Optical Ed Articles http://www.opticaled.com/articles Opticianry Articles Thu, 29 Dec 2011 01:17:23 +0000 en hourly 1 http://wordpress.org/?v=3.2.1 ABO Basic Review http://www.opticaled.com/articles/problem-solving/2011/08/abo-basic-review/ http://www.opticaled.com/articles/problem-solving/2011/08/abo-basic-review/#comments Tue, 02 Aug 2011 20:23:11 +0000 Brent McCardle http://www.opticaled.com/articles/?p=259 This is a review I did for the ABO

]]> http://www.opticaled.com/articles/problem-solving/2011/08/abo-basic-review/feed/ 0 Understanding Prism http://www.opticaled.com/articles/problem-solving/2009/11/understanding-prism/ http://www.opticaled.com/articles/problem-solving/2009/11/understanding-prism/#comments Wed, 18 Nov 2009 20:59:13 +0000 Brent McCardle http://www.opticaled.com/articles/?p=181 One area of Opticianry that seems to have more Opticians baffled is prism. I hope in this short tutorial to ease the misunderstanding of prism.]]> One area of Opticianry that seems to have more Opticians baffled is prism. I hope in this short tutorial it will ease the misunderstanding of prism.
This tutorial is not meant to understand every aspect of prism, it is meant to help you understand prism in your everyday setting as an Optician.
In order to help us we need to think of lens as two prisms. A lens with prisms base to base is a plus lens and a lens with prisms apex to apex is a minus lens.

This image shows a minus lens on the right and a plus lens on the left that are vertically oriented. This image is you looking at a patient, the left side is the patients right eye and the right image is the patients left eye. A minus lens has its optical center (OC) located where the apexes of the prism connect and a plus lens has its OC located where the bases of the prism connect.
This image will help us determine the direction of prism, Base Up(BU) and Base Down (BD).
Notice the pupil in the image, it is located below each OC. Erase the triangle that the pupil is not in or draw a new triangle.

Looking at this image where is the base in relation to the pupil. The triangle on the left the base is up and the triangle on the right the base is down. This is how we determine vertical base direction.
Now look at the same image, but with the pupils above the OC.

Again, erase the triangle that the pupil is not in and you will get the image below. Notice where the base is in relation to the pupil and you will see in the right eye (the triangle on the left side) the base is down and for the left eye (the triangle on the right side) the base is up.

Let us take a look at horizontal prism. The image below shows a pair of plus lenses(top) and a pair of minus lenses(bottom). The top image represents prism base out (BO). The bases are located to the outside of the pupil. The image on the bottom is base in(BI). The base is in in reference to our pupils

The next image is the opposite. The image on top is two plus lenses with BI, the bases of the prism are located in, in reference to the pupils. The image on the bottom is two minus lenses with BO, the bases are located out in reference to the pupil.
Remember, you only need to look at the triangle that the pupil is in.

When the bases of the prism are in the opposite direction they compound or add to each other. When the bases of the prism are in the same direction they will cancel or subtract from each other.
If you have vertical prism, i.e. BU or BD, BU and BU will cancel, BD and BD will cancel. BU and BD will compound
If you have horizontal prism, i.e., BO or BI, BI and BO will cancel, BO and BO will compound, and BI and BI will compound.
Example:

  • If you have 2^{\Delta}BU in the right eye and 2^{\Delta}BD prism in the left eye, you have a total of 4^{\Delta} diopters of prism.
  • If you have 1^{\Delta}BI in the right eye and 3^{\Delta}BO prism in the left eye, you have a total of 2^{\Delta} diopters of prism.

    • You will need to use cancelling and compounding prism when splitting prism for cosmetic reasons or when applying ANSI Standards.
    In order to apply ANSI Standards we need to find out the amount or prism per lens using Prentice’s Rule.
    P = c \times f
    P is the amount of prism
    c is the amount that the OC is from the pupil in centimeters
    f is the power of the lens
    Example:

    • If the Rx is:
      • OD -3.00 DS
      • OS -4.00 DS

    How much prism will the wearer experience if the patients PD is 63 mm, but they were made at 66 mm?

