# Optical Forums > Ophthalmic Optics >  Vertex distance question

## steff

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

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## HarryChiling

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.

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## steff

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

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## HarryChiling

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

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## steff

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

steff

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## HarryChiling

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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## steff

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

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## HarryChiling

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.

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## steff

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

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## HarryChiling

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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## 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?

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## HarryChiling

> 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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## Dave Nelson

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:

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## lensgrinder

> 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
> ...


 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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## Darryl Meister

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.

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## chip anderson

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

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## steff

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

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## Fezz

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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## Dave Nelson

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?" :Eek:

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## steff

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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## chip anderson

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

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## Dave Nelson

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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## steff

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

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## chip anderson

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

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## steff

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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## Dave Nelson

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.

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## steff

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

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## Dave Nelson

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?

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## steff

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

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## lensgrinder

> 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

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## HarryChiling

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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## Darryl Meister

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.

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## steff

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

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## HarryChiling

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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## chip anderson

http://www.opticalformulastutorial.com/

Chip:cheers:

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## HarryChiling

Great book chip, thanks.

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## steff

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

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## Dave Nelson

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?

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## steff

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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