# Optical Forums > Ophthalmic Optics >  Which formula to use?

## OdTech

Hello there i am having problems solving an Abo review question

I tried using the sagFormula and thick lens formula but still cant get the answer

What power lens is obtained if the
front surface power -  +16.00
back suface power -  -1.75
Center thickness - 8.0 mm
Diameter - 54 mm
Lens - CR-39 (n=1.50)

Answer is +15.62

Can't get the answer, please reply back
 :Confused: 

Also when doing Slab OFf/vertical imbalace problems, does lens style, Add power, and vertex has any importance ? for example:  patient is looking 5mm below, and problem shows vertex: 12 mm and FT25 lens style addpower +2.00.

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## Robert Martellaro

OdTech,

Try This...

http://www.optiboard.com/forums/show...&threadid=1783

Concerning slabs, add power and seg style are not a factor as long as they are the same in both eyes. 

In practical terms I would ignore vertex. I would also wait for a more definitive answer from one of the more knowledgeable opticians. 

Robert

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## Darryl Meister

> I tried using the sagFormula and thick lens formula but still cant get the answer


You are probably using the _exact_ thick lens formula while the question/answer is using the _approximate_ thick lens formula.

The _exact_ thick lens formula is:

_F_ = _F_1 / (1 - _t_/_n_ * _F_1) + _F_2

where _F_1 is the front surface power, _F_2 is the back surface power, _n_ is the refractive index, and _t_ is the center thickness in _meters_.

So, the exact answer would be:

_F_ = 16.00 / (1 - 0.008/1.500 * 16.00) + (-1.75)
_F_ = +15.74 D

The _approximate_ thick lens formula is:

_F_ = _F_1 + _F_2 + _t_/_n_ * _F_1^2

And the approximate answer would be:

_F_ = 16.00 + (-1.75) + 0.008/1.500 * 16.00^2
_F_ = +15.62 D

The exact formula is obviously more accurate, while the approximate formula is a little easier to use in some circumstances.




> Also when doing Slab OFf/vertical imbalace problems, does lens style, Add power, and vertex has any importance


If the add powers are the same for both eyes, these factors generally won't affect your vertical imbalance calculation for an ABO question, particularly if they have already provided you with a _reading level_ (or the distance below the optical centers that the wearer will use to read).

Best regards,
Darryl

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## OdTech

Wow Darryl Meister you real life saver.

Mr. Meister what about diameter? this what made me confused, that why i thought i needed sag formula.
By the way, the approximate thick lens formula
F = F1 + F2 + t/n * F1^2
do you divide by 2 or do you square F1.

Also is this is a british way since i 've got different but i guess similar.
I use it if the lens is 5mm more thicker

Dv=D1+D2+T/N * (D1)2     (D1) is squared.

T= meter
N= index of refraction
D1= Front surface
D2=Back surface
Dv or Dn= Nominal Power.

Thanx alot:cheers:

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## OdTech

Hello i've got another similarly challenging problem, need a little guidance.

A positive lens is to be made of glass having a refractive index of 1.55 and is required to have a focal length of 15cm. If the lems surfaces are equi-convex, the radii of curvature of the lens will be:

Ans: r1=+16.5cm
        r2=-16.5cm

Which formulas were used to get the answer?

My guess were #1 for sure was Lensmaker's Equation and maybe Surface Power Formula also Focal Length Formula.

Any aid is greatly appreciated

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## Darryl Meister

> Mr. Meister what about diameter?


You wouldn't need to know the diameter unless you were trying to calculate the center thickness, first (which they provided). Sometimes, part of solving these problems involves first determining what you do _not_ need to use in the formula. They may throw in extraneous information just to make you think or to even throw you off track, as in that slab-off question you asked.

Basically, if they provide you with a lot of details that you feel may be unnecessary, just ask yourself what you really need to solve the equation, and ignore the rest.




> do you divide by 2 or do you square F1.


Yes, the little carrot '^' symbol means "raise to the power of" or "exponent" in computer software and calculators and such. I didn't recall whether or not Steve's OptiBoard settings allow for superscripted numbers like exponents.




