# Optical Forums > Ophthalmic Optics >  Questions while studying for optometry boards

## Alvaro Cordova

Hey Darryl and optics junkies,
These questions have been posed to me and I'm stumped.

Q. The power of a lens in air is +5.00D.  What is the power of the lens in water (n=1.33).

The answer suppsosedly is +3.75
+5.00/1.33 = 3.75 ???  

My question is how would one arrive at this answer?  Isn't the lens power in water going to be dependent on the lens material?

2nd question is

Q. Two +8.00 Thin lenses are seperated by 10cm, what is the power of the equivalent power of the lens.

The answer they have is 9.6cm and have no idea how they get that.

I get 8.27D

Thanks ahead of time

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## HarryChiling

When a lens is in a material other than air the equation is

D=n/f

D= Power
n=index of surrounding material
f=focal length

so if we take the +5.00 lens in water

f=1.33/5 (focal length in water)

f=0.266 (focal length in water)

now we take the focal length and find the equivalent power in air

D=1/0.266 

D=3.75

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## Alvaro Cordova

> When a lens is in a material other than air the equation is
> 
> D=n/f
> 
> D= Power
> n=index of surrounding material
> f=focal length
> 
> so if we take the +5.00 lens in water
> ...



Thanks,  I completely forgot about this.

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## HarryChiling

The second question The power of the equivalent lens is found with the following equation

De=D1+D2-D1*D2*d

De=8+8-8*8*.1

De=16-6.4

De=9.6

I thiink what may be confusing you is the cm in the answer it is actually a measure in Diopters

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## Alvaro Cordova

> The second question The power of the equivalent lens is found with the following equation
> 
> De=D1+D2-D1*D2*d
> 
> De=8+8-8*8*.1
> 
> De=16-6.4
> 
> De=9.6
> ...



I understand what you are doing, but I'm not sure I've ever seen that equation before.  I got 8.27 using vergence.  

Is the equation you used De = D1 + D2 - (D1 * D2 (thickness in meters) / Index)

In my question above the thickness is 10 cm and the Index is 1.  Is that the right track?  

Well Harry, looks like I have to thank you yet again.  :bbg: You have been most helpful.  Have you started programming that Keating thing yet for faceform and panto?

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## HarryChiling

I got that equation from a Contact Lens Manual by CLSA and you are correct the last part  is d/n. I have not started to program the keating equation yet, but I will very soon.  I will post it when I do.

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## Alvaro Cordova

> I got that equation from a Contact Lens Manual by CLSA and you are correct the last part  is d/n. I have not started to program the keating equation yet, but I will very soon.  I will post it when I do.



Cool, what langauge are you writing it in?  I usually write everything in Java. I hate using javascript when it comes to highly mathematical projects.

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## HarryChiling

I made it in Javascript because the source is open and it is an easy language to learn and to tinker with so anyone can play with it.  Another advantage is anyone can add it to their own personal website.  I just finnished the keating equation and have added it to the zip file with all other equations that I have written scripts for. Enjoy

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## Darryl Meister

> Q. The power of a lens in air is +5.00D. What is the power of the lens in water (n=1.33)... The answer suppsosedly is +3.75


Actually, unless you've copied the question incorrectly, the _power_ of the lens in water should be 1.67 D (or None of the Above, apparently).

If you _have_ copied the question correctly, I would say that they miscalculated. You can multiply the power of the lens by an "index correction factor" of sorts, (1.500 - 1.333) / (1.500 - 1.000), in order to determine the power of the lens in water (this applies to a submerged lens, with water on both sides).

You can verify this for yourself by substituting (1.500 - 1.333) for (1.500 - 1.000) in any formula for the power of the lens, such as the lensmaker's equation:

*F = (n' - n) * (1 / r1 - 1 / r2) = (1.500 - 1.333) * (1 / r1 - 1 / r2)*

Note that the quantity (1 / r1 - 1 / r2) remains constant for the lens. Essentially, you are neutralizing the vast majority of the refracting power of the surfaces by submerging the lens in a medium whose index is approaching the index of the actual lens material.




> When a lens is in a material other than air the equation is... D=n/f


But the focal length f changes when you place the lens in water, for the reasons described above. You would find the new f by dividing n (1.333) by 1.67 D.