    • We know that the OC is a total of 3 mm off
    • We would divide that amount by two and get 1.5 mm for each eye. This tells us that the OC is 1.5 mm away from the pupil
    • Use Prentice’s rule to determine how much prism is present in each eye.
      • P = 0.15 \time 3
      • P = 0.45^{\Delta}OD
      • P = 0.15 \times 4
      • P = 0.60^{\Delta}OS
    • Now we determine base direction
    • Please note we did use the sign for the power, there is no need too


    The red dots represent the patients PD. The black lines represent where the OC is and the N is where the nose is, commonly referred to as the nasal. Notice that the PD is less than the OC, because the OC was made wider than the PD.
    Now look at the triangle that the pupil is in. The bases are located in on both eyes. Base in and Base in compound each other, so now you add the 0.6 and the 0.45 for a total of 1.05^{\Delta}.diopters of prism

    • If the Rx is:
      • OD +2.00 DS
      • OS +1.75 DS

    How much prism will the wearer experience if the patients PD is 60 mm, but they were made at 64 mm?

    • We know that the OC is a total of 4 mm off
    • We would divide that amount by two and get 2 mm for each eye. This tells us that the OC is 2 mm away from the pupil
    • Use Prentice’s rule to determine how much prism is present in each eye.
      • P = 0.2 \time 2
      • P = 0.4^{\Delta}OD
      • P = 0.2 \times 1.75
      • P = 0..35^{\Delta}OS
    • Now we determine base direction


    Again, look at the triangle that the pupil is in and notice that the base is out for both eyes. BO in both eyes will compound so we have a total of 0.75^{\Delta}
    If you have an Rx with cylinder you will need to find the power in the meridian in which you are looking for the prism. For example, if you are looking for horizontal prism you need to find the power on the 180^{th} if you are looking for the vertical prism you need to find the power on the 90^{th}meridian.
    If you are unsure how to find the power on these meridian, read my walk through on the oblique meridian formula

    ]]>     http://www.opticaled.com/articles/problem-solving/2009/11/understanding-prism/feed/ 0     Compensated and Effective Power http://www.opticaled.com/articles/problem-solving/2009/11/compensated-and-effective-power/ http://www.opticaled.com/articles/problem-solving/2009/11/compensated-and-effective-power/#comments Wed, 11 Nov 2009 21:36:19 +0000 Brent McCardle http://www.opticaled.com/articles/?p=167 When we compensate an Rx, we do so because the lens changes power as moves closer or farther away from the eye. We are making an Rx stronger or weaker depending on whether the lens is moved closer farther away.

    *Remember that a lens that is moved closer to the eye will gain in minus power and a lens that moves further from the eye will gain plus power.

    A glasses wearer tells you that her Rx is unchanged, but she wants newer, smaller glasses. Her old glasses fit at 13 mm, and the new ones that you fit her with are at 9 mm. If her right lens Rx is -12.00 DS, what will the compensated power for her new right lens be? Round to the nearest 1/8 diopter step.

    We need to solve for the compensated power which is going to compensate the lens when moved away from the face or towards the face.

    D_{new} = \frac{D_{rx}}{(1 - dD_{rx})}

    D_{new} is the new power

    D_{rx} is the sphere power

    d is the change in vertex in meters. It is negative if moved away from the face and positive if moved toward the face.

    D_{new} = \frac{-12.00}{[1 - (0.004)(-12.00)]}

    D_{new} = \frac{-12.00}{[1 - (-0.048)]}

    D_{new} = \frac{-12.00}{1.048]}

    D_{new} = -11.45\ D

    Since this is compensated we need to round this to a dioptric value. -11.50 D.

    What if the original Rx had cylinder? We need to put the Rx on the optical cross to find the major meridians of power.

    If the Rx is +13.00 – 3.00 X 180, they were refracted at 10 mm but the glasses fit at 13, what is the compensated Rx?

    The two major meridians of power are +13.00 X 180 and +10.00 X 90. Solve for each of these independently.

