Effective Power

**LadiDiabla** · 05-13-2008, 04:58 PM

I'm using "Pure Optics", one on their test questions is:

What is the effective power of a pair of -9.00 D lenses prescribed at a refracted vertex distance of 21 mm if the lenses are worn at a vertex of 8 mm?

a) -8.68 D
b) -9.32 D
c) -9.75 D
d) -8.25 D

They say the correct answer is b..........

actually asked 3 Doctors and none agreed either, so is the book wrong or are we crazy?

Need answer fast as ABO's are this Sunday

Thanks

**Fezz** · 05-13-2008, 05:12 PM

E-None of the above!

Or it could be.......

:cheers::cheers::cheers:

**Fezz** · 05-13-2008, 05:19 PM

What formula are you using?

**Andrew Weiss** · 05-13-2008, 05:21 PM

Sorry -- I gave you the power you'd need to grind into the lens to get the patient that -9.00 at 8mm. :o

I'd go with (c).

**FullCircle** · 05-13-2008, 05:22 PM

I'd say b. It can't be a or d as bringing a minus lense closer to the eye increases the effective power (for plus, the opposite is true). This leaves us with b or c. C seems like it's too much additional power (.75 of a diopter seems a bit much)

but, I could be wrong.:bbg:

**Fezz** · 05-13-2008, 05:27 PM

A minus lens will will lose plus power as it moves toward the eye, which effectively makes it a stronger minus.

**LadiDiabla** · 05-13-2008, 05:28 PM

have tried both.D=D/1+(h x D) and DxD
------
1000

sorry don't know how to do formulas on here

also believe if decreasing vertex on a minus rx it makes the power stronger

**Fezz** · 05-13-2008, 05:48 PM

Originally Posted by LadiDiabla

sorry don't know how to do formulas on here

Optiboard loudspeaker:

"Harry Chilling to the office please, Harry Chiling to the office. Thank you"

Hopefully Harry will chime in with his magic formula writing skills and straighten us out!!!

**finefocus** · 05-13-2008, 06:30 PM

yes, decreasing vertex always adds minus (or decreases plus, same thing)
rule of thumb: moving a 10.00 diopter lens 5mm gives change of .50 diopter - a 9.00 lens moved this far must produce well over .50 rx change

**Fezz** · 05-13-2008, 06:40 PM

De=New effective power
Dl=original lens power
d=change in vertex distance in meters

De= Dl
1+dDl

I keep getting -9.117.

What am I doing wrong here?

I'm trying!

**FullCircle** · 05-13-2008, 06:44 PM

Dc=compensated Rx
D=original Rx
d=vertex distance change
Dc=D/(1-d*D)
Note: change in vertex distance is in meters and the sign is negative if moving away from the eye and positive if moving towards the eye

From harry's site.

**FullCircle** · 05-13-2008, 06:48 PM

I got -8.05

-9.00/1- (-.0013*-9.00) =

-9.00/1.117 =

-8.05

But, I think I'm wrong.

**Fezz** · 05-13-2008, 07:02 PM

Originally Posted by FullCircle

I got -8.05

-9.00/1- (-.0013*-9.00) =

-9.00/1.117 =

-8.05

But, I think I'm wrong.

Remember, moving closer, increases minus power.

-9.00/1+(0.013*-9.00)=-9.117 is what I keep getting!

**HarryChiling** · 05-13-2008, 07:04 PM

-9.00 changes 4mm closer

1/-9.00 = 0.111 = -111mm

-111mm + 4mm = -107mm

1/-0.107 = -9.35

The first alarm bell that should have rung in your head is who would refract at 21mm vertex? I think Phernell meant to say 12mm which would make it a change from 12 to 8 of 4mm. The formula for compensation requires that you subtract the vertex difference from the focal length the effective formula requires that you add the difference in vertex distance. When in doubt:

convert you power into focal length of the lens in mm
then if you want the effective power (power that the lens will now be) add the vertex distance, if you ant the compensated power (then subtract the difference)
then convert back to power.

Good luck on your test.

**Fezz** · 05-13-2008, 07:13 PM

Harry,

Where is the -0.98 coming from in your equation?

Also..

Assume for test taking, that Phernell really did mean 21mm as written, what do you get?