> Also is this is a british way since i 've got different but i guess similar.


Yeah, it's really the same formula once you replace my carrot '^' with an exponent. They're just using different letters to represent the variables for the front and back surfaces and lens power.

Best regards,
Darryl

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## Darryl Meister

> My guess were #1 for sure was Lensmaker's Equation and maybe Surface Power Formula also Focal Length Formula


If they're ignoring center thickness, which they apparently are since no details were provided concerning the thickness or diameter of the lens, you would be perfectly correct.

You can verify their answer:

First calculate the lens power using the Focal Power formula:

_F_ = 1 / _f_

Where _f_ is the focal length in _meters_.

_F_ = 1 / 0.15
_F_ = +6.6666 D (the 6 repeats)

Now that you have the focal power, you can solve for the surface power using the Lensmaker's formula:

_F_ = _F_1 + _F_2

Where _F_1 and _F_2 are the front and back surface powers.

And, since you know that _F_2 is equal to _F_1 since the lens is "equi"-convex, it becomes:

_F_ = _F_1 + _F_1 = 2 * _F_1

Further, since you solved _F_ earlier,

6.6666 = 2 * _F_1

Which, after rearranging, gives you,

_F_1 = 6.6666 / 2
_F_1 = +3.3333 D

Finally you can calculate the radius of curvature using the Surface Power formula:

_r_ = (_n_ - 1) / _F_1

Or,

_r_ = (1.55 - 1) / 3.3333
_r_ = 0.165 m (or 16.5 cm)

Or, if you're feeling especially brave, you can use a variation that combines all three formulas:

1 / f = (n - 1) / _r_1 + (1 - n) / _r_2

Where _r_1 and _r_2 are the front and back radii of curvture in _meters_.

Had you plugged your numbers into this formula to see if both sides of the equation were equal, you could have skipped your other intermediate steps.

Best regards,
Darryl

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## OdTech

Hello and thanx for the aid clarifying the question. By the way where did you get the "variations that combine all three formulas"

I used 'Optical Formulas Tutorial' to get the formulas and basically uderstand it, but there isn't any formula that shows this particualr combination.

:cheers:

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## Darryl Meister

> By the way where did you get the "variations that combine all three formulas"


It's just a variation of the basic Lensmaker's formula that you're likely to find in some books on optics. You can actually come up with it (or "derive" it), yourself, by just substituting the formulas you already know.

Start with your basic Lensmaker's formula:

_F_ = _F_1 + _F_2

Now, you know that _F_, or focal power, is also equal to:

_F_ = 1 / _f_

Which, after substituting for _F_ in the Lensmaker's formula, gives you,

1 / _f_ = _F_1 + _F_2

You also know that the front surface power _F_1 is equal to:

_F_1 = (_n_ - 1) / _r_1

The same goes for the back surface power _F_2, except that you need to flip the (_n_ - 1) quantity around as part of the "sign convention":

_F_2 = (1 - _n_) / _r_2

So, after substituting for _F_1 and _F_2 in our new Lensmaker's formula, we have:

1 / _f_ = (_n_ - 1) / _r_1 + (1 - _n_) / _r_2

Think of it like:

{ 1 / _f_ } = { (_n_ - 1) / _r_1 } + { (1 - _n_) / _r_2 }

{ F } = { _F_1 } + { _F_2 }

Best regards,
Darryl

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## rsandr

While we are on formulae....

s=(y^2)F/(2000n-1)

so
 s(2000n-1)=(y^2)F

so
s(2000n-1)/y^2=F

Is this correct as i am having some trouble with excel trying to find surface powers from sag

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## Darryl Meister

> Is this correct as i am having some trouble with excel trying to find surface powers from sag


You're pretty close:

_F_ = [2000 * (_n_ - 1) * _s_] / _y_^2

Where _y_ is the semi-diameter (one-half diameter) of the lens in _millimeters_ and _s_ is the sag of the surface. Note that you subtract 1 from _n_ before you multiply it by 2000.

