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Q&A Thread for Optometric Curriculum Material

Discussion in 'Optometry' started by qwopty99, Dec 16, 2008.

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  1. qwopty99

    qwopty99 Optometrist
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    Questions pertaining to the body of knowledge of optometry can be posted here in this one thread.
     
    #1 qwopty99, Dec 16, 2008
    Last edited: Aug 4, 2009
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  2. KHE

    KHE Senior Member
    Optometrist SDN Advisor 10+ Year Member

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    Jesus....

    You can get an approximation of the power in the resulting meridian using a simple pythagorean equation. There is little clinical relevance to using sin2 angle + cos2angle =1. The two alphas in your example mean different things. The Falpha is referring to the POWER (not angle) in the meridian you wish to calculate. The sine alpha is the acute angle between the desired meridian and the axis meridian.

    Also, when you're subtracting angles, you should use a value of zero for axis 180 rather than the numerical value 180.

    I think that's right. My head hurts. Please don't post **** like that again!
     
    #2 KHE, Dec 16, 2008
  3. JMU07

    JMU07
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    Yeah, next time wait until finals are over!! :scared::D
     
    #3 JMU07, Dec 16, 2008
  4. gochi

    gochi
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    For pre-optos, this literally neiog2no291wnf5nwo.
     
    #4 gochi, Dec 17, 2008
  5. qwopty99

    qwopty99 Optometrist
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    I wasn't aware of this (but I'm not surprised that you can). I'm assuming the estimated power is the hypotenuse, and the two cylinders each contribute a sine and cosine component?
     
    #5 qwopty99, Dec 17, 2008
    Last edited: Jul 30, 2009
  6. JMU07

    JMU07
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    I think he's talking about power vectors here. Simple formulas you can use to convert from vector form to a prescription format. Makes it much easier to deal with oblique meridians!
     
    #6 JMU07, Dec 17, 2008
  7. eyestrain

    eyestrain Member
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    I do not understand a word posted on this thread. My solution to this problem: tobradex qid ou. You're welcome.
     
    #7 eyestrain, Dec 17, 2008
  8. qwopty99

    qwopty99 Optometrist
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    I've come across literature (advertising material of a specific lens company) recently showing treated CR39 has better UV protection than PC. Is this true? Is this what some of you have learned?
     
    #8 qwopty99, Dec 29, 2008
    Last edited: Jul 30, 2009
  9. eyestrain

    eyestrain Member
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    I assume both PC and UV CR39 do a fine job. I've never taken the time to research it, but I'm willing to bet the difference is negligible.
     
    #9 eyestrain, Dec 29, 2008
  10. qwopty99

    qwopty99 Optometrist
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    Hi folks.

    Please confirm.

    The item in the image is called a lens former. Correct?

    Thanks in advance.

    [​IMG]
     
    #10 qwopty99, Jan 25, 2009
  11. drbizzaro

    drbizzaro Varilux/Essilor Advocate
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    it's called a pattern blank
     
    #11 drbizzaro, Jan 25, 2009
  12. qwopty99

    qwopty99 Optometrist
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    thnx.
     
    #12 qwopty99, Jan 25, 2009
  13. qwopty99

    qwopty99 Optometrist
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    Hi

    Just curious. I'm aware that in previous generations of Transitions lenses, the "transitioning" part was a single layer in the lens (typically below the front surface).

    Are ALL modern models of Transitions lenses made in the same way? Or do "through and through" type of Transitions exist? i.e. do any Transitions lenses (for plastic lenses) exist where the transitioning material is embedded throughout the lens polymer?

    Thanks in advance.
     
    #13 qwopty99, Feb 21, 2009
  14. qwopty99

    qwopty99 Optometrist
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    Hi folks

    OK - no replies to my last question. What about the following one:

    Is there any professional terminology to describe lens styles? For instance, the following lens is quasi-rectangular - but it "jets out" superior-temporally. Is there a name for this? How would you describe this lens shape/style?



    Thanks in advance.
     
    #14 qwopty99, Mar 1, 2009
    Last edited: Apr 1, 2009
  15. armstrong77

    armstrong77
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    Just a Q regarding lensometry...

    I just recently listened to a lecture talking about lensometry of high-plus Adds. For High-Adds, you know how we have to find front-vertex power (FVP) of the add portion. Is the Add calculated by:

    1. (Add FVP) MINUS (Distance BVP)

    OR

    2. (Add FVP) MINUS (Distance FVP) ?

    I always thought it was #1 (since we are looking through the distance BVP, and I think we should be comparing THAT to the Add FVP), but the recent lecture I listened to seemed to suggest it was the front FVP that we subtract it from (but I have my doubts about the authority of the speaker). So which is it?

    Thanks.
     
    #15 armstrong77, Apr 1, 2009
  16. armstrong77

    armstrong77
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    I was wondering if anyone can point out what these bifocal styles look like:

    -sovereign
    -panoptik
    -kryptok
    -ultex
    -ribbon
    -perfection bifocal

    Thanks.
     
    #16 armstrong77, Apr 1, 2009
  17. armstrong77

    armstrong77
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    More on that lecture...

    The speaker was talking about PALs and was talking about appropriate/inappropriate lens styles. e.g. Aviators are not appropriate.

    The person said that lenses shouldn't be too big (i.e. neither A nor B boxing lengths should be too great) because it will lead the majority of the lens to have aberrations (i.e. the blending region will take up the majority of the lens, and therefore, not provide good optics).

    I haven't heard of "big lenses" being a contraindication for PALs. Is this in any way true?
     
    #17 armstrong77, Apr 1, 2009
  18. qwopty99

    qwopty99 Optometrist
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    I have one here:

    Nevermind - answered it!
     
    #18 qwopty99, Apr 4, 2009
    Last edited: Jul 30, 2009
  19. qwopty99

    qwopty99 Optometrist
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    I was wondering if anyone knows, or if there's any easily-accessible literature on the power-diagram of a PAL.

    Thanks.
     
    #19 qwopty99, Apr 30, 2009
    Last edited: Jul 30, 2009
  20. qwopty99

    qwopty99 Optometrist
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    One more.

    Does anyone know the average axial length of the eye? e.g. Gullstrands, or whatever.
     
    #20 qwopty99, Apr 30, 2009
  21. qwopty99

    qwopty99 Optometrist
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    Anyone?
     
    #21 qwopty99, Apr 30, 2009
    Last edited: Jul 30, 2009
  22. Penguin2012

    Penguin2012
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    The average axial length is 23mm I believe according to a study by Atchinson et. al in 2004 (eye shape in emmetropia and myopia).

    In terms of models, I am not sure about Gullstrand, I know Elmsley is 22.22 mm.
     
    #22 Penguin2012, Apr 30, 2009
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