> I made it in Javascript because the source is open and it is an easy language to learn and to tinker with so anyone can play with it.


But keep in mind that JavaScript is both the least secure solution and the solution least likely to work correctly in all browsers. Since JavaScript is also routinely disabled for these reasons, I generally recommend limiting its use to unimportant Dynamic HTML elements (drop down menus, roll-overs, etcetera). For simple lens calculations on a basic web page, it probably doesn't matter much, but a server-side language like Perl or JSP or PHP is really a better alternative if you're interested in robustness.

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## Alvaro Cordova

That's exactly what my gut was telling me.  We've done problems similar to this and remember it couldn't be so simplistic as that  (In the water goggles thread).  I did copy the question correctly and remember that I did get around a 1.6ish on the diopter assuming it was CR-39.  Thanks for weighing in.  The questions are from students who took the optometry board exam.  Obviously, the questions are a little off in wording and answers etc...  

I second that notion about javascript.  Javascript is a browser dependent language.  PHP and JSP are server-side and generally the safer way to go.  I personally go the Java route because I like object oriented programming.  Java is *great* for that.  Just to let you know, the transition from javascript to php should be easy since the langauges are weakly typed langauges.  (ie variables do not have to be declared in terms of a datatype)

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## Darryl Meister

> the transition from javascript to php should be easy since the langauges are weakly typed langauges.


If you've programmed in C for a while, you get into the habit of declaring and initializing everything, anyway. ;)

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## HarryChiling

The index correction factor you talk about introduces a variable that you are assuming.  The 1.50 index and the r1, what if the lens material is higher in index than the answer to the equation you gave.  This would change the assumed index would change the radius of the lens as well.  What we know is that no matter what medium the lens is in the radius of curvature of the lens never changes.  That equation will not work.  I am assuming that the answers you gave are right and you asked how would I come up with that.

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## HarryChiling

I am not a programmer, I like javascript and it is easy to duplicate and tinker with (I know some people like to consider their code as an asset, mine is free for knowledge).  I understand that it's implementation is not consistent across browsers but the 2 major ones seem to do fine.  I do not like server side scripts, I prefer that the host machine use its resources to compute than bogging down my server.  Look at google maps they use ajax wich is an implementation of javascript and xml.  If you are going to serve it over the web javascript is better, just my opinion.

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## Darryl Meister

> The index correction factor you talk about introduces a variable that you are assuming. The 1.50 index and the r1, what if the lens material is higher in index than the answer to the equation you gave. This would change the assumed index would change the radius of the lens as well. What we know is that no matter what medium the lens is in the radius of curvature of the lens never changes. That equation will not work.


Really, the equation will work, regardless of the lens. But, Yes, as you point out the refractive index variable would have to be adjusted for anything other than n' = 1.500. However, you'll find that not only is this a reasonable assumption, but that it's also a common assumption in these problems, as well as optical approximations, since it simplifies the math a bit. Still, to your point, it _was_ an _assumption_ on my part. I suspect that the person who wrote the original question didn't fully understand the principle involved, perhaps using the same equation you posted earlier, which is why he or she didn't specify a refractive index for the material.

If you are looking strictly at the _vergence_ of a wavefront, or the _reduced thickness_ of a plate with a different index of refraction, the expression l / n would work though. For instance, you could ask something like what the _reduced vergence_ of a wavefront would be in water if it measures 5.00 D in air (though I can't see much of a practical use for this).

Also, it doesn't really matter what R1 and R2 are, since they are constant between air and water, and would therefore factor out. For that matter, since no center thickness was provided, which by necessity assumes a "thin" lens, the quantity (1 / r1 - 1 / r2) would remain unchanged for a given refractive index and lens power, regardless of the individual surface radii.

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## Darryl Meister

> I am not a programmer, I like javascript and it is easy to duplicate and tinker with (I know some people like to consider their code as an asset, mine is free for knowledge).


If you are only reproducing basic optical formulas in JavaScript, I don't know that anyone would bother protecting his or her code, even if this were a possibility with JavaScript in the first place.




> I understand that it's implementation is not consistent across browsers but the 2 major ones seem to do fine.


