ABO Study Question

**miracle worker** · 05-11-2011, 08:25 AM

Hello everyone. I am a long time lurker to this board, and am in need of assistance. I am studying for the ABO, and have a practice question that is giving me a hard time. Things just do not seem to add up on this for me. The question is:

The optician has just neutralized an unknown power lens with test lenses. The test lenses used to neutralize the power in the vertical is a plus 1.00 and the test lensused to neutralize the power in the horizontal is a plus 1.50. (NOTE: These test lenses were used alone for each meridian, Not superimposed .) What is the Rx?
a. -1.50 +.50 x 180
b. +1.50 +1.00 x 180
c. -1.50 -.50 x 180
d. +1.00 +.50 x 90

If one of you wise opticians could walk me through this, I would appreciate it. I do understand that the test lens is supposed to zero out the lens being neutralized, but canot seem to come up with an answer listed.

**Uncle Fester** · 05-11-2011, 08:51 AM

Originally Posted by miracle worker

Hello everyone. I am a long time lurker to this board, and am in need of assistance. I am studying for the ABO, and have a practice question that is giving me a hard time. Things just do not seem to add up on this for me. The question is:

The optician has just neutralized an unknown power lens with test lenses. The test lenses used to neutralize the power in the vertical is a plus 1.00 and the test lensused to neutralize the power in the horizontal is a plus 1.50. (NOTE: These test lenses were used alone for each meridian, Not superimposed .) What is the Rx?
a. -1.50 +.50 x 180
b. +1.50 +1.00 x 180
c. -1.50 -.50 x 180
d. +1.00 +.50 x 90

If one of you wise opticians could walk me through this, I would appreciate it. I do understan that the test lens is supposed to zero out the lens being neutralized, but canot seem to come up with an answer listed.

Too visualize this make an optical cross (a one inch by one inch + ) on a piece of paper.

Use the info given and label the cross so the vertical (axis 90) is +1.00
The horizontal (axis 180) is +1.50.

Now if plus lenses were used to neutralize the lens what would the x90 and x180 power be?

(If you're stuck just say I'm stuck. And don't worry we all get stuck. Some of us even after 30+ years of doing this!)

**miracle worker** · 05-11-2011, 11:21 AM

Well, if I am using the test lenses I have to say that the lens being neutralized is -1.50 at the 180, and -.50 at the 90?

**Uncle Fester** · 05-11-2011, 12:20 PM

Originally Posted by miracle worker

Well, if I am using the test lenses I have to say that the lens being neutralized is -1.50 at the 180, and -.50 at the 90?

Note the test lens is not superimposed so the +1.00 vertical meridian is a -1.00.

Now make a new cross with x90 -1.00
and x180 a -1.50

Now can you see which answer is correct?

Hint- remember the full power of the cylinder is 90 degrees from the axis.

**mshimp** · 05-11-2011, 12:47 PM

+1.00 neutralizes the vertical (90 degrees) so it would be -1.00. +1.50 neutralizes the horizonal (180 degrees) so it would be -1.50. Of course the answer is in a plus cylinder form to confuse you. The answer is a. -1.50+.50x180. Remember the axis is always in relationship to the sphere power. When you transpose to the minus cylinder form you can "see" it better, -1.00-.50x90. Hope this helps and welcome to optiboard.

**Uncle Fester** · 05-11-2011, 12:51 PM

So much for the teachable moment...
(Note post #2 headline)

**miracle worker** · 05-11-2011, 03:13 PM

That's it. I see what I was doing wrong. Thanks for the lesson Uncle Fester!! I owe you one.

**fullclout** · 05-12-2011, 01:23 PM

Hi. Sorry if this is very basic, but what is a test lens? Is it like a trial lens? Are they used to neutralize glasses? I tried to follow the question/answer but I'm afraid I have not encountered this, maybe the wording is different from what I know. This makes me nervous because I will also be taking the ABO in November. Would someone be nice enough to explain this problem to me?

**Uncle Fester** · 05-12-2011, 02:04 PM

It's nothing special. It can be any lens which you know the power of. In school (Jurassic period) we used uncut lens blanks but trial lenses would also work (just smaller fields).

**fullclout** · 05-12-2011, 03:50 PM

So you're just holding a known power lens in front of an unknown power lens? How does that neutralize anything? What are you supposed to see? I don't get it.

**gmc** · 05-12-2011, 07:07 PM

You look through an unknown lens and you notice that the image you see through the lens is a little smaller than what you see with the naked eye. You move the lens to the right and notice the image you see through the lens also moves to the right. You rotate the lens and notice that the image doesn't get shorter or taller. You now know you have a minus spherical lens.

You go to your drawer of finished SV lenses and pull out a +0.50 sphere. When you hold the lenses together, you see that the image is a little larger, but still slightly minified. The image also still moves to the right when you look through the pair of lenses, although not as much as before. You pull out a +0.75 sphere and repeat the process. The image is still slightly minified and still has a little with motion. Next, you pull out a +1.00 sphere and repeat the process. This time the image isn't smaller and when you move the lens the image stays put. The +1.00 lens "neutralized" or canceled out the power of the unknown lens, so it must be a -1.00.

Just for kicks, you pull out a +1.25 sphere and hold it together with the -1.00 sphere, so you know you now have a total power 0f +0.25. When you look through this combo, the image is slightly magnified and when you move the pair of lenses to the right, the image you see moves to the left. This confirms that the power of the unknown lens is in fact -1.00.

Hope this helps.