# Optical Forums > Ophthalmic Optics >  Question in some old exercíces

## cafn

Hi there

First of all, sorry if this is in wrong place, I just registed in this foruns.

I've been reviewing some of the opthics exercices, and the result I get doesnt match the result I wrote first time I did it.).

If you guys could help, i would thanks a lot!

- Calculate the refraction index of a thin lens, which measures are : F1 =  +5,25D and F2 = -1,75D, being the power obtained by neutralization of +6,00D. The Sag meter used were calibrated to CROWN glass and possued a sistematic error of -0,25D.

- A diver normally uses glasses with a potency of -5,00D with a previous surface of +2,00D. The lens material got a 1,523 refraction index. If we addapt these lens to a diving glasses, which would the potency inside the water? How could we modify the posterior surface so the diver would be well corrected inside the water?

and 

- Calculate the border thickness of a lens with potency -16.00D and a diameter of 60mm. The previous surface radius is 523mm and the center thickness is 0.6mm for the CROWN glass, with n = 1.523


Thanks for the help!

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## MakeOptics

*- Calculate the refraction index of a thin lens, which measures are : F1 =  +5,25D and F2 = -1,75D, being the power obtained by neutralization of +6,00D. The Sag meter used were calibrated to CROWN glass and possued a sistematic error of -0,25D.*

The simple way to work this problem out is to first find out the clocked power = +5.25 -1.75 = +3.50

Now used the clocked power, the material power, and the clocked index to determine the material index

6.00 = +3.50 * (Nmat - 1)/(1.523-1)

Nmat = 1.8966 (this is the index after doing the math)

*- A diver normally uses glasses with a potency of -5,00D with a previous surface of +2,00D. The lens material got a 1,523 refraction index. If we addapt these lens to a diving glasses, which would the potency inside the water? How could we modify the posterior surface so the diver would be well corrected inside the water?
*
Using a thin lens equation:

F1 + F2 = Power

We know in air:

+2.00 + F2 = -5.00
F2 = -7.00

The F1 power modified for water (n = 1.33):

(1.523 - 1)/2.00 = (1.523 - 1.33)/F1water
F1water * 2 = 0.523/0.193
F1water * 2 = 2.7098
F1water = 2.7098/2
F1water = +1.35

Since the back of the mask is still a glass air interface we don't reqquire any compensations:

F1water + F2 = Power
+1.35 + F2 = -5.00
F2 = -6.35


*- Calculate the border thickness of a lens with potency -16.00D and a diameter of 60mm. The previous surface radius is 523mm and the center thickness is 0.6mm for the CROWN glass, with n = 1.523
*
First convert the radius to diopters.

F1 = (1.523 - 1)/0.523 = +1.00

Now lets figure out the back curve power, first lets get rid of the thickness in the equation

F1(air equiv) = 1/(f1 - (t/n))
f1 = 1/F1
f1 = 1/1 = 1

F1(air equiv) = 1/(1 - (0.0006/1.523))
F1(air equiv) = 1/(1 - 0.000394)
F1(air equiv) = 1/0.9996
F1(air equiv) = 1.00 (in this case the thickness did not effect the power much but since it was supplied I considered it, air equiv is just the air equivalent of the front surface power at the back surface or in essence converting a front surface and thickness into one variable to use in a thin lens formula)

F2 = Power - F1(air equiv)
F2 = -16.00 - 1
F2 = -17.00

Now lets figure the sag of the surfaces, first convert the back power to radius

r2 = (1.523-1)/17.00
r2 = 30.8mm

Now the sags of the surfaces:

sagFront = 523 - sqrt(523^2 - (60/2)^2)
sagFront = 523 - sqrt(273,529 - 900)
sagFront = 523 - sqrt(272,629)
sagFront = 523 - 522.1
sagFront = 0.9mm

SagBack = 30.8 - sqrt(30.8^2 - (60/2)^2)
SagBack = 30.8 - sqrt(948.64 - 900)
SagBack = 30.8 - sqrt(48.64)
SagBack = 30.8 - 7.0
SagBack = 23.8

Now we have all the variables to figure out the edge thickness on a 60mm blank 0.6mm thick in a 1.523 -16.00 lens:

Edge = sagBack + thickness - sagFront
Edge = 23.8 + 0.6 - 0.9
Edge = 23.5mm

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## optical24/7

> *- Calculate the refraction index of a thin lens, which measures are : F1 =  +5,25D and F2 = -1,75D, being the power obtained by neutralization of +6,00D. The Sag meter used were calibrated to CROWN glass and possued a sistematic error of -0,25D.*
> 
> The simple way to work this problem out is to first find out the clocked power = +5.25 -1.75 = +3.50
> 
> Now used the clocked power, the material power, and the clocked index to determine the material index
> 
> 6.00 = +3.50 * (Nmat - 1)/(1.523-1)
> 
> Nmat = 1.8966 (this is the index after doing the math)


Interesting MO. I learned the approximation formula from Darryl that went;

N = .53 X (tp/cp) + 1

N = index approximation
tp= true power
cp = clocked power

Using it I come up with;

.53 X (6.00/3.50) + 1 = 1.90857

Now this is nit picking because both essentially come up with a 1.9 index, and I see the difference is in using .523 instead of .53 but which formula do you feel is more accurate across the most power spectrums? Are most lens clocks calibrated for .523?

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## MakeOptics

> Interesting MO. I learned the approximation formula from Darryl that went;
> 
> N = .53 X (tp/cp) + 1
> 
> N = index approximation
> tp= true power
> cp = clocked power
> 
> Using it I come up with;
> ...


Good to see you posting on fun threads.  Your absolutely right in using 1.53.  The only reason for not using it here was the phrase calibrated for CROWN glass. 

Another reason you can see from the posters use of commas instead of periods and vocabulary they are from an overseas program.  In the USA we are more likely to see problems with results that are neat and clean, I never assume that with programs outside the USA.

Thanks for the double check always good to have a master check my work.

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## optical24/7

Ah, gotcha. I forgot about your attention to details. I missed the calibrated for crown reference. Thanks!

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## MakeOptics

> Ah, gotcha. I forgot about your attention to details. I missed the calibrated for crown reference. Thanks!


Thanks, your math is razor sharp and your cognitive abilities lend a confidence to the question that i am sure the poster appreciates i know i do.

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