Let me preface this by saying that I am VERY new to these forums so please forgive me if this has already been covered (although a search didn’t reveal anything).
I have read some individuals discussing the merits of having a shorter focal length between the light source and the fresnel that collects the light. The reasoning seems quite obvious – light decays as the square of distance. I other words, if a collimator with a focal length of 12” (and thus 12” from the light source) collects 500 lumens then a collimator with a focal length of 6” (and thus 6” from the light source) would collect 2000 lumens. The immediate implications seem obvious as a system can be designed to be brighter/have a less powerful (and less costly) lighting system.
This got me thinking, it is just as important to illuminate the LCD brightly as it is to illuminate the LCD evenly. Since we already know that a reduction in the collimating lens’s focal length will increase the amount of light gathered what are the prospects for even distribution of that light over the LCD? Unfortunately they are bad, very bad and the reasons for this are two-fold:
1 – As we are all aware, light that this shined on to a surface at a 90 degree angle will be concentrated on a smaller area than light shined at any other angle. We are all aware of this fact if for no other reason than the equator is hot and the poles are cold. The degree to which light intensity from a light sources drops based on the declination angle can be calculated with our good-old friend Mr. Sin.
Lets go to the diagram.
Givens:
- green things are collimating fresnel lenses
- fresnel lenses are 12” across (ok, so they are not to scale – sue me…)
- fresnel in F1 is 12” from the light source
- fresnel in F2 is 6” from the light source
Based upon these numbers, a little trig reveals that:
- Angle of ray F1b relative to fresnel F1 is 64 degrees
- Angle of ray F2b relative to fresnel F2 is 45 degrees
- And, obviously enough, rays F1a and F2a are at right angles to their respective lenses
So now lets determine the ratio of light that drops of from then center to the edge of the lens in both F1 and F2:
For F1 - (sin 90) / (sin 64) = 1.11
For F2 - (sin 90) / (sin 45) = 1.41
As expected lens F2 is 1.4 times brighter in the center than at the edge whereas F1 is only 1.1 times brighter in the center than at the edge. Remember, all this is due to the fact that when the lens is closer to the light source the rays strike the outer edges at greater angles of declination.
2 – The second factor that works against us is the same thing that allowed us to increase the brightness in the first place - Light intensity = k/r2 – light intensity decays as the square of distance. The problem stems from the fact that the lens-light distance difference ratio between outer-edges of the collimating fresnel and it’s center increases.
Lets go to the diagram.
Givens:
- green things are collimating fresnel lenses
- fresnel lenses are 12” across
- fresnel in F1 is 12” from the light source
- fresnel in F2 is 6” from the light source
From these givens we can determine F1B and F2B distance from the light source. Since I don’t feel like typing out the trig equations here are some numbers you are just going to have to trust:
- distance from light source to F1B is 8.5”
- distance from light source to F2B is 13”
So what does this mean? Well nothing until we plug them into our aforementioned little light intensity question. Since we are looking for a ratio here I am just simply going to assign k the arbitrary value of 1 (the number makes no difference here, believe me….):
For lens F1 - (1/122) / (1/132) = 1.17
For lens F2 - (1/62) / (1/8.52) = 2.01
Well…. what we see from these numbers is just what was expected. When the lens is 12” from the light source the center (F1A) receives 1.17 times more light than the edge (F1B). However, here is the real kicker - when the lens is 6” from the light source the center (F2A) receives more than twice as much light than the edge (F2B). Talk about un-even lighting! So what is the equation for determining this “difference ratio”? It’s late and I need to finish up so maybe if I have time later I’ll calculate it.
All right, I need to hit the sack so I’ll conclude by saying that when we combine these two factors we find that (in our above given examples) the final difference ratio between the two is quite large. Lets calculate the combined effect:
F1 – 1.17 * 1.11 = 1.30
F2 – 2.01* 1.41 = 2.83
When combined we see that F1 is 1.3 times brighter in the center than on its edge where as F2 is a WHOPPING 2.8 times brighter in its center than on its edge….
In conclusion (finally), moving the collimating lens further from the light source causes the amount of light entering it to fall off as the square of distance BUT we can also expect the “evenness” of light distribution into collimating fresnel (and ultimately the LCD) to INCREASE in a manner approximating a squared relationship relative to distance (I could calculate the exact equation but I am just too tired right now).
So what’s the lesson? There must be a happy medium somewhere…
Pete
I have read some individuals discussing the merits of having a shorter focal length between the light source and the fresnel that collects the light. The reasoning seems quite obvious – light decays as the square of distance. I other words, if a collimator with a focal length of 12” (and thus 12” from the light source) collects 500 lumens then a collimator with a focal length of 6” (and thus 6” from the light source) would collect 2000 lumens. The immediate implications seem obvious as a system can be designed to be brighter/have a less powerful (and less costly) lighting system.
This got me thinking, it is just as important to illuminate the LCD brightly as it is to illuminate the LCD evenly. Since we already know that a reduction in the collimating lens’s focal length will increase the amount of light gathered what are the prospects for even distribution of that light over the LCD? Unfortunately they are bad, very bad and the reasons for this are two-fold:
1 – As we are all aware, light that this shined on to a surface at a 90 degree angle will be concentrated on a smaller area than light shined at any other angle. We are all aware of this fact if for no other reason than the equator is hot and the poles are cold. The degree to which light intensity from a light sources drops based on the declination angle can be calculated with our good-old friend Mr. Sin.
Lets go to the diagram.