Also note that this is based upon the _approximate_ sag formula.

So, given a semi-diameter of 25 mm, a sag of 2.0 mm, and a refractive index of 1.500, your answer would be:

_F_ = [2000 * (1.500 - 1) * 2.0] / 25^2
_F_ = [2000] / 625
_F_ = 3.20 D

Best regards,
Darryl

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## rsandr

Just as i was about to go mad.

I was using half my intended diameter in my formula instead of half the area i was sagging.

Now whats the formula for comping the front surface due to thickness

Fe=F+(F/1000t)
Is this correct, i forgot to bring my books home.

Cheers,
Rick

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## rsandr

Or is it
Fe=F/1-tF with t in metres

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## Darryl Meister

> Now whats the formula for comping the front surface due to thickness


Usually, you would compensate it for thickness, refractive index, and the initial front surface power. The formula would look something like this:

_F_c = _F_1 / (1 + _t_/_n_ * _F_1)

Where _F_c is the compensated front surface power, _t_ is the center thickness in _meters_, and _F_1 is the nominal -- or initial -- front surface power.

This formula, which is a lot like the back vertex power formula, simply subtracts the gain in plus power produced by the form (front surface and thickness) of the lens from the front surface.

Best regards,
Darryl

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## OdTech

Hello to everyone who helps answer the q's. I comprehend the question but i don't the answer or the way it was shown

If you were to use a material of 1.530 to make a lens with a final power of +4.00-2.00x180 Diopters sphere on a 7.00 Base. What would the ocular curve be?
Ans: 3.00x5.00

Why would anyone give answer in this form? :Confused: 

Through trial and error without any known formulas I figured the answer was gotten by

(Base)7minus 4(sphere@180*)=3
7+[-2](cyl @90*)=5

Is there any formula for this?

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## Darryl Meister

> Is there any formula for this?


That's just your basic Lensmaker's formula:

Power = Front Curve + Back Curve

Or,

_F_ = _F_1 + _F_2

A lens with cylinder power has a _toric_ surface, which varies in surface power from the flattest point (the _sphere_ curve or _back base_ curve) to the steepest point (the _cross_ curve) 90 degrees apart. The difference in between between these two back curves is the cylinder power.

If you have a 7.00 front curve (base curve) and need a +4.00 sphere power, the back sphere curve is:

+4.00 = 7.00 + _F_2

_F_2 = +4.00 - 7.00

_F_2 = -3.00 D

You can simply add the cylinder power (-2.00 D) to this to get the cross curve:

-3.00 + (-2.00) = -5.00 D

Similarly, you can apply the Lensmaker's formula to meridian of the lens that contains the cylinder power to get the cross curve. For instance, a +4.00 -2.00 x 180 lens has +4.00 D through the 180 meridian and a +4.00 + (-2.00) = +2.00 D through the 90 meridian. So, you would use Lensmaker's formula to determine the cross curve through this meridian with a +2.00 D power:

+2.00 = 7.00 + _F_2

_F_2 = +2.00 - 7.00

_F_2 = -5.00 D

Combined with your back sphere curve of -3.00 D that you determined earlier, you have:

-3.00 D / -5.00 D as the back sphere curve and cross curve

Best regards,
Darryl

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## Oha

Darryl,

You amaze me with your answers, but then again, you've mazed me for many years.

Could you take a couple of paragraphs and remind us where you got to learn all this, what education you've had, and how we can get to be as smart as Darryl too (well, almost as smart).

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## OdTech

Mr Darryl greatly appreciate your help.

I have a q that requires Resultant Prism Formula and Resolving Prism. I can't solve it since it 'combined prism' not like separate eyes and prism direction (ex: OD pl 3^BU & 5^BI)

Resultant Prism:
P^2=H^2+V^2
Tan a = V/H

Resolving Prim:
V=(Prism)(sin a)
H=(P)(cos a)

The Question:
A lens surfaced with 3^DBU and IN @45* combined with 4^DBU and OUT @45* What is the amount and direction of resulting prism?

