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Thread: Vertex distance question

  1. #1
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    Vertex distance question

    Hi all,
    This is my first post, so please be kind (I graduated 5 years ago, so my ophthalmic optics knowledge is a little rusty).

    With the vertex compensation formula Dc=D/(1-dD) where Dc is the compenssated rx, D is the original rx and d is the vertex change, how would I adjust it in a non-air medium (eg if I submerged the lens in water)?

    Thanks in advance
    steff

  2. #2
    ATO Member HarryChiling's Avatar
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    If you break the equation down into it's parts it can easily be worked:

    Dc= Compensated Rx
    Dp= Dioptric Power
    Dv= Vergence Change
    nr= Index of Refracting Medium
    ni= Index of Incident Medium
    r= Radius of Curvature

    Dc=Dp/(1-dDp)
    Dc=Dp/1 - Dp/dDp
    Dc=Dp - 1/d

    Dc=Dp-Dv

    Dp=(nr-ni)/r
    Dv=ni/d

    Now that we hav the 2 components of our equation the Dioptric power minus the vergence change we can modify these

    Dc= (nr-1.33)/r - 1.33/d

    1.33 is aproximately the index of water.
    Last edited by HarryChiling; 03-14-2007 at 12:29 AM.
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  3. #3
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    Thanks!

    so does the first part of the last equation (n-1.33)/r come from the whole F=(n'-n)/r equation, and do I need to apply this formula if my compensated rx (Dc) is already in water, or does it just become
    Dc=D-1.33/d?

    Thanks

    steff
    Last edited by steff; 03-14-2007 at 12:29 AM.

  4. #4
    ATO Member HarryChiling's Avatar
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    I just edited the equation slightly to better illustrate, yes it is from the same equation you quoted. You do need to compensate the two components seperately. Otherwise the you are compensateing the vergence portion in water and taking that away from the lens part which is in air. Hope that makes sense.

    Keep in mind that the equation in the first post you quoted Dc=D/(1-dD) is a simplified version. It assumes that the surrounding index is air (n=1). When the lens gets submerged into water, you have to unsimplify your equation back to it's original form and take those assumptions of air being the surrounding medium out. Since both the dioptric power of the lens and the vergence change are both dependent upon the surrounding medium, they have to both be compesated for seperately. I have a slight issue with the way optics is being taught with over simplified equations, with assumptions built in. It cheats the user from seeing the dynamics of how the equation was derived and how it can b modified. I would suggest you check out Brent McCardles article on ray tracing, which is a great example of the dynamics behind some of the equations we use.

    http://onlineopticianry.com/wordpress/?p=108
    Last edited by HarryChiling; 03-14-2007 at 12:42 AM.
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    Thanks

    Thanks - I will. I think my little academic exercise just got a whole lot more complicated!

    steff

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    ATO Member HarryChiling's Avatar
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    Steff,

    Actually it should get easier from here, if you learn where the formulas come from you are going to have an easier time figuring out things. Your example was only complicate becas we were working the equation you were familar with backwards to get to a point where we could modify it. If I were to say he equation should be learned as:

    Dc=Dp-Dv

    thats a pretty simple formula to remember. Yet what most people will learn is:

    Dc=Dp/(1-dDp)

    I sometimes think that the formula looks more impressive, so this is the way some have chosen to teach it.
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    Ummm...

    Ok, can I run this by you?
    What I am trying to do (don't ask me why), is to find out what power (in air), a lens would have in order to compensate for the fact that when you are in water, you can't see due to the difference in refractive index between water and air.

    So, what I did was this...
    1st assumption: F(cornea in air)= 48D
    2nd assumption, radius of curvature of a cornea is about 7.8x10-3 m

    So, I figured out that:
    F(cornea) = (n'-n)/r
    48 = (n'-1)/0.0078
    n' (cornea) = 1.374

    If I submerge an eye in water
    F (equivalent cornea)+ = (1.374-1.33)/0.0078
    = 5.69D

    Therefore, to compensate for this, I need an extra 42.31D in the corneal plane.

    So, after this, what I was trying to do was extrapolate this out to a nominal vertex distance (eg 12mm), which is where I guess the
    Dc=(n-1.33)/r-1.33/d rule comes in, but I'm not quite sure how to apply it.

    Does any of this make sense, and have I made any glaring mistakes?

    thanks

    steff

  8. #8
    ATO Member HarryChiling's Avatar
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    I think I know what you are trying to do, but I don't think that you should be using that set of equations.