Givens:
- green things are collimating fresnel lenses
- fresnel lenses are 12” across (ok, so they are not to scale – sue me…)
- fresnel in F1 is 12” from the light source
- fresnel in F2 is 6” from the light source
Based upon these numbers, a little trig reveals that:
- Angle of ray F1b relative to fresnel F1 is 64 degrees
- Angle of ray F2b relative to fresnel F2 is 45 degrees
- And, obviously enough, rays F1a and F2a are at right angles to their respective lenses
So now lets determine the ratio of light that drops of from then center to the edge of the lens in both F1 and F2:
For F1 - (sin 90) / (sin 64) = 1.11
For F2 - (sin 90) / (sin 45) = 1.41
As expected lens F2 is 1.4 times brighter in the center than at the edge whereas F1 is only 1.1 times brighter in the center than at the edge. Remember, all this is due to the fact that when the lens is closer to the light source the rays strike the outer edges at greater angles of declination.
2 – The second factor that works against us is the same thing that allowed us to increase the brightness in the first place - Light intensity = k/r2 – light intensity decays as the square of distance. The problem stems from the fact that the lens-light distance difference ratio between outer-edges of the collimating fresnel and it’s center increases.
Lets go to the diagram.
Givens:
- green things are collimating fresnel lenses
- fresnel lenses are 12” across
- fresnel in F1 is 12” from the light source
- fresnel in F2 is 6” from the light source
From these givens we can determine F1B and F2B distance from the light source. Since I don’t feel like typing out the trig equations here are some numbers you are just going to have to trust:
- distance from light source to F1B is 8.5”
- distance from light source to F2B is 13”
So what does this mean? Well nothing until we plug them into our aforementioned little light intensity question. Since we are looking for a ratio here I am just simply going to assign k the arbitrary value of 1 (the number makes no difference here, believe me….):
For lens F1 - (1/122) / (1/132) = 1.17
For lens F2 - (1/62) / (1/8.52) = 2.01
Well…. what we see from these numbers is just what was expected. When the lens is 12” from the light source the center (F1A) receives 1.17 times more light than the edge (F1B). However, here is the real kicker - when the lens is 6” from the light source the center (F2A) receives more than twice as much light than the edge (F2B). Talk about un-even lighting! So what is the equation for determining this “difference ratio”? It’s late and I need to finish up so maybe if I have time later I’ll calculate it.
All right, I need to hit the sack so I’ll conclude by saying that when we combine these two factors we find that (in our above given examples) the final difference ratio between the two is quite large. Lets calculate the combined effect:
F1 – 1.17 * 1.11 = 1.30
F2 – 2.01* 1.41 = 2.83
When combined we see that F1 is 1.3 times brighter in the center than on its edge where as F2 is a WHOPPING 2.8 times brighter in its center than on its edge….
In conclusion (finally), moving the collimating lens further from the light source causes the amount of light entering it to fall off as the square of distance BUT we can also expect the “evenness” of light distribution into collimating fresnel (and ultimately the LCD) to INCREASE in a manner approximating a squared relationship relative to distance (I could calculate the exact equation but I am just too tired right now).
So what’s the lesson? There must be a happy medium somewhere…
Pete
Attachments
What happens when you put a condencer lens between the lamp and the fresnel? It's thicker in it's center, so it should dim the straight light a little bit. It also makes the distance between the fresnel and lamp shorter, and obviously depending on how it's been made condences the light more at the edges of the frensell. I'd say that a properly designed condencer lens solves are these problems.
I will assume you are talking about an aspheric condenser assembly. I did a bit of research on condenser lenses and I can't find anything that would lead me to believe they more evenly spread light over the surface of the collimating fresnel. My understanding is that they serve as a light gathering lens.
In order to evenly distribute light over the collimating fresenl the condenser would have to somehow divert a portion of the light in the center incrementally toward the outer portions. Moreover, if a condensing lens can somehow do this the condenser lens would then have to be matched to the size of the LCD so as to properly gradate the illumination.
I certainly won’t claim to be an optics expert so maybe someone can throw in their 2 cents worth.
Pete
In order to evenly distribute light over the collimating fresenl the condenser would have to somehow divert a portion of the light in the center incrementally toward the outer portions. Moreover, if a condensing lens can somehow do this the condenser lens would then have to be matched to the size of the LCD so as to properly gradate the illumination.
I certainly won’t claim to be an optics expert so maybe someone can throw in their 2 cents worth.
Pete
Ok, here is my idea as to how to use collimating fresnels with shorter focal lengths. I believe if we simple print a pattern such as shown in the attached image (although not nearly as pixilated!) onto an overhead sheet then, when placed between the collimator and light source, we can reduce the amount of light entering through the center and progressively allow more light through as we approach the edges.
While such an idea might seem wasteful in that we are blocking some of the light I believe that this will be more than offset with our ability to move the collimator closer to the light source.
While such an idea might seem wasteful in that we are blocking some of the light I believe that this will be more than offset with our ability to move the collimator closer to the light source.
Attachments
The "printed pattern" theory would work
The theory of using a "filter" to even out the distribution of light across the image has been used in some large format cameras to accomplish the same thing. Some super-wide angle lenses on these large format cameras have significant light falloff toward the edges and they are supplied with a matching filter which is darker in the middle and transparent toward the edge to even things out. Simple concept and it does work, although as stated, you are losing some of the light to pull this off.
The theory of using a "filter" to even out the distribution of light across the image has been used in some large format cameras to accomplish the same thing. Some super-wide angle lenses on these large format cameras have significant light falloff toward the edges and they are supplied with a matching filter which is darker in the middle and transparent toward the edge to even things out. Simple concept and it does work, although as stated, you are losing some of the light to pull this off.
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