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## Darryl Meister

> A lens surfaced with 3^DBU and IN @45* combined with 4^DBU and OUT @45* What is the amount and direction of resulting prism?


Well, if this is referring to one lens, the question is really flawed, since you can't have up and _in_ at 45 deg as well as up and _out_ at 45 deg. (Up and out in the right eye would have to be at 135 deg.) However, assuming that the question has just made an oversight in the necessary sign convetion, the easiest way to do this is to resolve both prisms into simple horizontal and vertical components. (You can also use something called the law of cosines from trigonometry for those familiar with it.)

Anyway, first resolve your first resultant prism into horizontal and vertical prisms using:

Vertical Prism = Resultant Prism * sin theta

Vertical Prism = 3.0 * sin 45 deg
Vertical Prism = 2.12 D Base Up

Horizontal Prism = Resultant Prism * cos theta

Horizontal Prism = 3.0 * cos 45 deg
Horizontal Prism = 2.12 D Base In

Note that since the initial resultant prism angle is 45 deg, the individual horizontal and vertical prism components are equal.

Then, resolve your second resultant prism into horizontal and vertical prisms:

Vertical Prism = 4.0 * sin 45 deg
Vertical Prism = 2.83 D Base Up

Horizontal Prism = 4.0 * cos 45 deg
Horizontal Prism = 2.83 D Base Out

Now, you need to add your prisms together.

The 2.12 D base in and 2.83 D base out prisms will cancel each other out to some extent: 2.83 - 2.12 = 0.71 D base out (since the base out prism is larger).

The 2.12 D base up and 2.83 D base up prisms will add together: 2.12 + 2.83 = 4.95 D base up.

Finally, you need to resolve these two prisms into another single, resultant prism and angle.

Resultant Prism = SQRT(Horizontal^2 + Vertical^2)

Resultant Prism = SQRT(0.71^2 + 4.95^2)
Resultant Prism = 5.0 D

Prism Angle = arctan (Vertical / Horizontal)

Prism Angle = arctan(4.95 / 0.71) = arctan(6.97)
Prism Angle = 81.8 deg

However, this is measured from the "out" side of the lens (between 90 and 180 deg), so you need to subtract this angle from 180 deg to get the actual angle: 180 - 81.8 = 98.2 deg. (This part of these problems can be a bit confusing without a picture of some sort.)

Anyway, let me know if this is one of your answers or not. ;)

If it makes any of you feel any better, you won't have to solve any problems like this on the ABO exam. ;)

Best regards,
Darryl

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## Darryl Meister

> Could you take a couple of paragraphs and remind us where you got to learn all this, what education you've had, and how we can get to be as smart as Darryl too (well, almost as smart).


Thanks, Oha. I have an AS degree in physics, but much of my optical and dispensing background is the result of over-preparing for the ABO's Master exam many years ago. While I got a great score on the exam, I was left knowing more than I ever really needed to about lenses and dispensing. ;) I also took the OAA's Refractometry course and other seminars and stuff, which I highly recommend. And, finally, a lot of my experience is due in no small part to my work in Technical Services and Technical Marketing at SOLA over the past seven and a half years, as well with various standards committees, like ANSI, and such.

But, you can spare yourself all that and just get a good book on dispensing and ophthalmic optics. A book like Brooks's _System for Ophthalmic Dispensing, 2nd Ed._ will teach you all you need to know about this stuff.

Best regards,
Darryl

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## OdTech

Darryl Thanx so much for aiding 
The answer will be 6.5 base UP and IN @74 degrees
By the way Are you kidding that i won't have this kind of q's on ABO. OPtimystically my Professors told me that ABO won't give q's that have toooo much math that too complex to answer.

By the way for q's that asks to find "Total Power at axis XX*"
I know I should transpose the Rx if its shown near or exactly 90 meridian but what if its shown in 180 axis still transpose?

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## Darryl Meister

> The answer will be 6.5 base UP and IN @74 degrees


Do you have the exact question handy? Based on the way you originally worded the question, the final prism couldn't be up and _in_, since the base out prism was larger than the base in prism.




