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


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.

[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

[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 = F1 / (1 - t/n * F1) + F2

where F1 is the front surface power, F2 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 = F1 + F2 + t/n * F1^2

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

[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:

[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

[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

[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 = F1 + F2

Where F1 and F2 are the front and back surface powers.

And, since you know that F2 is equal to F1 since the lens is "equi"-convex, it becomes:

F = F1 + F1 = 2 * F1

Further, since you solved F earlier,

6.6666 = 2 * F1

Which, after rearranging, gives you,

F1 = 6.6666 / 2
F1 = +3.3333 D

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

r = (n - 1) / F1

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) / r1 + (1 - n) / r2

Where r1 and r2 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

[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:

[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 = F1 + F2

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 = F1 + F2

You also know that the front surface power F1 is equal to:

F1 = (n - 1) / r1

The same goes for the back surface power F2, except that you need to flip the (n - 1) quantity around as part of the "sign convention":

F2 = (1 - n) / r2

So, after substituting for F1 and F2 in our new Lensmaker's formula, we have:

1 / f = (n - 1) / r1 + (1 - n) / r2

Think of it like:

{ 1 / f } = { (n - 1) / r1 } + { (1 - n) / r2 }

{ F } = { F1 } + { F2 }

Best regards,
Darryl

[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

[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

[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

[rsandr]

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

[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:

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.

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

[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?

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?

[Darryl Meister]

Is there any formula for this?

That's just your basic Lensmaker's formula:

Power = Front Curve + Back Curve

Or,

F = F1 + F2

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 + F2

F2 = +4.00 - 7.00

F2 = -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 + F2

F2 = +2.00 - 7.00

F2 = -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

[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).

[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?

[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

[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

[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?

[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

[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?

[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

[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?