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Thread: More course questions.

  1. #1
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    More course questions.

    1. A lens measure is placed on a copper tube and is then rotated untill its legs are at right angles to the axis of the tube. The reading given by the instrument reads +10.50. What is the external diamter of the tube in mm, if the measure was calibrated for glass of refractive index 1.64.


    not looking for the answer so to speak, more someone to explain the formulae i need to use and which ones to use, so i can figure it out myself.



    2. A lens measure reads -0.25 when placed on a flat surface and +3.00 on the surface of a lens. If the measure was calibrated for n =1.523, what is the curvature of the surface?

    Is this just 3.00 - 0.25 to get the actual power minus lens measure offset, and then put it into the radius of curvature formulae?

    (1.523 - 1)1000
    ---------------
    2.75

    ? or am i doing something silly

  2. #2
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    No. 1 - You are going to use the formula to convert diopters to the radius of curvature

    r = (nr - ni) / D

    You have the refractive index, and assume the incident is 1, now double your radius to get the diameter

    2 * r

    No. 2 - Your kinda on the right track, it takes a +0.25 to bring your lens clock back to zero or calibrated, so any measure you get add the +0.25:

    +3.00 + +0.25 = +3.25

    Think of it in 2 steps the first step being how much power to zero your instrument, then add that to the reading. The rest your on the money with.

  3. #3
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    soo

    r = (1.654 -1)1000
    --------------
    10.50


    r = 62.29mm

    r * 2 = 124.58mm


    and 2

    +3.00 + 0.25 = 3.25

    r = (1.523 - 1)1000
    --------------
    3.25

    r = 161 mm


    Thanks you

  4. #4
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    Bingo :cheers:

  5. #5
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    toric transposing,

    Thanks HarryChilling!!

    ok doing some revision, exams next month

    (A)
    I cant seem to get the same answers as my course notes for 2 questions out of 20.


    Transpose the following prescriptions to toric form with +4.00 BC

    (I)
    -5.50/+0.75x60

    i get

    +4.00 DC x 150/+4.75 DC x 60
    ----------------------------
    -9.50DS (notes say -10.25DS)

    Simply what do i need to add to +4.00 BC to get the working sph in this case -5.50. i get -9.50DS course notes say -10.25DS.


    (II).
    +3.75 / - 0.50 x 110

    i get

    +4.00 DC x 110 / +4.50 DC x 20
    ------------------------------
    -0.75DS (notes get -8.25DS)

    Simply what do i need to add to +4.00 to get the working sph in this case +3.25 after transposing to alternate sph/cyl. i get -0.75DS course notes say -8.25DS.

    So what am i doing wrong?


    (B)

    The Rx -4.00 DS / +1.50 Dc x27 is to be made up using a semi finished +1.50 sphere blank. Write out the power of the toroidal tool necessary to produce this lens.


    (C)
    A torric lens is to be made to the specification -2.00/+8.00x90. Owing to the limitations of available tools the front surface is ground +4.00DC x 180/+9.00Dc x 90. Which tool must be used for the back surface?


    Like normal, would rather a quick explanation of how to do them rather then just the answer, prefer to figure out the exact answer myself :)
    Last edited by Del; 05-07-2007 at 03:55 PM. Reason: added more!

  6. #6
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    Check this thread out

    http://www.optiboard.com/forums/showthread.php?t=18768

    Darryl gives a throrough break down of your exact question.

  7. #7
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    cheers

    the actual transposing of the Rx to toric form given a base curve or a sphere curve i can do, i know the steps easy as pie. Its just if i went wrong on the 2 questions in (A) then i just cant figure out what my mistake was.

    (I) -5.50/+0.75x60 with a +4.00 Base curve

    Step 1. If power sign of cyl does not match sign of base curve then transpose to alternate sph/cyl Rx (not needed as they are both +)

    Step 2. Write the Base curve as a cyl with its axis at right angles to the axis of the cyl in the sph-cyl Rx.
    +4.00DC x 150

    Step 3. Add the cyl of the Rx to the Base Curve, in order to obtain the Cross Curve. Write the Cross Curve as a cyl with the axis the same as the axis in the sph-cyl Rx.
    +4.75 DC x 60

    Step 4. To obtain the Sph curve, as yourself "What curve must be combined with the Base curve in order to obtain the sph component of the sph-cyl Rx. -9.50 DS

    +4.00 DC x 150/+4.75 DC x 60
    ----------------------------
    -9.50DS

    But the offical answer in the booklet says 10.25DS. What am i missing?