The _newer_ versions of the two major browsers generally have consistent performance. Earlier versions differed wildly in how they handled layers, events, objects, and so on. You had to worry about the browser, the browser version, the operating system, etcetera.

JavaScript programmers would have to add a lot of extra code to get cross-browser consistency for many DHTML applications. (Though simple math functions really haven't been an issue as far as I know.)




> I do not like server side scripts, I prefer that the host machine use its resources to compute than bogging down my server.


I think you would be more concerned with bandwidth nowadays, since that's what you generally _pay_ for (unless you own your own servers), and server-side scripts cut down on bandwidth usage. Besides, the code for something like a simple sag function is probably inconsequential for server-side processing. Of course, if you aren't using PHP or ASP or the like for anything else on your site, it might be a waste to have your server processing a bunch of pages for non-existent scripting tags.




> If you are going to serve it over the web javascript is better, just my opinion.


For this stuff, I agree that it probably doesn't matter either way. But if your page _needs_ scripts to do something important, most seem to agree that JavaScript -- which is often disabled -- is not the best solution. Moreover, as far as I know, _fewer and fewer_ sites are actually using JavaScript as a solution for anything more than basic stuff. Server-side lanugages like PHP, on the other hand, have increased dramatically in usage -- and are virtually a necessity for sites that interact with a database, process forms, and so on.

I originally used JavaScript for all of my optical calculators, and I'm glad that I switched to PHP. But if JavaScript is your web tool of choice, there's certainly nothing wrong with that.

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## HarryChiling

So then if we were to assume that your equation would be correct the 

D=(n-n')/r
5=(1.50-1)/r
r=.5/5
r=.1

then we would run it back through the equation with the known radius

D=(n-n')/r
D=(1.5-1.33)/r
D=.17/.1
D=1.7 Diopters

Lets say that the same 5 Diopter lens was made from a 1.74 index material

D=(n-n')/r
5=(1.74-1)/r
r=.74/5
r=.148

D=(1.74-1.33)/.148
D=.41/.148
D=2.77 Diopters

So if we were to use the index of the lens as a factor we are inroducing a variable that we don't know.  I had though of this and actually tried to work the true power formula to get the answer wich by your theory would work as well

Dwater/Dair=n-1.33/n-1

but no matter how I arranged the formula I would have to know the answer to the equation to find the index wich is the variable I would need to find the answer to the equation. (major chicken, egg thing)

After thinking for a while I came up with the focal length equation.  It was the first question asked but the second question I answered.  I thought of replying that the question was missing something.  But since I was given the answer I though how would I work the equation to come up with that answer.  If I was to truly do it the way that I think would be logical I would come up with yet still another answer.

D=n/f
5=1/f
f=1/5
f=.2

Then put it back through the eqaution changing the index.

D=1.33/.2
D=6.65 Diopters

I know that is wrong for obvious reasons.  I would still disagree with your equation, but it wouldn't be the first time I was wrong.

I hate server side languages.  It is a pain in the butt to test the code and it does have its own security issues, leaving you more vurnerable to dos attacks and other issues that come from poor implementation of code.   With javascript if it has any vulnerabilitys it affects the client not my server.  In my previous lab the computer that was networked with the generator did not have internet access for security and stability reasons.  Most labs that I have had the pleasure of checking out do not have internet access on the computer that communicates with the generator.  Javascript in this case can be imported on my thumb drive and used at the station closest to the equipment that I would need the equations at.  Javascript can also be used inside of flash files to create a more graphical interface for the calculators.  

Oh and browsers are free, you should not be using an older browser at this point and if you are UPGRADE.  I would even be willing to burn a few different options on a CD and mail them to whoever needs it.

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## Darryl Meister

> So if we were to use the index of the lens as a factor we are inroducing a variable that we don't know.


Please see my earlier post:

_But, Yes, as you point out the refractive index variable would have to be adjusted for anything other than n' = 1.500. However, you'll find that not only is this a reasonable assumption, but that it's also a common assumption in these problems, as well as optical approximations, since it simplifies the math a bit. Still, to your point, it was an assumption on my part._




> I would still disagree with your equation, but it wouldn't be the first time I was wrong.


Harry, "my equation" (known as the Lensmaker's formula) is a fundamental principle of optical physics, and has been for hundreds of years. ;)




> I know that is wrong for obvious reasons. I would still disagree with your equation, but it wouldn't be the first time I was wrong.


