}/(h*c).
Now, These photons exist within a cylinder of height c. Now, after one second, the number of photons hitting the emulsion is N(A) + N(B). Let the minimum number of photons necessary to cause the appropriate actinic reaction be a. Then, we have
a = {N(A)+N(B)}*T(exp)
where T(exp) is the exposure time, that is, the time necessary for the minimum number of photons to interact with the emulsion to cause the actinic reaction. As an aside, what happens if we continue to pour more and more photons into the emulsion? This would depend on the density of actinic sites within the emulsion, which we don't know. However, in the case of contra-propagating beams, we now have an expression for the exposure time in terms of the energy needed for a given actinic reaction.
Let the beams still be contra-propagating, but twist the plate. Now the beams impinge on the plate at the same angle, relative to plate normal, but of opposite sign. So, for example, if we twisted the plate by 30 deg, then the beams would impinge at +/- 30 deg. Would the exposure time change? Only by the time needed for the lower photons to reach the plate, that is, the delay = (d/2)*sin(theta), where d is the plate dimension and theta is the twist angle. This is insignificant, so effectively there is no change in the exposure time. What change then does such a twist cause? Well, the number of actinic reaction, or rather some actinic function of the spatial coordinates would change. Generally, the rate of ractions within the emulsion would be given by
a(x,y) = A*cos{K(x)*x + K(y)*y},
where K(x) and K(y) are the spatial frequencies of the Bragg planes.
By twisting the plate relative to the two beams, you've changed K(x) and K(y) only. However, the number of photons is still the same, and it's the number of photons that hit the plate/sec that causes actinic reactions and determines exposure.
I'd put some numbers in, but, you know, theorists hate numbers
N = (P*lambda)/(h*c).
So, for beam A, the number of photons/sec is
N(A) = {P(A)*lambda}/(h*c)
and, similarly, the number of photons/sec in B is
N(B) = {P(B)*lambda}/(h*c).
Now, These photons exist within a cylinder of height c. Now, after one second, the number of photons hitting the emulsion is N(A) + N(B). Let the minimum number of photons necessary to cause the appropriate actinic reaction be a. Then, we have
a = {N(A)+N(B)}*T(exp)
where T(exp) is the exposure time, that is, the time necessary for the minimum number of photons to interact with the emulsion to cause the actinic reaction. As an aside, what happens if we continue to pour more and more photons into the emulsion? This would depend on the density of actinic sites within the emulsion, which we don't know. However, in the case of contra-propagating beams, we now have an expression for the exposure time in terms of the energy needed for a given actinic reaction.
Let the beams still be contra-propagating, but twist the plate. Now the beams impinge on the plate at the same angle, relative to plate normal, but of opposite sign. So, for example, if we twisted the plate by 30 deg, then the beams would impinge at +/- 30 deg. Would the exposure time change? Only by the time needed for the lower photons to reach the plate, that is, the delay = (d/2)*sin(theta), where d is the plate dimension and theta is the twist angle. This is insignificant, so effectively there is no change in the exposure time. What change then does such a twist cause? Well, the number of actinic reaction, or rather some actinic function of the spatial coordinates would change. Generally, the rate of ractions within the emulsion would be given by
a(x,y) = A*cos{K(x)*x + K(y)*y},
where K(x) and K(y) are the spatial frequencies of the Bragg planes.
By twisting the plate relative to the two beams, you've changed K(x) and K(y) only. However, the number of photons is still the same, and it's the number of photons that hit the plate/sec that causes actinic reactions and determines exposure.
I'd put some numbers in, but, you know, theorists hate numbers
Re: Exposure time
It took me a couple of weeks to think about the subject. As Dinesh noted, it is really a tricky point!
As everyone here agrees that light power should be measured with the detector perpendicular to the beam, I tried to figure out how to explain it. After all, when the detector is not perpendicular to the beam, the light meter shows smaller numbers. Why is the photosensitive plate behavior different?
Dinesh tried to explain it using photons. I appreciate that, but as I do not understand photons, I try to avoid them whenever possible. So, here is my model what happens.
Let's have a plate covered with photosensitive emulsion. The emulsion is composed of e.g. gelatin and a very small amount of photosensitive grains. Let's assume that number of grains is very small, so that they do not occlude each other. In the picture, the cyan beam illuminates the plate. There are just four grains in the emulsion. The white lines behind the grains denote geometrical shadow, i.e. that some light is absorbed by the grains.
Suppose that the plate of area A = 1 cm2 is illuminated at normal incidence, light power is 1 mW/cm2, which means that the plate gets energy 1 mJ/s. As the number of grains is small, most of light just comes through the plate. Just a small fraction of light is caught by the grains. If the exposure time T is long enough, all of them get exposed.
Now, rotate the plate by 60 degrees, which means that the plate receives just half of the energy:
However, the grains still occupy the same area as seen by the light source. Although the energy that passes through the plate is just 0.5 mJ/s now, is is used more efficiently by the grains as they seem to be more dense as seen by the light source. Hence, exactly the same exposure time T is needed to expose them.
What do you think? Is that model reasonable?
As everyone here agrees that light power should be measured with the detector perpendicular to the beam, I tried to figure out how to explain it. After all, when the detector is not perpendicular to the beam, the light meter shows smaller numbers. Why is the photosensitive plate behavior different?
Dinesh tried to explain it using photons. I appreciate that, but as I do not understand photons, I try to avoid them whenever possible. So, here is my model what happens.
Let's have a plate covered with photosensitive emulsion. The emulsion is composed of e.g. gelatin and a very small amount of photosensitive grains. Let's assume that number of grains is very small, so that they do not occlude each other. In the picture, the cyan beam illuminates the plate. There are just four grains in the emulsion. The white lines behind the grains denote geometrical shadow, i.e. that some light is absorbed by the grains.
- grains1.png (4.99 KiB) Viewed 3963 times
Now, rotate the plate by 60 degrees, which means that the plate receives just half of the energy:
- grains2.png (6.01 KiB) Viewed 3963 times
What do you think? Is that model reasonable?
Re: Exposure time
Yes, I think so.lobaz wrote:What do you think? Is that model reasonable?
Well, one reason may be that the ccd is flat. So, any light hitting a specific ccd site, will record normal to the surface of the ccd. In the case of grains, these are round, so any light from any direction is perpendicular to the grain.lobaz wrote:After all, when the detector is not perpendicular to the beam, the light meter shows smaller numbers. Why is the photosensitive plate behavior different?
Re: Exposure time
And probably the electron-hole production in a ccd has a preference for the incident light's electric field oriented parallel to the surface. Whereas even if the state change of the grains has a preferred electric field orientation, the grains themselves would be randomly oriented in the medium.Din wrote:Well, one reason may be that the ccd is flat. So, any light hitting a specific ccd site, will record normal to the surface of the ccd. In the case of grains, these are round, so any light from any direction is perpendicular to the grain.lobaz wrote:After all, when the detector is not perpendicular to the beam, the light meter shows smaller numbers. Why is the photosensitive plate behavior different?