    We are talking about an emmetrope whose eye can be simulated with lets say a 60 D lens with a screen 16.67mm from the lens to simulate the image formed on the retina

    D=1/f
    f=1/D
    f=1/60
    f=0.01667

    so then if we were to go backwards and try to find out what power the eye would need to be in order for the rays to emmerge from the eye and be parallel we would

    D=1.33/f
    D=1.33/0.01667
    D=79.78 Diopters

    which minus our original power of 60 diopters would be 19.78 D of additional power needed under water to simulate the same vergence as an emmetrope in air. I think ths is what you are looking for.
    Last edited by HarryChiling; 03-14-2007 at 03:45 AM.
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    So that would be 19.78D at the corneal plane? Would I then need to compensate for the fact that you'd be mounting this at some sort of reasonable vertex distance? And if so, do I then need to compensate for the fact that instead of having an eye/air/lens/air interface, I have an eye/water/lens/water interface?

    What we're trying to do is (theoretically at the moment) build a pair of goggles that you could wear, with water on either side of the lens, but be able to see in water (eye/water/lens/water interface), instead of traditional goggles which are an eye/air/lens/water interface..

    steff

  10. #10
    ATO Member HarryChiling's Avatar
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    I keep questioning wheather or not the equations being used are going to be correct so, when I get a chance I will ray trace it, which I think in this situation is going to be the best idea.
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    I doubt if there is any gain in making vertex calculations. If the refraction in air is at the spectacle plane, then the corrected equivilent should be just as valid. The correction would need to be to the radii of the lens surfaces to account for the surrounding medium index, as well as the neutralization of the anterior corneal surface by the water. Since both water and corneas have the same index of refraction, the anterior corneal surface is neutralized, and should cause a hyperopic shift. Very high index lenses may need to be employed to make the lens/water interface index differential greater, so less extreme radii would need to be employed on the lenses. I am curious though, why you are interested in such a set-up?

  12. #12
    ATO Member HarryChiling's Avatar
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    Quote Originally Posted by Dave Nelson
    I doubt if there is any gain in making vertex calculations. If the refraction in air is at the spectacle plane, then the corrected equivilent should be just as valid. The correction would need to be to the radii of the lens surfaces to account for the surrounding medium index, as well as the neutralization of the anterior corneal surface by the water. Since both water and corneas have the same index of refraction, the anterior corneal surface is neutralized, and should cause a hyperopic shift. Very high index lenses may need to be employed to make the lens/water interface index differential greater, so less extreme radii would need to be employed on the lenses. I am curious though, why you are interested in such a set-up?
    I had a feeling someone would say that. I am still curious as to the math and will take the time to work the problem out, when I get a chance. Their are other issues involved that would make it more impossible to do as well, like the transmission of light in water, and the transmission through a higher index lens such as you mentioned. It should be fun to work it out as an exercise.
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    It does make for an interesting topic though. I'll watch for your ray tracing. But what about myopia induced by crossed (corneal and lenticular) cylinders? What about astigmatism induced by the neutralization of the corneal toricity? Oh, I'm getting a headache.:hammer:

  14. #14
    Master OptiBoarder lensgrinder's Avatar
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    Quote Originally Posted by steff View Post
    Ok, can I run this by you?
    What I am trying to do (don't ask me why), is to find out what power (in air), a lens would have in order to compensate for the fact that when you are in water, you can't see due to the difference in refractive index between water and air.

    So, what I did was this...
    1st assumption: F(cornea in air)= 48D
    2nd assumption, radius of curvature of a cornea is about 7.8x10-3 m

    So, I figured out that:
    F(cornea) = (n'-n)/r
    48 = (n'-1)/0.0078
    n' (cornea) = 1.374

    If I submerge an eye in water
    F (equivalent cornea)+ = (1.374-1.33)/0.0078
    = 5.69D

    Therefore, to compensate for this, I need an extra 42.31D in the corneal plane.

    So, after this, what I was trying to do was extrapolate this out to a nominal vertex distance (eg 12mm), which is where I guess the
    Dc=(n-1.33)/r-1.33/d rule comes in, but I'm not quite sure how to apply it.

    Does any of this make sense, and have I made any glaring mistakes?

    thanks

    steff
    To me it looks like you are on the correct path. So you need to add 42.31 D to the Rx. Let us say that the Rx is -10.00 refracted at 14mm and the goggles sit at 10mm the compensated power in water will be
    f1 = n/D

    1.33/-10 - 0.004 = -0.129
    1.33/-0.129 = -7.75

    Now compensate the corneal power you need so adjust 10mm in water.
    1.33/42.31 + 0.001 = 0.0324
    1.33/0.0324 = +41.00 D

    Now add the two powers together this from -7.75 D + 41.00 D = +33.25

    You would make a +33.25

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    Master OptiBoarder Darryl Meister's Avatar
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    Dave is spot on.