> By the way Are you kidding that i won't have this kind of q's on ABO.


No. Generally, you'll just have simple addition/subtraction problems and maybe a simple question or two on Prentice's rule and vertex distance compensation. There is very little math beyond that level. But you should definitely understand the principles involved.




> By the way for q's that asks to find "Total Power at axis XX*" I know I should transpose the Rx if its shown near or exactly 90 meridian but what if its shown in 180 axis still transpose?


I don't know that I entirely understand your question, but it sounds like you're trying to calculate the amount of cylinder power present at a specified meridian of the lens. For something like this, you would use:

Power = Sphere Power + Cylinder * (sin theta)^2

Where theta is the angle between the axis of the prescription and the meridian of interest.

For instance, given a prescription of -2.00 -1.00 x 50, the power along the 0-180 meridian (50 - 0 = 50 deg) is:

Power = -2.00 + (-1.00) * (sin 50)^2
Power = -2.00 + (-0.59)
Power = -2.59 D

While you won't need to know the formula for the ABO, you should know that 30 deg from the axis of a cylinder you get 25% of the cylinder power, at 45 deg you get 50%, at 60 deg you get 75%, at 90 deg you get 100%, at 120 deg you get 75%, at 135 deg you get 50%, at 150 you get 25%, and -- back to the axis -- at 180 deg you get 0%. You should also be able to put a prescription on an _optical cross_, and to take it back off in either plus or minus cylinder form.

Best regards,
Darryl

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## OdTech

Hello Darryl, thanks for helping. still have a few q's

1)You are fitting a patient with a stock frame
DBL=24 A= 44 skull temples=5.5inch DistPD=62mm
If the Rx isn't decentered Q: How much prism(^) is induced(created)?

Rx OU -2.25-1.50x045
answer: 0.9^BI

2)An Uncut lens must be at least___mm in diameter to allow edging of a 54mm round lens decentered 3.0mmIN
ANSwer: 60

Book Formula: MLB=ED+(2)(Dec)+2  Answer come out 62mm
#1] My approx Formula: MLB=ED+(Dec x2) Answer come out 60mm
#2] Another approx Formua: MLB=(ED+Dec)x2= answer subtract original ED( In oval,round, eyesize is equal to ED sometimes)

An Uncut lens must be at least___mm in diameter to allow edging of a 48mm round lens decentered 4.0mmIN
ANSwer: 56
You can use#1,2 Formulas to get the answer.

MY Q is Which Formula is authentic enough to use for anything.?


3) The lensmeter reads:

Distance: -2.00+1.00x180
Intermediate: -.50+1.00x180
Near: +1.00+1.00x180

What is Near ADD? answer: +/-3.00

How do you get near add?

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## Darryl Meister

> 1)You are fitting a patient with a stock frame


First calculate the decentration. For a binocular PD measurement, the decentration Dec is given by:

Dec = (A + DBL - PD) / 2

Dec = (44 + 24 - 62) / 2
Dec = 3 mm per eye

Now, you will have to determine the approximate power through the horizontal (0 - 180) meridian of the lens. Since the cylinder axis is 45 deg, which is 45 deg away from the 0 - 180 meridian, we know that 50% of the cylinder is in effect. (See my previous post.) So you want to add 50% of the cylinder to the sphere power to determine the power through the horizontal meridian:

Power = -2.25 + 0.5 * -1.50
Power = -3.00 D

Next, calculate the prism using Prentice's rule.

Prism = Power * Dec / 10

Where Dec is the decentration in _millimeters_. We'll ignore the sign (+/-) of the power for now. So you have:

Prism = 3.00 * 3 / 10
Prism = 0.9 D per eye

Now, since the lenses were _not_ decentered, the optical centers are outside of the pupil centers. Which, for a _minus_ lens -- whose thickness increases away from the optical center, means that the wearer will experience base _in_ prism.




