Will you actually have to calculate power in any meridian for your ABO exam? Absolutely! It’s almost guaranteed.  The properties of sphere and cylinder power is how optical lenses work, and it’s fundamental for understanding your job. But what sort of questions will these be?  What can you expect?

First of all, there are always the easy ones…

Basic toric transposition

All tests have some super easy questions.  After all, everyone has to get at least ONE answer right!  So there are always “give away” questions.  I estimate it’s usually about 10%.  Out of 100 questions, 10 will extremely easy.  You might see this obvious question concerning cylinder and total power calculation:

If the Rx is: -1.00 -2.00 x 180, what is the total power on the 180° meridian?

The answer, of course, is -1.00

Then they might get a tiny bit more daring and make you work a little bit:

If the Rx is: -1.00 -2.00 x 180, what would be the total power in the 90° meridian?

               — And, of course, the answer is: -1.00 +(-2.00) = -3.00!

There may even be some questions along the same line that try to make the concept look difficult without actually being difficult at all:

If your Rx is -3.00 -1.50 x 60, what would be the power of the sphere in the 150° meridian?

                — Still simple transposition! -3.00 + (-1.50) =  -4.50

All those should be SUPER simple for you since you do this transposition every time someone brings in a plus cylinder Rx.

By now you should be aware that:

  • The sphere power of every prescription is the power in the meridian of the axis.
  • The full power of the cylinder is felt 90° away from the axis.
  • The total power of the Rx is the sphere power at 90° away from the axis.

None of you should have any problem with this because it is all routine and everything involves something 90° away.

It’s when we want something other than 90° away that the fun starts!

Basic power in any Meridian

You might see a question or two that look like this:

If the patient has an Rx: -2.00 -2.00 x 180, what would be the total power at axis 60?

Here it would be helpful to be able to remember that ¾ of a cylinder’s power is felt 60° away from its axis.

75% of -2.00 would be -1.50.

The sphere (-2.00) added to 75% of the cyl (-1.50) would equal -3.50.

There is a handy dandy formula and chart that you can see on last week’s post ABO study: calculating the power in any meridian.  If you use that you can’t go wrong.

There are any number of twists this type of question can take:

  • They can start out with an axis of 30°, 45°, or 60° and ask you to calculate the power in the 90° or 180°.
  • They can ask you what the cylinder power, not the total power, is in axis 30°, 45°, or 60°.
  • They can ask you what is the percent of the cylinder power in axis 30°, 45°, or 60°.
  • They can ask you for either the cylinder power or total power in axis 120°, 135°, or 150°.
  • They even can start with an oblique axis and then ask you to calculate power for some axis 30°, 45°, 60°, 120°, 135°, or 150° away.

All you need to be able to ace these questions is to figure out if you’re using ¼, ½, or ¾ of the cylinder.  All you have to do is memorize the formula.

Challenging Power in any Meridian Questions

But… the ABO/NCLE test designers just might get sneaky and decide to play “schmarty pants” with the optician.  You might get a question like this one:

If the patient’s Rx is -1.00 -4.00 x 60, what would be the  power of the Rx in the 75° meridian?

This one is nasty in two ways.  It doesn’t start out with a 90° or 180° axis, and it doesn’t use the 30°, 45°, or 60° difference.  Yet, it isn’t really that hard!

If you analyze the problem, you can figure out that if a difference of 30° would be one quarter of the cylinder power, 15° would be one eighth of the cylinder power!  The cylinder power would be -0.50 in the 75° meridian and the total power would be -1.50.

But… what if the cylinder doesn’t break down so evenly?

Given the Rx -7.50 +3.00 x 145, what would be the approximate power of the cylinder at axis 180°?

Whew!  Approximate power!  They wouldn’t really ask that, would they? (I mean, come on!)

Unfortunately, the answer, of course, is yes.  I made this question up myself, but I’ve seen questions that ask for approximations on the practice tests in some optical text books.

There really isn’t any type of question that they couldn’t ask you.  If every exam most likely contains about 10% super easy questions, then every test should also contain about 10% super hard ones!

So, for every 100 question test, about 10%, or 10 questions, would be extremely challenging.  This might be one of them.

But here, too, it’s not really that bad. You just have to reason it out.  They want axis 180° and they give you 145°. The difference between 180° and 145° is 35°.

30° away from the axis, only a quarter of a cylinder is felt.
45° away from the axis, one half of the cylinder is felt.