    (II).
    +3.75 / - 0.50 x 110 with +4.00 Base Curve

    Step 1. +3.25/+0.50 x 20
    Step 2. +4.00 DC x 110
    Step 3. +4.50 DC x 20
    Step 4. -0.75DS

    i get

    +4.00 DC x 110 / +4.50 DC x 20
    ------------------------------
    -0.75DS

    course notes say -8.25DS.



    As for (B) and (C) This may sound stupid but knowing the basic rules and being able to apply them without an example for me is kinda hard. Thats the one thing i hate about a overseas course. No 1 to 1 with a instructor, i know soon as i see an example i will kick myself for being a bit slow. But still would like to see an example, dosent have to be an answer to either B or C just something that i can look though and understand.

  8. #8
    Rising Star lensgrinder's Avatar
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    Del double check your course notes, the transposition looks correct to me.

  9. #9
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    hmm ok

    well double checked and so emailed the course provider, so will see if they get back in touch with me.


    Any help available for B and C?

  10. #10
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    Quote Originally Posted by Del
    (B) The Rx -4.00 DS / +1.50 Dc x27 is to be made up using a semi finished +1.50 sphere blank. Write out the power of the toroidal tool necessary to produce this lens.
    Since no material index is given I am assuming that the blanks Front Curve is the true front.

    Then place the Rx on the Optical Cross to find the powers in both meridians

    Then use the Back Vertex power formula to figure out the curve needed for the back curve

    Dback = Dpower - [Dfront / (1 - (Dfront * thickness/index))]

    Figure this out for both your meridians and whalla you have your back curves, now you can convert these curves into the tools index

    Dtool = [(ntool - 1) / (nmaterial - 1)] * Dback

    Whalla, now you have the tool curves. If you know your pad thickness you can also factor that in by converting the power from diopter to radius and taking the thickness off the radius and then converting back to diopters.

    Quote Originally Posted by Del
    (C) A torric lens is to be made to the specification -2.00/+8.00x90. Owing to the limitations of available tools the front surface is ground +4.00DC x 180/+9.00Dc x 90. Which tool must be used for the back surface?
    Same as above except for now instead of using the same value for Dfront for both your meridians you will be using different Dfront because the front curve is toric.

  11. #11
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    Since no material index is given I am assuming that the blanks Front Curve is the true front... Then place the Rx on the Optical Cross to find the powers in both meridians... Then use the Back Vertex power formula to figure out the curve needed for the back curve
    Dback = Dpower - [Dfront / (1 - (Dfront * thickness/index))]
    Generally, if neither a refractive index nor a center thickness is provided, you must assume that the calculations are for a thin lens and that the surfaces curves are provided in the index of the material. Fortunately for Del, this is just a simple toric transposition problem, similar to the earlier problems.

    The Rx -4.00 DS / +1.50 Dc x27 is to be made up using a semi finished +1.50 sphere blank. Write out the power of the toroidal tool necessary to produce this lens.
    It's best to begin by placing each principal meridian power onto an optical cross:

    -4.00 D @ 27 and -2.50 @ 117

    Then subtract the base curve (+1.50) from each principal meridian:

    -4.00 - 1.50 = -5.50 D @ 27
    and
    -2.50 - 1.50 = -4.00 D @ 117

    Consequently, the rear tool curves would be:

    -4.00 D / -5.50 D
    Darryl J. Meister, ABOM

  12. #12
    ATO Member OptiBoard Bronze Supporter HarryChiling's Avatar
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    Thanks Darryl,

    You get cool points for that.

  13. #13
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    Always glad to help.
    Darryl J. Meister, ABOM

  14. #14
    Rising Star Bezza's Avatar
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    Also it is important to note that with tool power problems compensation for pad thickness is often required. Since the question does not ask for it it most likely isn't required however it may at least prudent to mention that your answer has not allowed for this compensation.

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