> 2)An Uncut lens must be at least___mm in diameter ... MY Q is Which Formula is authentic enough to use for anything


For a single vision lens, the minimum blank size MBS is given by:

MBS = 2 * Dec + ED

Use this formula for questions. Formulas with this and + 1 mm or + 2mm are simply adding an extra 1 or 2 mm for an edging allowance. While this makes good sense in practice, forget this allowance for tests unless the problem mentions it.




> 3) The lensmeter reads: ... How do you get near add?


Subtact the distance reading sphere power from the near reading sphere power:

Add = Near Sphere - Distance Sphere

Add  = +1.00 - (-2.00)
Add = +3.00 D

Note that when you subtract a negative number, you are basically adding it.

Best regards,
Darryl

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## OdTech

Darryl I think you are mistaken in #1 , So far you got the right Dec but the way you got the 

"Now, you will have to determine the approximate power through the horizontal (0 - 180) meridian of the lens. Since the cylinder axis is 45 deg, which is 45 deg away from the 0 - 180 meridian, we know that 50% of the cylinder is in effect. (See my previous post.) So you want to add 50% of the cylinder to the sphere power to determine the power through the horizontal meridian:

Power = -2.25 + (0.5 * -1.50)
Power = -3.00 D"

I fixed you formula with Parenthesis

Orig Rx not transposed!!!
OU -2.25-1.50x45
45*= 50% of cylinder
Now take the "Cyl" and multiply by "%" then the "answer" add to sphere, now you got the required answer.

-1.50 x .50=-0.75
-2.25 + (-0.75)= -3.00 D

Greatly appreciated the help

Also Does ABO use only 1995 ANSI STANDARDS?

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## Darryl Meister

> I fixed you formula with Parenthesis


The order of operations in mathematics requires that you multiply before adding, so the parentheses aren't necessary. ;)

Best regards,
Darryl

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## Darryl Meister

> Also Does ABO use only 1995 ANSI STANDARDS?


Unfortunately, I don't even know that the ABO could tell you for certain which revision of the standard they are using...

Best regards,
Darryl

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## rsandr

> Usually, you would compensate it for thickness, refractive index, and the initial front surface power. The formula would look something like this:
> 
> Fc = F1 / (1 + t/n * F1)
> 
> Where Fc is the compensated front surface power, t is the center thickness in meters, and F1 is the nominal -- or initial -- front surface power.


Hi darryl,
Am i correct that the above formula is the effective power of the front surface at the back surface?
If i am trying to determine necessary back surface powers and i am given the front surface could i use:-

Fc=F1/(1-t/n *F1) and then subtract the powers i need in each meridian from this


as i asssume the effective power increases with thickness.

Regards,
Rick

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## Darryl Meister

> Am i correct that the above formula is the effective power of the front surface at the back surface?


It's actually correcting for (or neutralizing) the effective power at the back surface. The effective power at the back surface is given by your other formula:

F_1_ = F_1_ / (1 - _t_/_n_ * F_1_)

So, given a lens with a +8.00 D front curve, 1.500 refractive index, and 6 mm (0.006 m) center thickness, your effective power at the back surface becomes:

F_1_ = 8.00 / (1 - 0.006/1.500 * 8)
F_1_ = +8.26 D

Note that the extra +0.26 D of plus power is due to the gain in power caused by the thickness and form of the lens. You would simply add the back surface power to the effective power of the front surface in order to determine the actual _back vertex power_ of the lens (that is, the power you would measure with a lensometer or focimeter).

So, if the lens has a back surface of -4.00 D, the back vertex power F_v_ becomes:

F_v_ = 8.26 + (-4.00)
F_v_ = +4.26 D

Now, the _compensated_ front surface power for that nominal +8.00 D front curve is:

F_1_ = F_1_ / (1 + _t_/_n_ * F_1_)

F_1_ = 8.00 / (1 + 0.006/1.500 * 8)
F_1_ = +7.75 D

This formula basically reduces the front surface power enough to make the actual effective power of the front surface equal to the original uncompensated -- or nominal -- power (i.e., +8.00 D) after that gain in power caused by the thickness and form of the lens.

