Results 1 to 6 of 6

Thread: Abo help

  1. #1
    OptiBoard Novice
    Join Date
    Sep 2020
    Location
    Louisiana
    Occupation
    Optical Retail
    Posts
    2

    Abo help

    Has anyone here taken the ABO exam recently? There are 2 things I can't figure out how to get the correct answer and my boss hasn't taken the exam in years and is on vacation, is anyone here able to help me with some example problems? I would be forever grateful!

  2. #2
    OptiBoard Apprentice Optical Roy's Avatar
    Join Date
    May 2019
    Location
    Nacogdoches Texas
    Occupation
    Dispensing Optician
    Posts
    28
    It's been many moons since I took the exam, but what are the questions?
    Roy W. Jackson, Sr. ABOC

  3. #3
    OptiBoard Novice
    Join Date
    Sep 2020
    Location
    Louisiana
    Occupation
    Optical Retail
    Posts
    2
    It’s asking for prism induced for -6.50 +1.50 x30 patient pd =64 fitted pd =68
    options are:
    2.45 BI
    2.45 BO
    1.225 BO
    1.225 BI

    I’ve tried figuring it out but I’m not sure how to get the correct answer? If you could help I’d appreciate it. Thank you

  4. #4
    Master OptiBoarder
    Join Date
    May 2000
    Location
    Fayetteville, NC, USA
    Occupation
    Dispensing Optician
    Posts
    1,890
    Find the power in the horizontal meridian (0-180). Then simply use Prentice's Rule to find the correct answer, P = hD.

  5. #5
    Master OptiBoarder OptiBoard Silver Supporter
    Join Date
    Feb 2016
    Location
    usa
    Occupation
    Dispensing Optician
    Posts
    494
    You can use the power in any axis formula, but I think the ABO basic will always have questions at axis 30(120), 45(135), 60(150), 90, or 180. They want to you know the precent of cylinder present at each of these. So 25% of the cylinder power is present 30(120) degrees away from the axis, 50% at 45(135) degrees away, 75% 60 degrees, and all of the cylinder power is present 90 degrees away from the axis. You should still learn the cylinder power in any axis because it is simple and handy, but I don't think you get a calculator with trig functions for the test.

    So the cylinder power present at 180, 30 degrees away the axis, is 25% of the full Cylinder power. .25*1.50= +.375. Add that to the sphere power -6.50 +.375 = -6.125. This is the power at 180. 4mm of decentration. .4cm*-6.125= -2.45. This is a minus lens with the OC displaced temporaly, so base in prism.

    Cyl power in any meridian for future use is:

    Dc'= Ds + Dc (sin²θ)


    Dc´= the power in the meridian in question (in this case it is 180°)
    Ds = the sphere power of the Rx
    (-6.50)
    Dc = the cylindrical power (+1.50)
    θ = the difference between the axis of the Rx and the meridian in question (180 – 30° = 150°) or (30-0=30) or (30-180=-150) or (0-30=-30)
    Dc' = -6.50+1.50(sin 150)²

    Note it doesn't matter if you use 0 or 180 and it doesn't matter which meridian you subtract from which, you will get the same answer. The sin of 30 is the same as the sin of 150, -30, and -150, and squaring the answer will remove the negative. So if you are looking for power at 180 you can just use the axis of the cyl and save some steps.


  6. #6
    Master OptiBoarder OptiBoard Silver Supporter lensmanmd's Avatar
    Join Date
    Oct 2016
    Location
    Maryland
    Occupation
    Optical Wholesale Lab (other positions)
    Posts
    961
    Quote Originally Posted by wmcdonald View Post
    Find the power in the horizontal meridian (0-180). Then simply use Prentice's Rule to find the correct answer, P = hD.
    +1
    I bend light. That is what I do.

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
OptiBoard is proudly sponsored by:
Younger Optics and Vision Equipment