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Thread: Prisms and decentration questions.

  1. #1
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    Prisms and decentration questions.

    I am doing a distance leaning optical technicians course and i am having some problems fully understanding the whole issue of the direction of base when figuring out prisms and decentration.

    1)
    A lens of power -4.00DS / +2.00 DC x90 is decentred 4mm IN and 2 mm UP on a right eye. What horizontal and vertical prisms will be produced.

    My answer
    Power Meridans
    along axis. Sphere power only F @ 90 = -4.00

    power merdian 90 to axis - sphere plus cyl -4.00 + 2.00 = -2.00


    F @ 90 = -4.00
    F @ 180 = -2.00


    P @ 90 = C x F = 0.2 x 4 = 0.8^
    P @ 180 = C x F = 0.4 x 2 = 0.8^

    As F @ 90 is negitive the prism base will be in the opposite direction as the decentration ie. Base out.

    AS F @180 is negitive. the prisim base will be in the opposite direction as the decentration. ie Base out

    0.8^ BASE Down and 0.8^ BASE Out.


    2)What prism will there be at a point 4mm IN and 2mm UP on the same Rx as in question 1? Answer in horizontal and vertical meridians.

    I tryed to do this one and got the same answer as the first, feeling very mnixed up i attemped the rest and at this stage have no idea what is going on!!

    3) You wish to produce a prism of 3.00^ base up on Rx +5.00DS/+2.00DC x 180 by decentration. state the amount of decentration and its direction for a left eye.

    4) How much would you have to decentre a lens -2.50 DS / -2.50 DC x 90 to produce a prism of 5^ base IN and 2^ Base up for a right eye?

    5) How much prism will there be at a point 6mm UP and 2 mm IN on Rx. +2.00DS/-3.00 DC x180, right eye, answer as a single resultant prism.


    Anyone able to explain how to do these more so then actually doing them <defeats the purpose if i cant understand the "how" part. My Thanks

  2. #2
    Master OptiBoarder lensgrinder's Avatar
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    Question 1 and 2 have the same values(4mm in and 2mm up)

    Question 3

    Power at 90 is +7.00 now use Prentice's rule:

    P=cD
    3 = c X 7
    c= 3/7 = 0.43 or 4.3mm up

    Question 4, just use Prentice's rule in each meridian.

    Question 5
    How much prism will there be at a point 6mm UP and 2 mm IN on Rx. +2.00DS/-3.00 DC x180, right eye, answer as a single resultant prism.

    Find prism in each meridian using Prentice's Rule:

    P180=0.2 X 2 = 0.4^
    P90 = 0.6 X 1 = 0.6^

    Now draw a right trinagle on a piece of paper. Put 0.6 on the vertical line and 0.4 on the horizontal line.

    Now use pythagoras theorum to find the resulting prism.

    P = square root(H^2 + V^2)
    where p is total prism
    h is horizontal prism
    v is vertical prism

    P = square root( 0.4^2 + 0.6^2)
    P = square root(0.16 + 0.36)
    P = 0.72^

    Now using geometry find the angle. Use opposit over adjacent or tangent
    tan (a)=0.6/0.4
    tan(a) = 1.5
    now take the inverse tangent
    a = tan-1(1.5)
    a = 56.3 degrees

    So you have 0.72^ @ 56.3 degrees, since it is in quadrant I there is no need to do anything with the angle (see attachment)
    Attached Thumbnails Attached Thumbnails Prism.bmp  
    Last edited by lensgrinder; 04-23-2007 at 01:40 PM. Reason: Not Finished

  3. #3
    Master OptiBoarder
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    Lensgrinder has provided a great post.

    You may want to try a fellow optiboarder's web site. Daryl Meister has a great website with online calculators and loads of other stuff. www.opticampus.com . I know it may not help with the theory, but, it may at least give an answer to check your work. Good luck.

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