surface power of thick lenses....help!!!

[Scott Carpenter]

I am now two weeks before finals!! I am working on previous exam papers and
wonder if you would mind looking at a question for me?

It is "A meniscus lens of back vertex power +12.00D has a nominal
magnification of 2.75 times. The axial thickness of the lens is 9mm and the
refractive index 1.50. Calculate the surface powers of the lens.".

Can you possibly help? It is probably fairly obvious how to do it and I
just can't see it; I think it is to do with the difference between nominal
and surface powers. However without having one surface power I am having
problems. If you can shed any light on it I'd be most grateful at this
stressful time!

thanks in advance, Scott

[Darryl Meister]

Hi Scott & Gillian,

Good to hear from you guys, again. Thanks for posting this one. The ABDO courses always have challenging questions. I've run through the solution for you, below.

Knowns:
Nominal Magnfication (M) = 2.75x
Back Vertex Power (FV) = +12.00 D
Thickness (t) = 9 mm (0.009 m)
Refractive Index (n) = 1.500

Unknown:
Front Curve (F1) = ?

Reasoning:
The "nominal magnification" of a lens is given by its "equivalent power" divided by 4. (I can post a further explanation of why this is so, should anyone be interested.) By the way, you guys are lucky that I have a British reference book on Optics... We don't call that formula "nominal magnfication" over here, so I couldn't find it at first! Now, the "equivalent power" is the important part here. We know the "back vertex power" of the lens, already. Using our formula for nominal magnification (M), the equivalent power (FE) would be:

FE = 4 * M = 4 * 2.75
FE = +11.00 D

We can now add FE = +11.00 D to our list of unknowns. The problem now is to determine what the front curve (F1) of the lens is, given our known values. It just so happens that the back vertex power (FV) of a lens is related to its equivalent power (FE) by:

FV = FE / (1 - t/n * F1)

It should be readily apparent that we have values for all of these variables, except for F1 -- which is our unknown. Rearranging the equation to solve for F1 gives us:

F1 = (FE - FV) / (-t/n * FV)
F1 = (11.00 - 12.00) / (-0.009 / 1.500 * 12.00)
F1 = -1 / -0.072
F1 = +13.89 D

Consequently, the front curve of the lens should be +13.89 D. In addition, the back curve will have to be -3.15 D to produce the desired back vertex power. Let me know if any of your possible choices are close to this.

Best regards,
Darryl

[This message has been edited by Darryl Meister (edited 06-09-2000).]

[Scott Carpenter]

Thank you very much, Darryl. We are now going to place our combined 2 brain cells together and digest the explanation above.
Scott & Gillian ( UK )

[Note: I removed the re-quote of Darryl's complete message to save space on the server. - Steve]

[This message has been edited by Steve Machol (edited 06-09-2000).]

[Darryl Meister]

No problem. As always, I'm glad to help. Would you guys mind posting some of the other challenging questions that you've run across from these exams? Perhaps people who have recently taken the ABO and/or Advanced ABO exams can gleen a better appreciation for the kind of demands placed upon opticians in other countries...

Best regards,
Darryl