**rbaker** · 05-13-2008, 07:19 PM

Kinda seems like all high myopes are screwed.

**Diane** · 05-13-2008, 07:21 PM

Originally Posted by HarryChiling

-9.00 changes 4mm closer

1/-9.00 = 0.111 = -111mm

-111mm + 4mm = -107mm

1/-0.98 = -9.35

The first alarm bell that should have rung in your head is who would refract at 21mm vertex? I think Phernell meant to say 12mm which would make it a change from 12 to 8 of 4mm. The formula for compensation requires that you subtract the vertex difference from the focal length the effective formula requires that you add the difference in vertex distance. When in doubt:

convert you power into focal length of the lens in mm
then if you want the effective power (power that the lens will now be) add the vertex distance, if you ant the compensated power (then subtract the difference)
then convert back to power.

Good luck on your test.

After working the problem with the numbers you originally posted, I looked up the question in Phernell's book and the refracted vertex listed was 12 mm and worn vertex was 8 mm.

The simple formula works on this as well. It is...

Per mm = D2
1000

Where:
D = Dioptric power

Diane

**lensgrinder** · 05-13-2008, 07:23 PM

A lens gains plus power as it is moved away from the eye and looses plus power as it moves closer to the eye.
You can use the approximation since it is the ABO

$Power_{approx} = \frac{dD^2}{1000}\\ Power_{approx} = \frac{13 \times 81}{1000}\\ Power_{approx} = 1.05D$

The approximate effective power is -10.05D

Full Circle
You have the formula correct, but 0.013 M is positive since you are moving closer.
$D_{effective}= \frac{-9.00}{1 + (0.013 \times -9.00)}\\ D_{effective} = \frac{-9.00}{0.883}\\ D_{effective} = -10.19 D$

You could also convert the power to a focal length and add the vertex distance.
$\frac{1}{-9}\\ -0.111\ m \ or\ -111\ mm\\ -111 + 13 = -98\\ \frac{1}{-0.098}\\ -10.20$

**HarryChiling** · 05-13-2008, 07:23 PM

Originally Posted by Fezz

Harry,

Where is the -0.98 coming from in your equation?

Also..

Assume for test taking, that Phernell really did mean 21mm as written, what do you get?

That was me using the numbesr given assuming he meant 21mm then it would be a 13mm vertex change:

-111mm + 13mm = -98mm

1 / -0.098 = -10.20 which isn't an option, that's why I went back and tried it using 12mm as the vertex assuming that it was probably a typo and whalla the correct answer. Nice catch Fezz. ;)

**LadiDiabla** · 05-13-2008, 07:25 PM

I have the 2nd edition 2006 and it is 21mm...........maybe a typo? If so that is where the confusion is coming in

Finally I'm crazy your are cming up with the same answer as me..............Thanks for all the great help

**HarryChiling** · 05-13-2008, 07:26 PM

Originally Posted by Diane

After working the problem with the numbers you originally posted, I looked up the question in Phernell's book and the refracted vertex listed was 12 mm and worn vertex was 8 mm.

The simple formula works on this as well. It is...

Per mm = D2
1000

Where:
D = Dioptric power

Diane

Thanks Diane I had a feeling that the refracted vertex was 12mm but I don't actually have the book, wow check out all the great minds that this person got working on his problem. You better pass the ABO, looks who rooting for you.

**Fezz** · 05-13-2008, 07:33 PM

I'm getting a headache! Where is my Cabernet and Fiouricets?

:cheers::cheers::cheers::cheers:

**Fezz** · 05-13-2008, 07:47 PM

Oh, by the way....

Welcome to Optiboard LadiDiabla!

Please forgive us for not saying it to you before this!!

We get a little excited sometimes!!!

:cheers::cheers::cheers::cheers::cheers::cheers:

**FullCircle** · 05-13-2008, 07:50 PM

welcome to the board!

I'm just tickled that I was close to the correct answer via process of illimination.

**LadiDiabla** · 05-13-2008, 07:53 PM

Thank You!!!!!!! You've all been great. I've had the book for 6 weeks and the answer was driving me crazy. Just glad to know I wasn't wrong with the specs given.

35 yrs in the field and can't believe I have to take this test. I think it would be easier if I was a newbie.