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Thread: Vertex distance question

  1. #26
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    Chip, I think you may be misunderstanding what free diving is. It is a sport where people breath-hold dive to extreme depths, often to more than 200 feet. Pressure increases one atmosphere every 33 feet or so, so the divers must use valuable lung capacity to equalize the pressure in the mask, thus having less oxygen to continue the dive. If I understand this correctly, Steff is trying to find a way to eliminate the airspace altogether, and replace it with water, which is non-compressable, and neutral with the surrounding pressure.

  2. #27
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    Dave, YAY! Thank you. That's exactly what I'm trying to do. You put it so much more eloquently than I was attempting to do

    steff

  3. #28
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    I did a little searching on google for liquid-chamber goggles for keratitis sicca, and found this. Try googling "liquivision" they are fluid filled goggles that have small very high plus lenses incorporated. (just what I predicted:bbg: ) and they are specific to free diving. Have you heard of them?

  4. #29
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    Yep, I have, but... the reason this all sort of started, was the guy who produces them seems to think that there is no equivalent dioptric reading for the lens, as it's made for water, not air. I think he's wrong. It may be a very high plus lens, but you can still vert it...

    Then it all got a little more in depth, and I started wondering if I could figure it out for myself (without just borrowing a pair of them, and chucking them under my vert).

    steff

  5. #30
    Master OptiBoarder lensgrinder's Avatar
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    Quote Originally Posted by Darryl Meister View Post
    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.
    Could this compensation be found the way I found it in my post(not using the -10.00D since she is emmentropic)? Using 42.31D as the loss in corneal power.

    f1 = n/D
    f1 = 1.33/42.31 = 0.031M focal length in water
    Add this to your vertex distance. Let us use 10mm
    D = 1.33/(0.031+ 0.01) = 32.44D

  6. #31
    ATO Member HarryChiling's Avatar
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    Ok, so I have been working on it a bit I think I might have a solution for you.

    If a emmetrope in air sees 20/20 and the cornea is 48D like you mentioned previously, and the index is 1.374 as mentioned previously then

    The object at 6m (or 20feet) has the effective vergence of -0.17D

    D=1/-6=-0.166

    So the light entering the cornea has a vergence of -0.17

    So the light entering the cornea + the coneal power gives us the total power through the system

    Dtotal=+48.00 + (-0.17) = 47.83 is the vergence leaving the cornea plane, since this is an emmetrope the focal point is going to fall on the retina, so:

    fcornea1.374/47.83=28.7mm

    So now we have our system, the light leaves an object at 6000mm (6m or 20ft) and is refracted at the cornea plane then comes to a focus 28.7mm past the corneal plane. if we were to equate the cornea to a lens and find the radius we would be able to accurately determine the power of the cornea under water.

    Dcorneaair=(1.374-1)/48.0=7.79mm radius

    Since the radius doesn't change we can now find the power in water

    Dcorneawater=(1.374-1.33)/.00779=5.65D

    So now we know that the power of the cornea underwater is 5.65D and the light entering the cornea under water is:

    Dobjectunderwater=1.33/-6=-0.22D

    So the power leaving the cornea is

    Dcorneawaterout=5.65 + (-0.22)=5.43D leaving the corneal plane. Form here we now that we need an additional:

    Dlens=47.83-5.43=42.4D at the corneal plane to make up for the loss.

    This is where we compensate the lens for the vertex change:

    1.33/42.4=31.4mm focal distance lens we are moving it 10mm in front of the eye so 41.4mm focal length which is the equivalent of a lens of power:

    1.33/0.0414 = 32.13D lens

    About the same thing as lensgrinder, so I would assume that it should be correct, although I took the long way around and factored the objects vergence into the equations.
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  7. #32
    Master OptiBoarder Darryl Meister's Avatar
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    For a "typical" eye, the cornea has an anterior radius of 7.8 mm and a refractive index of 1.376. This means that a typical corneal surface produces around (0.376 - 1) / 0.0078 = 48.20 D of power in air; though some of this is neutralized by the negative power of the posterior surface. This also means that, when submerged in water, the power of the anterior corneal surface drops to about:

    (1.376 - 1.333) / 0.0078 = 5.51 D

    So, the spectacle lens would probably have to make up the difference at the spectacle plane (corrected for the index of water). But keep in mind that this assumes the wearer happens to have an exactly "typical" cornea. Without taking K readings and such, this would all be speculation.
    Darryl J. Meister, ABOM

  8. #33
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    Wow. Awesome. Thanks guys... now... has anyone had any experience with grinding a lens that sort of power? So far, the research I've managed to do (in between life), is that a plano convex lens would be the best configuration for this (reducing spherical abberation)? Any comments or disputes to this? Hmmm... I'll do some research and see what I can find so I'm not just leaching answers off you.

    steff

  9. #34
    ATO Member HarryChiling's Avatar
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    I would look into fresnel type optics. I don't see anyone even atempting to grind this kind of power into a lens, and to make a convex plano lens you would be talking about a front curve of over +30.00, it would be redonkulous.
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  11. #36
    ATO Member HarryChiling's Avatar
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    Great book chip, thanks.
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  12. #37
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    OK, After a bit of thinking, I was wondering, can I do this?

    We need 42.4D at the corneal plane

    I put a +20.00D soft (large diameter) contact lens on the eye, therefore needing an extra 22.4D at the corneal plane.

    1.33/22.4=59.4mm focal distance lens we are moving it 10mm in front of the eye so 69.4mm focal length which is the equivalent of a lens of power:

    1.33/.0694 = +19.1D lens, which I could grind in a lenticulated lens?

    I think there's a flaw in my thinking cause I've effectively created a telescopic system, but I'm not sure...

    steff

  13. #38
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    There is still the issue with fluid neutralization of the lens surfaces immersed in water, including the anterior surface of the contact lens. The contact lens will not likely remain in the eye while immersed in water anyway. If you want to talk telescope, there may be a way of building a Galilean system where the refracting surfaces are enclosed in an air filled tube, which would have to be resistant to pressure, and have a very short tube length, likely less than an inch or so. The front and rear of the scope would be plano, and the piece would be mounted on a mask faceplate, with the ocular lens placed very close to the cornea. Picture a smaller version of the B&L contact lens inspection device.
    Steff, what is your deepest dive?

  14. #39
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    OK,
    I've had a play, and a soft contact lens doesn't seem to fall out when diving.

    Any thoughts on how I can figure out the effective power I need for a contact lens/ spec lens combination for these (getting more complicated every time I look at them) goggles?

    Dave, my deepest dive yesterday was 26m (constant weight ie no sled, kickign up and kicking down). 30m is my goal at the moment

    steff

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