# Q&A Thread for Optometric Curriculum Material

#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
Questions pertaining to the body of knowledge of optometry can be posted here in this one thread.

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#### KHE

##### Senior Member
15+ Year Member
Hi folks

I've started a number of threads since the summer about questions relating to optometric pot-pourri (e.g. optyl/zyl, the UV thread, RMS error). I thank those who have responded - and in effort to clean up the threads a bit, I'll post (and invite others to post) questions pertaining to the body of knowledge of optometry here in this one thread.

My question today...

With regards to astigmatic lenses and the issue of calculating power in an oblique meridian (i.e. what is the power at 180 for a -2.00 -1.00 x 030 lens?), the equation I learned for power at 180 is:

F(alpha) = F(sphere) + F(cylinder)*sin(alpha)^2

I hope you can make out what I just wrote. The alpha/sphere/cylinder associated with the "Fs" in the equation are subscripts, whereas the alpha in the "sin" is the alpha value that is part of the equation. My question is what is the definition of alpha? I have,

alpha = angle between axis meridian and desired meridian.

Is it the acute angle between the two? Or is it the axis value subtract the desired meridian value? Or the desired meridian value subtract the axis value? In the last 2 cases, the value could be greater than 90 degrees, and can be negative (but I guess negative doesn't matter since the sine will be squared. The >90 might not matter either since sine is sinusoidal). So my question is, for desired power at 180, and axis 30, is alpha:

the acute angle = 30 degrees? (desired meridian drawn towards axis)
the acute angle = -30 degrees? (axis drawn towards desired meridian)
30-180 = -150 degrees?
180-30 = 150 degrees?

As I said, due to the squarring of the sine, and the sinusoidal nature of the sine function, they may all give the same answer (I think they do), but what is the definition of alpha? If you had to tell someone to calculate this, would you tell them to find the acute angle, or to simply subtract the desired meridian from the axis (or vice versa).

Thanks.

Jesus....

You can get an approximation of the power in the resulting meridian using a simple pythagorean equation. There is little clinical relevance to using sin2 angle + cos2angle =1. The two alphas in your example mean different things. The Falpha is referring to the POWER (not angle) in the meridian you wish to calculate. The sine alpha is the acute angle between the desired meridian and the axis meridian.

Also, when you're subtracting angles, you should use a value of zero for axis 180 rather than the numerical value 180.

I think that's right. My head hurts. Please don't post **** like that again!

#### JMU07

##### Full Member
10+ Year Member
5+ Year Member
I think that's right. My head hurts. Please don't post **** like that again!

Yeah, next time wait until finals are over!!

#### gochi

##### Full Member
10+ Year Member
Hi folks

I've started a number of threads since the summer about questions relating to optometric pot-pourri (e.g. optyl/zyl, the UV thread, RMS error). I thank those who have responded - and in effort to clean up the threads a bit, I'll post (and invite others to post) questions pertaining to the body of knowledge of optometry here in this one thread.

My question today...

With regards to astigmatic lenses and the issue of calculating power in an oblique meridian (i.e. what is the power at 180 for a -2.00 -1.00 x 030 lens?), the equation I learned for power at 180 is:

F(alpha) = F(sphere) + F(cylinder)*sin(alpha)^2

I hope you can make out what I just wrote. The alpha/sphere/cylinder associated with the "Fs" in the equation are subscripts, whereas the alpha in the "sin" is the alpha value that is part of the equation. My question is what is the definition of alpha? I have,

alpha = angle between axis meridian and desired meridian.

Is it the acute angle between the two? Or is it the axis value subtract the desired meridian value? Or the desired meridian value subtract the axis value? In the last 2 cases, the value could be greater than 90 degrees, and can be negative (but I guess negative doesn't matter since the sine will be squared. The >90 might not matter either since sine is sinusoidal). So my question is, for desired power at 180, and axis 30, is alpha:

the acute angle = 30 degrees? (desired meridian drawn towards axis)
the acute angle = -30 degrees? (axis drawn towards desired meridian)
30-180 = -150 degrees?
180-30 = 150 degrees?

As I said, due to the squarring of the sine, and the sinusoidal nature of the sine function, they may all give the same answer (I think they do), but what is the definition of alpha? If you had to tell someone to calculate this, would you tell them to find the acute angle, or to simply subtract the desired meridian from the axis (or vice versa).

Thanks.

For pre-optos, this literally neiog2no291wnf5nwo.