    D_{new} = \frac{13.00}{[1 - (-0.003)(13.00)]}

    D_{new} = \frac{13.00}{[1 - (-0.039)]}

    D_{new} = \frac{13.00}{1.039}

    D_{new} = +12.51

    D_{new} = \frac{10.00}{[1 - (-0.003)(10.00)]}

    D_{new} = \frac{10.00}{[1 - (-0.03)]}

    D_{new} = \frac{10.00}{1.03}

    D_{new} = +9.71

    So we have +12.51 X 180 and +9.71 X 90. If you pull this off of the optical cross you will have +12.50 -2.75 X 180

    When we talk about effective power we are referring to the power the wearer will feel if the lens is moved closer or farther away.
    The formula is the same except that the numerator is now [(1+dD_{rx})]

    Using the first example of the -12.00 moved 4 mm closer to the eye. What would the effective power be.
    D_{new} = \frac{-12.00}{[1 + (0.004)(-12.00)]}

    D_{new} = \frac{-12.00}{[1 + (-0.048]}

    D_{new} = \frac{-12.00}{0.952}

    D_{new} = -12.61\ D

    So the power that the wearer will experience if a -12.00 D lens is moved closer to the eye is -12.61 D.

    ]]>     http://www.opticaled.com/articles/problem-solving/2009/11/compensated-and-effective-power/feed/ 0     Oblique Meridian http://www.opticaled.com/articles/problem-solving/2009/11/oblique-meridian/ http://www.opticaled.com/articles/problem-solving/2009/11/oblique-meridian/#comments Wed, 11 Nov 2009 04:23:35 +0000 Brent McCardle http://www.opticaled.com/articles/?p=141 1. The approximate power on the vertical meridian for the Rx +3.00 -1.50 x 102 is ______. Give the power rounded to two decimal places, not to 1/8 diopter steps.

    Vertical meridian is the power in the 90th meridian. First we need to subtract our axis(102 in this problem) from the meridian we are looking for(90 in this problem). This will be \alpha in our oblique meridian formula.

    \ F_{\theta} = [(\sin \alpha)^2 \times F_{cyl}] + F_{sph}

    \ F_{\theta} is the power in the meridian you are trying to find
    \alpha is the difference in the axis and the meridian you are trying to find
    \ F_{cyl} is the cylinder of the prescription
    F_{sph} is the sphere power of the prescirption

    \ F_{\theta} = [(sin \12)^2 \times (-1.50)] + 3.00

    \ F_{\theta} = [0.0432273\ \times\ -1.50] + 3.00

    \ F_{\theta} = -0.06484095 + 3.00

    \ F_{\theta} = +2.94\ D

    So you have a total power of +2.94 D on the 90th meridian

    2. If the Rx reads -10.25 -2.50 x 040, what is the approximate power on the horizontal meridian? Give the power rounded to two decimal places, not to 1/8 diopter steps.
    Horizontal meridian is the power on the 180th meridian. In this case you can subtract 40 from 180 or 0 from 40 since it is either 0 or 180.
    \ sin(40)^2 = 0.413176
    sin(140)^2 = 0.413176

    \ F_{\theta} = [(\sin \40)^2 \times (-2.50)] + (-10.25)

    \ F_{\theta} = [0.413176 \times (-1.50)] + (-10.25)

    \ F_{\theta} = [-0.619764] + (-10.25)

    \ F_{\theta} = -10.87\ D

    ]]>     http://www.opticaled.com/articles/problem-solving/2009/11/oblique-meridian/feed/ 0     Back Vertex Power http://www.opticaled.com/articles/problem-solving/2009/11/back-vertex-power/ http://www.opticaled.com/articles/problem-solving/2009/11/back-vertex-power/#comments Tue, 10 Nov 2009 19:36:47 +0000 Brent McCardle http://www.opticaled.com/articles/processing/2009/11/92/ 1. If a polycarbonate lens (n = 1.586) is made with a front surface power of +10.25, a back surface power of -3.00, and a center thickness of 4 mm, what power will the lensmeter show? Give the power rounded to two decimal places, not to 1/8 diopter steps.