Anything in between 30° and 45° would be between ¼ of the cylinder (+0.75) and ½ the cylinder (+1.50).

35° really isn’t that much more than 30°, so you would be looking at a cyl value that is just a little larger than +0.75.  Chances are the answers will be something like:

  1. +0.87
  2. +0.62
  3. +1.12
  4. +1.87
  5. +1.37

Since +0.87 is just a little bigger than +0.75, that would be the obvious answer!  If you went the wrong way and went just a little bit smaller, you’d get +0.62.  Otherwise, the answer list contains no +1.00 to screw you up.  It HAS to be +0.87.

I truly doubt that ANY question you get will be so ambiguous that you can’t reason it out.  ONE answer will be the best answer.  You can count on it!

But is this all the types of questions you might see on cylinders and meridians?

Not on your life!  The ones we’ve covered so far are academic, theoretical.  We have yet to cover the REAL world practical examples, the sort of stuff that affects your day to day job…

Practical Applications for Power in Any Meridian

I can just hear the protests right about now…

WHEN would we use the power in any meridian in our day to day jobs?

Well… when we’re calculating Prentice’s Rule for prism tolerance!

Think about it.  If the Pd is off and we want to ascertain whether or not the job passes inspection, we apply Prentice’s Rule:

Δ = (Power)(decentration) ÷ 10

But what power?  And what decentration?  It all depends upon what type of prism, doesn’t it?

If we are considering horizontal prism, we want to focus on the power in the 180° meridian.

If we are concerned with vertical prism, we are looking at the power in the 90° meridian.

And THAT is the real reason for understanding toric transposition.  In order to calculate for prism in the 180° or 90° meridian, we have to know the power in that meridian.

So, you might very well see a question like this:

The job you are final inspecting is 4mm narrow in the
Pd.  If the Rx is:
+1.50 -4.00 x 30
+1.00 -3.00 x 45

what would be the total prism effect and would the job
pass ANSI standards?

The first thing we do here, since we are looking for horizontal prism, is to calculate the power at 180°.

For the OD, at 30° away from the axis, the power in the cylinder is only ¼.  -4.00 ÷ 4 = -1.00.   If we add a -1.00 to a +1.50, the power in that meridian is only +0.50.  Applying Prentice’s Rule: (power) +0.50 x (decentration) 2 (half of 4) ÷ 10 = 0.10Δ BI.

For the OS, at 45° away the power in the cylinder is half. -3.00 ÷ 2 = -1.50.  If we add -1.50 to a +1.00 sphere the power in that meridian is +0.50.  Applying Prentice’s Rule: +0.50 (power) x 2 (decentration) ÷ 10 = 0.10ΔBO

0.10ΔBI in one eye and 0.10ΔBO in the other eye gives you NO prism effect, and, of course, the job would pass inspection.

Here’s another one:

The OC of the right lens is 2mm above.  The OC of the left lens is 4mm above.  Given the following Rx, how much vertical prism has been induced, and is it within the limits of ANSI tolerances?
-4.00 + 3.00 x 30
-2.00 + 3.00 x 60

Here, since we need to calculate vertical prism, we are looking for the power at axis 90°.

For the OD: 30° is 60° away from 90°.  +3.00 (¾) = +2.25.  -4.00 +2.25 = -1.75.  This OC is 2mm above the geometric center. (-1.75 x 2) ÷ 10 = 0.35ΔBD.

For the OS: 60° is 30° away from 90°. +3.00 (¼) = +0.75.  -2.00 +0.75 = -1.25.  This OC is 4mm above the geometric center.  (-1.25 x 4) ÷10 = 0.50ΔBD.

Now… when it comes to combined vertical prism, if the bases are going the same direction, we subtract the values.  0.50ΔBD – 0.35ΔBD = 0.15ΔBD.

Yes, the job passes inspection since the tolerance for vertical prism is 0.33Δ.

Let’s hope you really don’t have a question like that!  But you might… It’s good to be prepared for anything!

Practice what you’ve learned about power in any meridian

Since you know you’ll be asked questions like these, the best thing you can do is practice.

At www.passyouropticalboards.com there are thousands of questions in the database and plenty of questions on toric transposition.  The questions also come with explanations that help you learn,  just like how I explained step-by-step how to do the calculations here.

Check us out and ACE your optical boards!