What this basically means is that if you made a lens blank with a +7.75 D front curve, but called it a +8.00 D front curve, you could use a simpler formula (i.e., the Lensmaker's or thin lens formula) to determine the back curve instead of the more complicated back vertex power formula:

F_2_ = F_v_ - F_1comp_

So, say you needed to produce a +4.00 D sph power. You would just subtract the compensated front curve from this value instead of using a complex formula:

F_2_ = 4.00 - 8.00
F_2_ = -4.00 D

Now, assuming that your lens would have roughly a 6-mm center thickness, you can verify that your actual vertex power will equal the desired power with that -4.00 D back curve using the back vertex power formula:

F_v_ = F_1_ / (1 - _t_/_n_ * F_1_) + F_2_

F_v_ = 7.75 / (1 - 0.006/1.500 * 7.75) + (-4.00)
F_v_ = 8.00 + (-4.00)
F_v_ = +4.00 D

So, in essence, use of a "compensated" front curve was simply to reduce the math involved. By changing the front surface slightly, the manufacturer could save you the trouble of figuring this stuff by hand back before the days of computer software and cheap calculators. This used to be done in the past quite a bit to simplify surfacing calcuations, but it isn't really done anymore. Consequently, you will rarely, if ever, do this sort of thing in practice, but understanding the principle will certainly help you develop a more solid intuition in ophthalmic optics.

Best regards,
Darryl

[Had to edit a positive/negative sign typo in my last formula]

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## OdTech

Hello guys i can't help myself but notice one thing from my review questions and the one that were sent to me by one of the optiboarders for which THANKS SO MUCH!!!
Well to the point ALL or MOST of questions are in one book which i assume you all have as reference and troubleshooting guide
"SYSTEM OF OPHTHALMIC DISPENSING" 3rd edition, by BROOKS and BORISH.

After each chapter there are exercise questions and believe it these are exact questions i've got from my review questions
at least 3 to 4 questions per topic.

NOte: To all neophytes and novices taking the ABO or being employed, know the questions as a back of your hand and  you won't regret at all.
:drop: :idea: :cheers:

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## Darryl Meister

Yes, the vast majority of information you will need for the ABO exam is in _System for Ophthalmic Dispensing_. Many (if not most) dispensing programs at various schools of opticianry and optometry use this book, and every career optician should own a copy, in my opinion. The lastest edition I am aware of is the second edition, and it adds a great deal of information over the first edition. Butterworth-Heinemann publishes this book, but you could also order it from the OAA at one point under the title _Professional Dispensing for Opticianry_. I have yet to find a more accurate, well-written, or thorough source of information on ophthalmic dispensing.

Best regards,
Darryl

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## Judy Canty

Has anyone read fellow OptiBoarder Phernell Walker's newly published Pure Optics ?

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## OdTech

Hello Darryl 
My questions is Does astigmatism has an axis? for this particualr kind of example

The refractive anomaly indicated by the Rx +1.75-2.75x90 is

a. Compound hyperopic astigmatism
b. Compund myopic
c. Simple myopic 
d. Mixed 

The answer with out transposing is "D"
The way i came up with the answer is I know that mixed astg is in Rx "+" "-"  signs; but not according to the axis shown.

Do all of this a - d astgms have any axis?

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## Darryl Meister

> My questions is Does astigmatism has an axis


Yes. Astigmatism is essentially caused by a surface with a curvature in one direction (or meridian) that differs from the curvature in another direction (or meridian). For instance, a _basketball_ would be _spherical_ in shape, since the curvature is the same in every direction. An eye with a "spherical" (at least centrally) cornea would not have astigmatism.

A _football_, on the other hand, is an example of an _astigmatic_ (or _toric_) surface. It is flatter along its length (one meridian) and steeper/rounder along its diameter (the other meridian). A cornea with this kind of shape would have astigmatism, which would be equal to the difference in curvature between these two meridians.