You know _your_ equation is wrong for obvious reasons, but don't agree with _my_ equation, which should be right for obvious reasons? ;) After all, the refractive power of the surface is due to the _difference_ in refractive index between the lens material and the surrounding medium.




> Then put it back through the eqaution changing the index.


But, again, the focal length f also changes once you place the lens in water. You can't solve for D in your equation without first calculating the new power of the lens.




> I hate server side languages.


Virtually any large-scale site runs a server-side language, including _this_ site. Just about any site with a database or a content management system uses some sort of server-side language, from Amazon.com to the smallest blog site.




> It is a pain in the butt to test the code and it does have its own security issues, leaving you more vurnerable to dos attacks and other issues that come from poor implementation of code. With javascript if it has any vulnerabilitys it affects the client not my server.


Yes, JavaScript exposes the _user's_ computer to security issues. A server-side language primarily exposes your web server's computer to these issues. Which do you think your _visitors_ are more worried about? And, again, are your web sites sitting on your own servers or someone else's?




> Oh and browsers are free, you should not be using an older browser at this point and if you are UPGRADE. I would even be willing to burn a few different options on a CD and mail them to whoever needs it.


That's your opinion, Harry, not necessarily the opinion of every one of the millions of Internet users out there. One of the principal rules of web design has always been to design your site to be compatible with the largest audience possible (hence, phrases such as "web safe color palettes" and "cross-browser scripts"). Besides, you are not going to personally mail out hundreds of thousands of CDs to users who haven't bothered to upgrade their browser version or whose machines can't necessarily support the latest and greatest software. ;)

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## Darryl Meister

> I will say one thing the book is very in depth and does make for a great read. I received it yesterday and have 3-4 chapters left. That is probably about 400 pages read and understood in one day. I would definately rate it as one of the best books will be adding to the library


By the way, I also took the liberty of checking Keating's book for you, since I know you recently purchased a copy. I found his version of my "index correction factor" equation on page 130, if you want to have a look at it.

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## HarryChiling

Page 30 in keatings book gives a great example , but what I keep trying to say is that in the original question we were not told the index of the material the lens is made of.

I am not argueing that the lens makers equation is false, I would hope you don't think of me as that stupid.  I am just argueing that how would we apply it to this problem with 2 unknows.  I have read through keatings book and if you look at the equation on the same page 130 example 7.34 it gives you the very same equation that I stated in one of the previous posts.




> Dwater/Dair=n-1.33/n-1


Also if you were to look at those examples in every case you are given the index of the material.  If you were to use this equation or the lensmakers equation it would not matter we still don't know for certain what the material of the lens is.  The problem does not give enough parameters to determine an answer.

Original Problem



> Q. The power of a lens in air is +5.00D.  What is the power of the lens in water (n=1.33).
> 
>  The answer suppsosedly is +3.75
>  +5.00/1.33 = 3.75 ???  
> 
> My question is how would one arrive at this answer? Isn't the lens power in water going to be dependent on the lens material?


As for the second part of the question. Yes, the power of the lens in the water is directly related to the index of the material.  If the lens were made of diamond I would be close to right (3.83 Diopters)

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## HarryChiling

By the way Darryl thanks for the book and I think we might have to agree to disagree on this equation, you are still the smartest guy I know. :cheers:

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## Darryl Meister

> but what I keep trying to say is that in the original question we were not told the index of the material the lens is made of... I am just argueing that how would we apply it to this problem with 2 unknows.


Yes, unless you _assume_ a refractive index, you cannot answer this question. As I said, I don't believe the person who wrote this question fully understood the principle involved. Or perhaps he or she inadvertently miscommunicated the intent.




> it gives you the very same equation that I stated in one of the previous posts. _Dwater/Dair=n-1.33/n-1_


Actually, I first posted that expression in my initial response to Al (with the index of 1.500). So, you have that much more reason to agree with me now. ;) If you have (or at least assume) a refractive index for the lens material, this is the only equation you need. But, again, without a refractive index, it cannot be answered. In any event, even if a refractive index had been provided, the answer almost certainly wouldn't have been correct, given the circumstances.




