Poly and acuity loss?

[eye1u2see2020]

Hi all, I have a quick question or verification. The lead lab tech had told me that people will lose 2 lines of acuity by just using poly. Is there any information out there that will show that this is not correct. I have worked in this field for 15 years and have never heard this. I know that there is a higher abberation value and that poly is not the best optically, however I don't think that you lose 2 lines of acuity. Please help me to disprove this theory.
Thank you
Rick

[Judy Canty]

I've been wearing poly for YEARS with no loss of acuity.

[hcjilson]

I have 2 pr of identical Rx's in Definity, both in the same 3pc mounting. One is in poly and one trivex. I went to poly for the transitions which is not yet available in Trivex Definity's. (2nd quarter and still counting!) They are both excellent but the fact of the matter is that my Trivex's are "clearer" My Rx is 5 and just over +5 in the distance with marginal astigmatic correction, and a 3 add. In terms of acuity I don't think I drop a line with poly but things seem a lot sharper with Trivex. It's nothing I can verbalize which is why I hesitated to bring it up, but my vision is better through Trivex. I also own a pair of 1.67's and same thing applies. My Trivex is clearer.

Pete and I have been going around the Trivex vs poly circle for years and he may be tempted to say that I've just bought the hype, but I KNOW how I see and it has nothing whatever to do with hype. Acuity loss?? I don't think so or marginal at best. I really don't know what to make of it!

[chip anderson]

Some folks think there is a lot of loss with Poly, some folks can't tell the difference, some think they are the same.
Same with AR some folks think it's wonderfully better, most folks can't tell the difference (at least from the patient's side of the lenses.)
But I can absolutely find you any number of opticians and quite a few patients that will swear that their being able to "prove" any of the above statements is true.

Chip

[hcjilson]

I can sure tell the difference driving at night, between AR and non AR. Of course, Chip hasn't experienced that yet because he is much younger than I, and his cataracts haven't developed yet!!! :):):):)

[MarcE]

Your lab tech is full of crap.
However, when you look off center in a low abbe material (poly), there is distortion. This distortion is proportional to the Rx power, and proportional to the distance off center. This distortion is not obvious at all to most of the population. The higher the Rx, the more noticable the distortion. I see hyperopes notice this more than myopes because the add power increases the absolute power of the Rx.

From the previous post, HC's total near power is over +8.0, and when reading he is looking off center by at least 16mm. That is why he notices that Trivex is clearer.

You can see this in a lensometer. When checking the add in a high hyperope in poly or 1.67 I notice that even when I get the mires as clear as they can be, they still aren't exactly clear.

In another thread, drK reported that 80% of the Rxs are between +/- 3.00. So poly should not be a problem for 80% of the population.

[Happylady]

Your lab tech is full of crap.

I agree!

Two lines? What a ridiculous thing to say. I can see the 20/15 line on the eye chart. So if I look at it with glasses made with polycarb I should only be able to see the 20/25 line if he is right.

It is easy to prove him wrong. Look at the eye chart through a standard cr39 lens and then look at it through a poly lens.

I honestly can't tell the difference when I look through poly, 1.6, 1.67, or regular plastic. I have glasses in both poly, 1.6, and cr39 right now and I see no difference. I know that is not true for everyone, though.

[HarryChiling]

Yes at a certain Rx a patient will lose visual acuity, if we were to look at a line on the chart generally a 0.25D change will equate to one line on the chart. We know that poly has an n=1.589 and an abbe=30 so lets break it down:

abbe = (n_yellow-1)/(n_blue-n_red)

so pluggin in our givens:

30 = 0.589 / (n_blue-n_red)
(n_blue-n_red) = 0.589 / 30
(n_blue-n_red) = 0.01963

Now we can make a fairly accurate assumption that the wavelengths for the red and the blue fall equadistance from the yellow wavelength except on opposit sides of the spectrum so then we can make the assumption that the indices are also equadistant from the yellow index. So lets put that word problem into a formula so that we can reduce the above even further (I used the average but it may be more accurate to use the mean instead, I don't know how much accuracy would be lost but you could always tweak the formula around here to see what the outcome would be):

n_yellow = (n_blue+n_red) / 2

Now if we were to solve the above equation for both red and blue:

n_blue = 2*n_yellow - n_red

and

n_red = 2*n_yellow - n_blue

Now we can use these formulas in the top equation to further reduce to get our indices for red and blue:

2*n_yellow - n_red - n_red) = 0.01963
2*n_yellow - 2*n_red = 0.01963
n_yellow - n_red = 0.01963 / 2
1.589 - n_red = 0.009817
n_red = 1.579

Now to solve for our blue index:

n_blue - (2*n_yellow - n_blue)= 0.01963
n_blue - 2*n_yellow + n_blue= 0.01963
2*n_blue - 2*n_yellow = 0.01963
n_blue - n_yellow = 0.01963 / 2
n_blue - 1.589 = 0.009816
n_blue = 1.599

Now we know our blue and our red indices in the poly medium and we know we don't want a greater than 0.25D of difference between the two in a lens. So let's set that up as an equation:

D_blue - D_red = 0.25

Alright now we need to break that down even further into a radius of measure if we are to get a constant:

[(n_blue - 1) / radius] - [(n_red - 1) / radius] = 0.25
(n_blue - 1) - (n_red - 1) = 0.25 * radius

Subbing out our known values:

(1.599 - 1) - (1.579 - 1) = 0.25 * radius
0.599 - 0.579 = 0.25 * radius
radius = 0.02 / 0.25
radius = 0.08m

So now we need to know what power so we are going to use our reference wavelength in our radius to diopter equation:

D = (1.589 - 1) / 0.08
D = 7.36

So now you know that to get a 0.25D difference or the equivalent of one line of visual acuity loss the power of the lens would need to be 7.36D, of course prism is another question all together and would require a whole nother post. ;)

[drk]

Yes at a certain Rx a patient will lose visual acuity, if we were to look at a line on the chart generally a 0.25D change will equate to one line on the chart. We know that poly has an n=1.589 and an abbe=30 so lets break it down:

abbe = (n_yellow-1)/(n_blue-n_red)

so pluggin in our givens:

30 = 0.589 / (n_blue-n_red)
(n_blue-n_red) = 0.589 / 30
(n_blue-n_red) = 0.01963

Now we can make a fairly accurate assumption that the wavelengths for the red and the blue fall equadistance from the yellow wavelength except on opposit sides of the spectrum so then we can make the assumption that the indices are also equadistant from the yellow index. So lets put that word problem into a formula so that we can reduce the above even further (I used the average but it may be more accurate to use the mean instead, I don't know how much accuracy would be lost but you could always tweak the formula around here to see what the outcome would be):

n_yellow = (n_blue+n_red) / 2

Now if we were to solve the above equation for both red and blue:

n_blue = 2*n_yellow - n_red

and

n_red = 2*n_yellow - n_blue

Now we can use these formulas in the top equation to further reduce to get our indices for red and blue:

2*n_yellow - n_red - n_red) = 0.01963
2*n_yellow - 2*n_red = 0.01963
n_yellow - n_red = 0.01963 / 2
1.589 - n_red = 0.009817
n_red = 1.579

Now to solve for our blue index:

n_blue - (2*n_yellow - n_blue)= 0.01963
n_blue - 2*n_yellow + n_blue= 0.01963
2*n_blue - 2*n_yellow = 0.01963
n_blue - n_yellow = 0.01963 / 2
n_blue - 1.589 = 0.009816
n_blue = 1.599

Now we know our blue and our red indices in the poly medium and we know we don't want a greater than 0.25D of difference between the two in a lens. So let's set that up as an equation:

D_blue - D_red = 0.25

Alright now we need to break that down even further into a radius of measure if we are to get a constant:

[(n_blue - 1) / radius] - [(n_red - 1) / radius] = 0.25
(n_blue - 1) - (n_red - 1) = 0.25 * radius

Subbing out our known values:

(1.599 - 1) - (1.579 - 1) = 0.25 * radius
0.599 - 0.579 = 0.25 * radius
radius = 0.02 / 0.25
radius = 0.08m

So now we need to know what power so we are going to use our reference wavelength in our radius to diopter equation:

D = (1.589 - 1) / 0.08
D = 7.36

So now you know that to get a 0.25D difference or the equivalent of one line of visual acuity loss the power of the lens would need to be 7.36D, of course prism is another question all together and would require a whole nother post. ;)

One of your assumptions is incorrect...one quarter diopter does not equal one line of visual acuity.

It's more like 0.50-0.75.

So, in the Chilinguerian fashion, add fudge factor 0.62 / 0.25 = 2.5
2.5 x 7.36D = 18D

As you can see, the conclusion is practically the same: it's a non-issue.

P.S. I enjoyed your color-coding. I think your method for determining dioptric difference between long and short wavelengths was quite nice. I think, though, that there was a methodology flaw...

When refracting with a glass lens and a monochromatic chart, the "correct" or "reference" refractive correction is arrived at, we agree to assume.

If the lens is fabricated in that power, the "dioptric dispersion" you calculated above will occur at black-white interfaces and "veil" the edges with chromaticity, which can perceivably affect acuity (although it is not completely safe to assume...).

But with a spread between red and blue of even 1/2 diopter, the observer would only see one half of that interval...with a white-on-black edge there would be only the red fringe visible (but not the blue, which would occur looking through the lens on the opposite side of the optical center).



So, to summarize the fudge factors, you'd have to further double even what you did above, making it ~28D!

[HarryChiling]

One of your assumptions is incorrect...one quarter diopter does not equal one line of visual acuity.

It's more like 0.50-0.75.

So, in the Chilinguerian fashion, add fudge factor 0.62 / 0.25 = 2.5
2.5 x 7.36D = 18D

As you can see, the conclusion is practically the same: it's a non-issue.

Thanks for the catch and yes it does change things in the above equation but the fact still remains that there will be a differnce of 0.25 between the red and blue wavelengths or across the visible spectrum in a 7.00 lens. If we were to look at the TCA in the same lens:

TCA = 7.36 / 30
TCA = 0.25

In Mohamed Jalies book he refers to a TCA of 0.10 as being significant and as you could see this power has 2 1/2 times that amount of aberration. I have been over this before but a lens power with TCA of 0.10 in poly would be:

Power = 0.10 * 30
Power = 3.00D

It's not to say it's gonna effect everyone but powers above a 3.00 could potentially affect some, just to add a side note. I wear a -5.00 in a poly and it has never bothered me, matter of fact if I look I can see color finges when I look at edges of contrasting colors like black/white. It has never bothered me and I also would point out that in certain powers a degree of TCA will be present no matter what, so there is probably nothign that can be done in some cases but the moderate to high range shoudl have an appropriate material. I have drunken from the Fezz juice lately and have been dabling in the Trivex and must say I am impressed so far and the price seems more reasonable lately so I think it is something that should be a great tool in situations like this.

[dbracer]

These posts are excellent.

It's been awhile since I given thought to such matters and you've made me delve into the literature and refresh myself on these matters.

To the original question the answer is short. Poly has poorer optics. Whether it bothers depends on the SRx and the patient, but it's seldom 2 lines of VA - not that it couldn't be in specific situation.

The various Trivex's are good, but the index separates them from "hi-index" lenses.

Once again drk and Harry, dang good posts.

Respectfully,
dbracer

[HarryChiling]

Thanks for the data Darryl always precise and on time, not bad on the estimate though I was off by 0.001 from your estimate. I love it when I throw a dart and it lands close.:bbg:

Now TCA and LCA would they have the same effects on VA? If so then the TCA computed in the above would be equivalent to a VA of 20/28, right?