Now, since a surface with astigmatism has meridians that vary in power, rotating a lens with astigmatism will rotate these different meridians. This means that lenses or surfaces with astigmatism have an orientation. Consequently you need to specify an _axis_ (or angle) in order to describe how far the lens has been -- or needs to be -- rotated. With spectacle prescriptions, the axis corresponds to the _sphere_ meridian of the lens.

If you rotate a spherical lens, on the other hand, it makes no difference since all of the meridians have the same power. Consequently, a spherical lens does not really have an orientation (at least until you start adding prism and such).

Back to the football-basketball analogy... A basketball will still look like it is in the same position after you spin (or rotate) it, while a football lying on its side will obviously look like it is in a different position after you spin it, because it has an orientation.

As far as your problem goes, you really need to understand optical crosses in order to answer this one. The axis of the prescription has nothing to do with the powers or signs of the meridians, just their orientation.

With this prescription,

+1.75 -2.75 (the axis is unimportant)

One meridian (the sphere meridian) is equal to +1.75 (_hyperopic_), while the other meridian (the cylinder meridian) is equal to +1.75 + (-2.75) = -1.00 (_myopic_). Since one meridian is positive and the other negative, it indicates _mixed_ astigmatism.

Had both meridians been positive, you would have compound hyperopic astigmatism, and vice versa for two negative meridians. If one meridian is positive and the other plano (such as +1.00 -1.00, which is also Pl +1.00), you would have simple hyperopic astigmatism, and vice versa for negative and plano meridians.

Best regards,
Darryl

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## OdTech

Hello i wonder does moderate to high amounts of cylinder could present such symptoms as dizziness and headache, blurry vision.
Sphere isn't strong but the cyl is

Also when the customer comes in with rx
requesting to wear a bifocal lens
-3 sph
-1 sph
addd +3.75
St25 lens

Doesn the optician tell the customer
A. The benefits and how to adapt to the lens
B. Tell the customer about the posssibility of vertical imbalance.

Additionally

An individual plays Piano wears bifocal lens but only thing is wrong is reading music notes has a little problems

Do you give

A. Intermediate single vision
B. Raise the bifocal
C. Distance single vision
D. Near Single Vision

My theoretical approach woulld say Raise bifocal for sure since he comfortable playing the piano looking through the segs if i raise the segs he will be able to see the music notes.

Please correctl me if am wrong

Cosnequently a customer comes in and tell you " Distance objects are large  and very round than near"

I consider the problesm as The Dr gave lots of cylinder.(my guess)

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## chip anderson

Sean:  

Been there done that with a lot of piano/organ players.  The "best solution" get patient to sit at his piano, as though he were playing same.  Have someone else measure the distance from the eye to the music.   Make glasses (probably SV) but possiblly a bifocal, that focus exactly that length.    

Chip

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## OdTech

Hello Does FDA states certain amount of years a practitioner should keep the patients Rx let say

2, 3,7 years? :Confused:

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## Darryl Meister

I'd have to double check to be certain, but I leave you are required to keep medical records for seven years by the FDA.

Best regards,
Darryl

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## OdTech

Interesting, are you implying that Rx is considered as a medical document of some sort, although its just some numbers and MD reccomendations. :Rolleyes:

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## Darryl Meister

> Interesting, are you implying that Rx is considered as a medical document of some sort, although its just some numbers and MD reccomendations


Yes, the Rx is a medical document. But it's not really a copy of the Rx that's important, but rather the examination record, itself. The prescription is just one component it. Seven years came to mind, but I am not certain of that figure, and it could very well depend upon individual state requirements, too.

Also, on a related note, the FDA requires that records be kept by the seller documenting the impact resistance of prescription eyewear for three years, as well.

Best regards,
Darryl

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## Shwing

I may be wrong (I've been know to be wrong) but up here (on most maps we're 'up') the record must be kept for 7 years.  I also believe this is modelled after the US.

Reason?  Taxes, of all things.  You must be able to produce documentation related to your taxes to Revenue Canada/ IRS in case of audit, going back 7 years.  Therefore, as many people will write off their specs, you would need to keep the receipts.  And of course it goes the other way, in that the prescriber would need to maintain the records for the very same reason.

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