> you are still the smartest guy I know


Quit. You're going to make me blush over here. :o

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## QDO1

> Yes, unless you _assume_ a refractive index, you cannot answer this question. As I said, I don't believe the person who wrote this question fully understood the principle involved. Or perhaps he or she inadvertently miscommunicated the intent.
> 
> 
> 
> Actually, I first posted that expression in my initial response to Al (with the index of 1.500). So, you have that much more reason to agree with me now. ;) If you have (or at least assume) a refractive index for the lens material, this is the only equation you need. But, again, without a refractive index, it cannot be answered. In any event, even if a refractive index had been provided, the answer almost certainly wouldn't have been correct, given the circumstances.
> 
> 
> 
> Quit. You're going to make me blush over here. :o


 Ive looked at the maths, and the question can not be solvedproperly with out the index of the lens.  Ive looked in terms of vergances, the lensmakers formula, etc.. Furthermore, if we assume a thick lens we cant work it out without a thickness either.  On the assumption of thin lenses, we only need the index

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## feel_love217

hiii dude!!
ur question is been already answered by a genious but i still want to learn or share that we cannot work out the power of the lens in water if we dont know the actaul refrective index of lens in air.. frm my point of view we got to have a radius of curvature of the lens in air to work out wht it will be like in water, but the way u guyz do it is bit diff then i hv learnt. this is my first time to c this site and i m absolutely thrilled to see that such kind of society exit and is just by fluke i came across but this is excellent i will be on net all the time. thanx for taking time to read..

Amit..

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## Darryl Meister

Welcome to the 'Board. ;)

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## feel_love217

thanx darryl, i m delighted to c something usefull today on internet.

i absolutely love optics as u guyz do

cherrs!!!!

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## feel_love217

hello harry!!

 i hv read all ur thoughts regarding working of power in diff medium and i hv given my thought regarding that. sorry its bit late to talk on that but this is a first time i hv know sumthing like this kinda society exists.
i just want to share thoughts and whts ur approach regarding this..

regards
Amit..

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## feel_love217

hello everyone!!! :Cool:  

just want to share my thought abt working on power of lens in diff medium and
pls correct me if im wrong as i hvent read any book b4 riting this..

1) radius of curvature stays the same when it is in a diffrent medium. well this is not a assumption but it does stays the same.

in order to work out the the power of lens in water we need to know the radius of curvature in air by the formula:--r=(n'-n/F)
a) r=radius of curvature of lens assumed thin but still can be worked for thick lens with centre thickness, in that case we will work out r1 and r2
b)n'=the refractive index on the lens
c)n=the refrective index of surroundings or medium
d)F=the power of the lens

2)if we know the radius of curvature of any given lens than we can simply drop the lens inside the water, the radius of curvature will not change as radius of curvature is solid and even refractive index of lens will not change(unless we are tired and break the lens or water leak inside the lens)

3) when the lens is droped inside the water, the ray of light will not behave in same fashion as it was doing in air, the reason been water will change its path.

4)as the refractive index of air is 1 which is becoz its ability to change the velocity of light and its properties such as slight bending of light.

5)due to the refractive index of air which is 1.333 and is higher than the air it will bend more light thus light entring more denser medium hv diff focal properties

6) as at intial stage when we wer in air and lens was bringing all the light to a point focus by its ability(pls assume the lens is a best form lens) and that distance frm the lens we called a focal lenght and describe it as dioptre(the actual ability of the lens)

7)as lens are designed to bring rays to point focus this is same lens will try to do the same regardless of any medium and now as ray of light will travel in water it will hit the first surface of lens where the light will find the same curvature but as it will leave the lens it will find a diff medium ie water becoz it will bend more and thus the focal lenght will change and hence the power will change automatically becoz now same lens hv diff ability to bring light into a point focus

8)the way we can find this is by same formula but this time F=(n'-n)/r

where n' will be water ie 1.33 and n will be the refractive index of the lens.

that's what i think abt this quiz!!

thanx for taking out time to read this

pls share ur views

Amit jain (F.B.D.O, smc tech);)

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## mjacob

I just started working at an Optical as a tech.  I'm really curious onthe procedures to pre-test a patient? Will anyone help me??

Thanks,

Mary Jacob

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## HarryChiling

If you can be more specific I can probably help.

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