    Remember when using the lensmeter we will check the back vertex power, so we use the back vertex power formula
    B_v = \frac{D_1}{1 - (\frac{t}{n})(D_1)} + D_2\\

    B_v=\frac{10.25}{1- (\frac{0.004}{1.586})(10.25)} + (-3.00)

    \ B_v=\frac{10.25}{1 -(0.002522068095839)(10.25)} + (-3.00)

    \ B_v=\frac{10.25}{1 -(0.02585119798234975)} + (-3.00)

    \ B_v= \frac{10.25}{0.97414880201765025} + (-3.00)

    \ B_v =10.52 + (-3.00)

    \ B_v =+7.52
    2. For a crown glass lens with a front surface power of +15.00, a back surface power of -6.00, and a thickness of 6.5 mm, what power will the wearer see? For crown glass, n = 1.523. Enter just the sign and power.

    The wearer will always look through the back vertex power, so we use the back vertex power formula.
    \ B_v=\frac{15.00}{1- (\frac{0.0065}{1.523})(15.00)} + (-6.00)

    \ B_v=\frac{15.00}{1- (\frac{0.0065}{1.523})(15.00)} + (-6.00)

    \ B_v = \frac{15.00}{1 - 0.064018384767} + (-6.00)

    \ B_v = \frac{15.00}{0.935981615233} + (-6.00)

    \ B_v = 16.03 + (-6.00)

    \ B_v = +10.03

    ]]>     http://www.opticaled.com/articles/problem-solving/2009/11/back-vertex-power/feed/ 0     Selecting the Correct Lens Blank http://www.opticaled.com/articles/processing/2009/06/selecting-the-correct-lens-blank/ http://www.opticaled.com/articles/processing/2009/06/selecting-the-correct-lens-blank/#comments Tue, 23 Jun 2009 03:26:50 +0000 Brent McCardle http://www.opticaled.com/articles/?p=3     Determining the correct lens blank is the first step to surfacing a lens.  This is a two step process, the fist step is to find the correct base curve and the second step is to determine the needed blank size. The base curve is the curve from which all other curves will be determined.  This its [...]]]> Determining the correct lens blank is the first step to surfacing a lens.  This is a two step process, the fist step is to find the correct base curve and the second step is to determine the needed blank size.

    The base curve is the curve from which all other curves will be determined.  This its usually the front curve.  Today’s labs will generally cut all curves on the back surface so if this is the case then your base curve is on the front surface.  If you happen to be cutting the front surface of a lens then your base curve is the back surface.  For our purpose we will call the base curve the front curve of the lens.

    Many surfacing laboratories use a computer or a base curve chart.  The base curve chart is different for each manufacturer.  You could have the same Rx on a 6.00 base for one manufacturer and another manufacturer will call for a 5.50 base curve.  The reason for this is manufacturer X says that Rx Y will work best on a Z base curve.  This is because they have tested there lens and found that if you put Rx Y on base curve Z it will reduce marginal astigmatism and curvature of field.  The bottom line is, try not to deviate from the manufactures recommendations.

    It is OK to change the base curve in order to fit into a frame, but try to keep this to a minimum and do not change the curve by more than a diopter.

    If the base curve chart indicates two different base curves and the difference between these base curves is less than 2.00 D then you may use the same base curve for each eye.  Use the rules below.

    Base Curve Rules

    •Both Lenses Plus- Use the steeper (highest numerical value) base curve of the two.
    •Both Lenses Minus- Use the flatter (lowest numerical value) base curve of the two.
    •One Lens Minus, One Lens Plus- Use the steeper base curve of the two.

    Keep in mind that this does not apply to progressives or any type of ashperic lens.  When dealing with these types of lenses it is best to always use manufacturers recommendations.  Also note that changing the base curve to fit into the wrapped frames will result in the patient having trouble seeing clearly.  It is beyond the scope of the paper as to the reason.  If you do come across these frames and want to put lenses into them use an outside lab, because there are advanced calculations that go into the process of making these types of glasses.