[Darryl Meister]

The lead lab tech had told me that people will lose 2 lines of acuity by just using poly.

I actually have a table of visual acuity values as a function of lateral chromatic available online at Chromatic Aberration Course. This table is based on a study done by Meslin & Obrecht, "Effect of Chromatic Dispersion of a Lens on Visual Acuity."

And you can calculate lateral chromatic aberration (C) by dividing the prism (P) produced at a given point through the lens (using good ole' Prentice's rule) by the Abbe value (v) of the lens material:



So, at 10 mm (1.0 cm) from the optical center, a +5.00 D polycarbonate lens, with an Abbe value of 30, will produce the following lateral chromatic aberration:





which corresponds to a visual acuity of roughly 20/25 in Meslin and Obrecht's study...

Thanks for the data Darryl always precise and on time, not bad on the estimate though I was off by 0.001 from your estimate. I love it when I throw a dart and it lands close.

Very true. ;)

The differences would become more significant as you approach the far ends of the spectrum, particularly in the blue end. Across the central wavelengths, the refractive index varies nearly linearly and, for most ophthalmic lens materials at least, the difference between the refractive index of the blue hydrogen line and the red hydrogen line is relatively small.

Now TCA and LCA would they have the same effects on VA? If so then the TCA computed in the above would be equivalent to a VA of 20/28, right

Lateral chromatic aberration is generally considered much more problematic. Keep in mind that the eye actually suffers from roughly 1 diopter of axial or longitudinal chromatic aberration. Also, since axial chromatic aberration may play a role in the accommodative mechanism, the visual system probably chooses the best color focus for a given object.

[dbracer]

As it turns out, the refractive index in the blue end of the spectrum changes more rapidly than the refractive index in the red end...

Now we're getting ridiculous.

When the formulas start looking like hieroglyphics and the one-up-manship looks like the last round of the Olympic tryouts, it's time to look at your motives.

Enough is enough, and it can be said a lot simpler.

Respectfully,
dbracer

[Darryl Meister]

When the formulas start looking like hieroglyphics and the one-up-manship looks like the last round of the Olympic tryouts, it's time to look at your motives

I was sharing information that I found to be very interesting, myself, including the results of a clinical study that speaks directly to the problem. Although I've been sharing this kind of information with fellow OptiBoarders for over 13 years now, I do understand that not everyone shares a passion for the "details" or more pedantic side of this stuff. As for the formula, I have no doubt that it did not look like hieroglyphics to Harry, who may very well find a generalized equation for refractive index useful for a number of applications; I actually developed it for a computer program that must convert between helium d and mercury e refractive index values for power calculations in different countries.

[HarryChiling]

I was sharing information that I found to be very interesting, myself, including the results of a clinical study that speaks directly to the problem. Although I've been sharing this kind of information with fellow OptiBoarders for over 13 years now, I do understand that not everyone shares a passion for the "details" or more pedantic side of this stuff. As for the formula, I have no doubt that it did not look like hieroglyphics to Harry, who may very well find a generalized equation for refractive index useful for a number of applications; I actually developed it for a computer program that must convert between helium d and mercury e refractive index values for power calculations in different countries.

You know I definately appreciated that, I was almost positive that the relationship wasn't linear and for an example here I figured a estimation would suffice, but like you I program and when the computer is doing the heavy lifting you tend to throw big weight at it.