    For a thin lens, the change in power needed to maintain the same focus in water would probably be given by something as simple as:

    Power in Water = (Material Index - 1) / (Material Index - 1.333) * Power in Air

    Which means that you could effectively ignore radius for now (at least until you get until high plus lenses). But, as Dave noted, the eye loses a significant amount of power due to the loss of an air-cornea interface, which would require a great deal of compensation at the spectacle plane.
    Darryl J. Meister, ABOM

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    Just to complicate your lives: Fresh or Salt-Water?

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    Hi all,
    THanks for all the great responses, you are really making me think. What we're trying to do is make a pair of freediving goggles. The problem when you're freediving is that you use quite a lot of air to equalise, but in competition, you have to dive with a mask. If you fill the mask up with water, then you don't need to equalise, but you can't see where you're going... it's really just an academic exercise for me to see if it's possible (and dredge up some of my rather rusty ophthalmic optics skills), rather than any desire to 'develop' anything really.

    The lens would be immersed in saltwater.

    I am assuming that the subject is emmetropic.

    Dave, I'm a little confused, as the subject is emmetropic, and we're talking about effectively recreating the power of the anterior cornea, doesn't that mean that the power is actually not in a spectacle plane, but in a corneal plane?

    Aaagh! :-) I'm in way over my head

    steff
    Last edited by steff; 03-15-2007 at 07:14 PM.

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    In over my head!

    I believe that I am way out of my league here. But, I have a question and feel free to laugh, giggle, and mock me at will.


    Does pressure on the eyeball and its effect on the overall visual system come into play here? The reason I ask is that I have dealt with several military types who have informed me of such things. You know, the deeper you go, the more pressure on the body(eye), and possibly change the shape/refractive error of the globe.

    I know that pilots experience this at extreme take off "g" forces.

    Ok, go ahead and laugh.;)

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    I should have thought of free diving. I am aware they generally use very low volume masks for the reasons you have stated.
    While it does indeed make for some interesting discussion, My understanding of the sport of free diving makes me question if the emmetrope is in need of any further correction, since they only need to see depth markers as they descend and ascend, and the vision, while not great, should be adequate. They employ rescue divers at certain intervals as well, and I believe use lights.
    As a former skydiver, base jumper and scuba diver, I have one question for your free-diver:
    "are you crazy? diving to 300 odd feet while holding your breath?"

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    Ah...

    actually, I am the freediver... (or one of them, anyway). There is a certain meditation type quality to it, which I don't get from any other sport (rock climbing, whitewater kayaking)... You're not holding you're breath, you're choosing not to breathe. There's a difference... trust me ;-)

    If you fill your mask with water, you really can't see much... not even the line

    steff

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    You must have read some of Greg Iles stuff to come up with all that freediving meditation stuff. Emetropes don't need any correction at all, much less addittional in the water. Some filters might help but no Rx needed just a reasonably good surface that will keep the water out.

    Chip
    Last edited by chip anderson; 03-15-2007 at 10:25 PM. Reason: One L too many

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    Since the cornea provides the greatest amount of refraction in the visual system, its neutralization by the surrounding medium would have to, as you already indicate, be replaced. This would amount to a very strong lens at the spectacle plane. Since the correcting lens is also immersed in water, its value in air would be partly neutralized by its immersion in water, so it would need to be increased by a further large amount, and I suspect the lens strength would greatly exceed any practical use it may have for your endeavor. (I suggest you discontinue your venture so your poor parents can sleep at night.);)

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    Chip, I'm afraid I don't understand. I'm not trying to induce a correction underwater, I'm trying to neutralise the fact that you've got water surrounding the anterior cornea, not air. Conventional goggles provide a water tight (ish) seal so that the anterior cornea still has an cornea/air/lens/water interface. What I'm trying to figure out if it is possible or not is to make a pair of goggles that have a cornea/water/lens/water interface.

    Does that make sense, or am I just babbling?

    Not sure who Greg Illes is, but if you want to point me in a direction that is useful, I would be grateful.

    steff
    Last edited by steff; 03-15-2007 at 09:57 PM.

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    Greg Iles is an author in Natchez who writes mysteries. One of his lead charaters is a free diver, swimmer who meditates under water in one of his novels.
    For your goggle a man in Cinncinatti, Kenneth Swanson once with a few friends developed a scleral contact lens with a free air space for this purpose or at least to eliminate goggles. You might see the CLSA members list for his address, I am sure he will be willing to share his findings if he is still around.

    Chip
    Last edited by chip anderson; 03-15-2007 at 10:25 PM. Reason: One L too many

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    Chip,
    That sounds interesting, however in competition, you aren't allowed to dive with CLs. Also, I wonder if there is a problem with an inablility to equalise the air space between the contact and the cornea.

    Interesting though - I've found a few papers on scleral lenses doing the same thing as I'm thinking of, however I did have my heart set on doing the equations for goggles (not so much for the end result, but for the process).

    Thank you for the input though.
    steff

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