    What if we do not have a base curve chart or a computer?  We will use Vogel’s Rule to determine the base curve.  Before we use Vogel’s Rule we need to find the spherical equivalent of the lens.  The spherical equivalent is the average of the powers in the two major meridians.

    Spherical Equivalent

    sph_eq

    *Please note that the spherical equivalent for a sphere is the sphere.

    Vogel’s Rule

    Plus Lenses

    plus-form

    Minus Lenses

    minus-form

    Example:

    What is the base curve needed for a +6.00 DS?

    • We know it is a plus powered lens
    • We know if is a spherical

    ex1-1

    ex1-2

    • The base curve needed for a + 6.00 DS lens is a +12.00 Base.

    Example:

    • What is the base curve needed for a -2.00 -1.00 X 90?
    • We know we have a minus lens
    • We know we have cylinder, so we need to find the spherical equivalent

    ex2-1

    ex2-2

    Now that we know how to find a base curve of the lens, we need to know what diameter of lens to pull. We are going to use the minimum blank size (MBS) formula, this formula gives us the minimum blank needed for the particular frame that was chosen.  In order to find the MBS we need to know the decentration.  Decentration is moving the lens so the OC is placed over the patient’s eye.  Although this is done on the finishing side of the lab we need to know the decentration to find the MBS.

    Decentratioin

    total_dec

    dec_per

    MBS

    mbs_total

    mbs_per

    Example:

    What is the minimum blank needed for the following:

    A= 50

    B = 40

    ED = 50

    DBL = 16

    PD = 54

    mbs_img

    mbs_img2

    The image above shows a frame with the same frame dimentions as in the example above and a 70 mm lens blank super imposed over the lens.  The OC of the lens is decentered in 6 mm (decentration per lens).  If we measure from the optical center of the lens to the furthest point of the frame, we will get a 31 mm radius.  Since a lens blank is round we double 31 mm and we get 62 mm (the two millimeters in the MBS formula is for any error that may occur).  If you use the numbers above you will see that the MBS needed is a 62 plus the  2 mm for error.

    Please note surfacing labs will use a smaller lens blanks than what the minimum calls for.  They will grind the OC off center in order for the lens to make cut-out.  If this is done then the minimum blank needed is the ED plus two millimeters, no need to account for decentration since it was ground in.  Please note that this formula is only 100% accurate if the ED lies on the 180° line.

    Putting it all together

    Example:

    What is the minimum blank and base curve needed for the following:

    A = 56                        Rx

    B = 40                        OD -2.00 -1.00 X 45

    ED = 58                     OS +2.00 -2.00 X 135

    DBL  = 15

    PD = 65

    The right eye needs to be put on a 4.75 base curve and the left eye needs to be put on a 7.00 base curve and we need at least a 66 mm blank.

    ]]>     http://www.opticaled.com/articles/processing/2009/06/selecting-the-correct-lens-blank/feed/ 0 Ray Tracing of Thick Lenses http://www.opticaled.com/articles/ray-tracing/2007/04/ray-tracing-of-thick-lenses/ http://www.opticaled.com/articles/ray-tracing/2007/04/ray-tracing-of-thick-lenses/#comments Sun, 01 Apr 2007 22:36:17 +0000 Brent McCardle http://www.opticaled.com/articles/?p=57 The purpose of this article is to build on what was discussed inRay Tracing of Thin Lenses. If you have not read that article it would be wise to do so. This article assumes a basic understanding of mathematical concepts and an understanding of some optics.]]> The purpose of this article is to build on what was discussed inRay Tracing of Thin Lenses. If you have not read that article it would be wise to do so. This article assumes a basic understanding of mathematical concepts and an understanding of some optics.

    Tracing a thick lens is different because the ray of light is going to be refracted at more than just one meridian. A Thick Lens is defined as a lens in which the separation between the two surfaces is too great to be ignored. For example the lens power measured from the front of the lens will be different than if it were measured from the back, due to the thickness.