[Darryl Meister]

You know I definately appreciated that, I was almost positive that the relationship wasn't linear

Oddly enough, I searched through a dozen different optical engineering textbooks and countless websites when I was trying to find a reasonably accurate dispersion equation for calculating the refractive index of a lens material at an arbitrary wavelength, and I couldn't find one single example that relied only on the Abbe value and mean refractive index. So, even if the other OptiBoarders who frequent the Ophthalmic Optics forum don't necessarily appreciate this stuff or find it especially interesting, the next poor guy out there trying to solve a similar optical engineering problem will have at least one hit show up in his Google search. ;)

[HarryChiling]

That's funny a few months back a thread got me thinking about the same thing and I had a hard time finding a formula for it as well I searched high and low and through various books. I am sure I could have found it in the OSA's Book, but I had already wasted too much effort on it at the time and just thought I'd move on.

I know how difficult a pull that one was and believe me if I could give you more than one positive in a row I would.

[Darryl Meister]

I know how difficult a pull that one was and believe me if I could give you more than one positive in a row I would

Well, I certainly appreciate the positive feedback.

I ran across a similar problem trying to find a good algorithm for calculating the dominant wavelength for a set of CIE color coordinates a few months ago, while finishing up that Spectacle Optics program. There are literally dozens of forums out there with posts asking for the very same solution, and I never found a single response with a complete numerical recipe. And it took a fair bit of effort to put one together.

So, to dbracer's point, perhaps nowadays I'm just looking for an excuse to post this stuff...

[HarryChiling]

So, to dbracer's point, perhaps nowadays I'm just looking for an excuse to post this stuff...

Better than an excuse not to. I saw you've been updateing your site, anything new.

[Fezz]

Hey...

Does this thread come with a study guide? Cliff notes? Instructions? A interperter?

;):cheers::bbg::cheers::shiner:

[DragonLensmanWV]

Hey...

Does this thread come with a study guide? Cliff notes? Instructions? A interperter?

;):cheers::bbg::cheers::shiner:

How about a "For Dummies" book?

Here's one observation about ABBE that I experience on a daily basis.
If I look at a fluorescent light fixture on the ceiling ( so what if I spend hours doing that? :D) through the edge of my lenses (Hoya 1.70) I can see at the top of the fixture the red fringe fading to yellow and at the bottom of the fixture I see blue fringe fading to black. Now looking at pictures or TV through the edge (the effects are less in the central portion) I see feature edges reproduced in a different color just offset from the original like if your newspaper's register is off so people look like thay have four eyes. Now with the Hoya 1.70 there is more color blur at the blue end than the red end. With 1.67, I have only a very small area that is usable and everything looks fringed and the lenses are basically useless for me, so I counsel against 1.67 for powers that exceed + - 4D. Of course, below that range there is no need for the 1.67, so my usage of 1.67 is zero.

[Darryl Meister]

Does this thread come with a study guide? Cliff notes? Instructions? A interperter?

In all fairness, it has always stated right at the top of this forum, Not for the faint of heart!

In all my years of moderating this forum, I've never seen so much balking at a little high-school-level algebra! For shame! ;)

I've decided to start a separate thread for dispersion equations because 1) I think this information is too valuable not to make available to those who might find it useful and 2) I do not want this thread on chromatic aberration in polycarbonate further derailed.

Harry, I would encourage you to post your own "first-order" (linear approximation) dispersion equation in this thread as well.

[Fezz]

Just having a little fun Darryl!

Just having a little fun...thats all!

:cheers::cheers::cheers:

[dbracer]

In all fairness, it has always stated right at the top of this forum, Not for the faint of heart!

In all my years of moderating this forum, I've never seen so much balking at a little high-school-level algebra! For shame! ;)

I've decided to start a separate thread for dispersion equations because 1) I think this information is too valuable not to make available to those who might find it useful and 2) I do not want this thread on chromatic aberration in polycarbonate further derailed.

Harry, I would encourage you to post your own "first-order" (linear approximation) dispersion equation in this thread as well.

Ah heck Darryl, you're probably right.

I probably should keep my comments to myself since I like the formula stuff and will probably follow you to your new thread.

I can't be too smart. I'm a dang optometrist. Go figure. (That's a rhetorical remark, not a command).

dbracer,
The guy with the big mouth or at least too fast of keys