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  1. #1
    OptiBoard Apprentice
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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 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.
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  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 HarryChiling's Avatar
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    Bingo :cheers:
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  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 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.
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