    A couple of points to note, all rays that are traced will be paraxial, close to the axis. Usually opticians tend to give a positive radius to the front curve and a negative radius to the back curve, because the powers are positive and negative respectively. We need to understand that a positive radius is defined as the center of curvature being to the right of the lens and a negative radius is when the center of curvature is to the left.


    Since we are dealing with thick lenses we need to know where to measure the focal length, object distance, and image distance. We will not measure these items from the center of the lens or from the lens vertices’s, but rather from the principal planes (). Principal planes are where refraction is assumed to occur

    Now we need to find the equivalent power of the lens. The equivalent power is assigning one power to a two lens system. For example, if you read a thick lens in the focimeter on the front and then read the back there would be two different powers. The equivalent power is the position of the principal planes with respect to the first and second focal points. Once we find the equivalent power we can then find the principal planes.

    We will explain in more detail later, just understand that

    is measured from the front vertex and
    is measured from the back vertex. Also understand that
    and
    are measured from
    and
    respectively.

    Formulas to trace a thick lens

    Radius of Curvature

    is the radius of curvature in Meters

    is the refractive index of the the refracting side of the surface

    is the refractive index of the incident side of the surface

    D is the power of the surface in Diopters

    Front Vertex Power

    is the front vertex power

    is the power of the front of the lens

    is the power of the back of the lens

    is the thickness in Meters

    is the refractive

    Back Vertex Power

    is the back vertex power

    is the power of the front of the lens

    is the power of the back of the lens

    is the thickness measured in Meters

    is the refractive index

    Primary Focal Point ()

    is the primary focal point measured in Meters

    is the front vertex power

    Secondary Focal Point ()

    is the secondary focal point measured in Meters

    is the back vertex power

    Equivalent Power

    is the equivalent power

    power of the front of the lens

    power of the back of the lens

    is the thickness measured in Meters

    is the refractive index

    Equivalent Focal Point()

    is equivalent focal point measured in Meters

    is the equivalent power

    Image Distance

    Linear Magnification

    Image Height

    Now we are armed with the information we need to trace a thick lens.

    Example: A plastic( n = 1.50) lens with a thickness of 20mm is surrounded by air. It has a front power of +20.00D and a back power of -3.00D. If and object is placed 100mm in front of the lens where does the image form, how much magnification is present, and is the image inverted or erect. Find the principal planes, equivalent power, linear magnification, image distance, and image height.

    Step

    1: Find the radius of curvature for both surfaces.

      * Note the refracting ray is in air and the incident ray is in the material

    Step 2: Find the back and front vertex powers.

    Step 3: Find the equivalent power.

    Step 4: Find the primary, secondary, and equivalent focal points.

    Now
    we are ready to draw the lens, optical axis, principal planes,
    ,
    and
    .
    First you need to draw the optical axis, grab a compass and draw
    (25mm).

    Second from the apex of the 1st surface
    measure and draw a point that is equal to the thickness of your
    lens(20mm) and with your compass draw
    (167mm)
    from that point.

    Now that we have the lens drawn we need to mark our
    primary focal point, secondary focal point, and our principal planes
    on our optical axis.

    We are now ready to finish drawing our optical axis,
    but we need to know how long to draw it. We know that our object is
    located -100mm from the front vertex, however we will measure it from
    p1.

    Take the difference between f1 to front vertex and p1 to
    f1. Then subtract that number from your object distance.

    -58 – (-56) = -2

    -100 – (-2) = -98

    -98 is the distance from p1 to your object or your
    object distance. Now we will use our equivalent power to find the
    image distance

    So the image is located131mm from p2. Now let us find
    linear magnification.

    So now we will trace three rays to show that our image
    ends up 131mm from p1, it will be inverted, and it will be magnified.
    We know this because of the result of our linear magnification
    formula.

    Optical Formulas Tutorial, 2nd Ed., Stoner, Perkins, Ferfuson
    ]]>     http://www.opticaled.com/articles/ray-tracing/2007/04/ray-tracing-of-thick-lenses/feed/ 0