#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
You can get an approximation of the power in the resulting meridian using a simple pythagorean equation.

I wasn't aware of this (but I'm not surprised that you can). I'm assuming the estimated power is the hypotenuse, and the two cylinders each contribute a sine and cosine component?

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#### JMU07

##### Full Member
10+ Year Member
5+ Year Member
I wasn't aware of this (but I'm not surprised that you can). I'm assuming the estimated power is the hypotenuse, and the two cylinders each contribute a sine and cosine component?

I think he's talking about power vectors here. Simple formulas you can use to convert from vector form to a prescription format. Makes it much easier to deal with oblique meridians!

#### eyestrain

##### Member
10+ Year Member
I do not understand a word posted on this thread. My solution to this problem: tobradex qid ou. You're welcome.

#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
I've come across literature (advertising material of a specific lens company) recently showing treated CR39 has better UV protection than PC. Is this true? Is this what some of you have learned?

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#### eyestrain

##### Member
10+ Year Member
Hi folks

Back in the day when going through optom school, I clearly remember one of the bottom lines in ophthalmic optics was that polycarbonate (PC) had the best UV protection of all lenses, exceeding that of UV-treated CR39.

I've come across literature (advertising material of a specific lens company) recently showing treated CR39 has better UV protection than PC. Is this true? Is this what some of you have learned?

I assume both PC and UV CR39 do a fine job. I've never taken the time to research it, but I'm willing to bet the difference is negligible.

#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
Hi folks.

The item in the image is called a lens former. Correct?

#### drbizzaro

10+ Year Member
Hi folks.

The item in the image is called a lens former. Correct?

it's called a pattern blank

10+ Year Member
7+ Year Member
thnx.

#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
Hi

Just curious. I'm aware that in previous generations of Transitions lenses, the "transitioning" part was a single layer in the lens (typically below the front surface).

Are ALL modern models of Transitions lenses made in the same way? Or do "through and through" type of Transitions exist? i.e. do any Transitions lenses (for plastic lenses) exist where the transitioning material is embedded throughout the lens polymer?

#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
Hi folks

OK - no replies to my last question. What about the following one:

Is there any professional terminology to describe lens styles? For instance, the following lens is quasi-rectangular - but it "jets out" superior-temporally. Is there a name for this? How would you describe this lens shape/style?

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#### armstrong77

##### Full Member
10+ Year Member
Just a Q regarding lensometry...

I just recently listened to a lecture talking about lensometry of high-plus Adds. For High-Adds, you know how we have to find front-vertex power (FVP) of the add portion. Is the Add calculated by:

1. (Add FVP) MINUS (Distance BVP)

OR

2. (Add FVP) MINUS (Distance FVP) ?

I always thought it was #1 (since we are looking through the distance BVP, and I think we should be comparing THAT to the Add FVP), but the recent lecture I listened to seemed to suggest it was the front FVP that we subtract it from (but I have my doubts about the authority of the speaker). So which is it?

Thanks.

#### armstrong77

##### Full Member
10+ Year Member
I was wondering if anyone can point out what these bifocal styles look like:

-sovereign
-panoptik
-kryptok
-ultex
-ribbon
-perfection bifocal

Thanks.

#### armstrong77

##### Full Member
10+ Year Member
More on that lecture...

The speaker was talking about PALs and was talking about appropriate/inappropriate lens styles. e.g. Aviators are not appropriate.

The person said that lenses shouldn't be too big (i.e. neither A nor B boxing lengths should be too great) because it will lead the majority of the lens to have aberrations (i.e. the blending region will take up the majority of the lens, and therefore, not provide good optics).

I haven't heard of "big lenses" being a contraindication for PALs. Is this in any way true?

10+ Year Member
7+ Year Member
I have one here:

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#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
I was wondering if anyone knows, or if there's any easily-accessible literature on the power-diagram of a PAL.

Thanks.

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#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
One more.

Does anyone know the average axial length of the eye? e.g. Gullstrands, or whatever.

#### qwopty99

##### Optometrist
10+ Year Member
7+ Year Member
I was wondering if anyone can point out what these bifocal styles look like:

-sovereign
-panoptik
-kryptok
-ultex
-ribbon
-perfection bifocal

Thanks.

Anyone?

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#### Penguin2012

##### Full Member
10+ Year Member
The average axial length is 23mm I believe according to a study by Atchinson et. al in 2004 (eye shape in emmetropia and myopia).

In terms of models, I am not sure about Gullstrand, I know Elmsley is 22.22 mm.