Transpose a prescription written in plus cylinder form to minus cylinder form as follows:

 1. Add the sphere and cylinder powers to determine the new sphere power.
2. Change the sign of the cylinder.
3. Change the axis by 90 degrees.

Example:

Transpose -3.00 +2.00 x 30

1. Add the sphere and cylinder powers to determine the new sphere power.

(-3.00) + (+2.00) = -1.00

2. Change the sign of the cylinder

-2.00

3. Change the axis by 90 degrees.

120

The transposed prescription is:
-1.00 -2.00 x 120

write the rx for the follwing true power

(REMEMBER BRIAN STROME ) 

TP+6.00@90  +4.75@180

(-6.00+4.75=-1.25)

RX +6.00-1.25X90

TP-10.00@180  -6.50@090 

(+10.00-6.50=+3.50)

RX -10.00 +3.50X180

TP +1.00@168 -0.50@078

(-1.00-0.50=-1.50)

RX +1.00 -1.50X168

TP plano@045 -3.00@135

(0.00 -3.00=-3.00)

RX PLO -3.00X045

TP -3.25@090 PLANO@180

(+3.25 0.00=+3.25)

RX-3.25 +3.25X090

TP -3.25@135 -4.50@045

(+3.25-4.50=-1.25)

RX -3.25 -1.25X135

TP -8.00@055 -3.50@145

(+8.00-3.50=+4.50)

RX -8.00+4.50X055

determine the true power of these lenses

(REMEMBER BRIGE AND LOVE STORE )MOVE AXIS 90DGRESS

RX +6.00-1.25X008

+6.00@008(BRIGE)

+6.00-1.25(LOVE STORE)=+4.75@098

RX -10.00+3.50X096

-10.00@096

-6.50@006

RX +1.00-1.50X168

+1.00@168

-0.50@078

RX PLANO-3.00X045

PLO@045

-3.00@135

axis NEVER TO PASS 180

so if axis is 90 add 90 (90+90)=180

if axis is 135 (take the first two number add togther (1+3=4)axis now is 45

if axis is 101(take the first two number add togther

(1+0=1)axis is now 11

if axis is 163(take the first two number add togther

(1+6=7)axis is now 73

A+DBL=FPD

FPD-PPD/2=HORIZONTAL DEC

B/2-OCHT=VERTICAL DEC

2XHIGHEST DEC+ED=MBS

A+DBL =FPD

FPD-(NEAR)PPD/2=HORIZONTAL

B/2-SEG HT =VERTICAL

2XHIGHEST DEC+ED=MBS

if signs are the same in the sph and cly(rx or true power)add them together and keep  the sign

if sign are not the same in the sph and cly(rx or true power )subtract the different between the too and keep the sign to the biggest number

We are going to take a look at the prescription +1.50 -0.50 x090. What are the major power/meridian combinations?

Well, the +1.50 and the axis 090 go together. +1.50 -0.50 = +1.00, and 090 + 90 = 180, so the other leg of the optical cross is +1.00 at 180.

OK, in one major meridian we have a power of +1.50, and in the other major meridian we have a power of +1.00. In-between those two meridians the power on the lens is not really good, accurate power, but what is there ranges in-between +1.00 and +1.50.

If this lens were the one we are discussing, then the meridian with the +1.00 power would have that nice clear line focus at one meter, [since a +1.00D power has a focal length of 1/1D = 1m -- remember?] and the meridian with the +1.50 power would have that nice clear line focus at 0.67m [1/1.5 = 0.666 . . .]

If I had that +1.50 -0.50 x 180 lens outside in the nice bright sun I would not get a round image of the sun that I could use to start a fire. Instead, when I held it one meter from the paper I would have a line image, and when I held it 2/3 meter from the paper I would get another line image perpendicular to the first line image. What would I have in-between these two distances? I would have a varying amount of fuzzy image, going from the first clear line, to elliptical in one direction, through round, to elliptical in the other direction, to the other clear line.

There is no good power between the two major meridians. That is why, when you look at a spherocylindrical lens in the foci meter, you do not get clear images for powers between the sphere lines and the cylinder lines.

The sun is a round object, but the images that we get from the toric lens are line images and, at one point, a fuzzy round image. The spherocylindrical lens distorts the image of the sun. At one particular length, however, the image of the sun IS round, even though it is fuzzy. This distance corresponds to what we call the circle of least confusion which is really the position of no distortion. This focal length can be converted to a power. The power that we get is called the spherical equivalent of the lens. We can find the spherical equivalent by taking the average power on the lens: the average of the two major meridian powers.

OK. So. The lens in our example has +1.50 in one meridian and +1.00 in the other. The average of +1.50 and +1.00 is [(+1.50) + (+1.00)] [image]2 = +1.25. So the focal length of +1.25 is the spherical equivalent of the lens, the focal length corresponding to the power of +1.25 is 0.8m, so the circle of least confusion is located at 0.8m from the lens.

Another Example:
+1.00 -3.00 x 90

• okay the +1.00 and the 90 go together.
• at 180 the power is -2.00
• so the average is [(+1.00) + (-2.00)] [image]2 = -0.50
• so our spherical equivalent is  -0.50
   

   

[image]
Optical cross - used to diagram prescriptions
 
 
 

Proof of Prentice's rule.

[image]

Distance(cm) X power = Prism , All it is is a way to determine the amount of prism you have induced taking into account the amount of power and the distance it is from where you actually need that power's optical center.
You figure you have someone wearing a +4 sph. OU with a pupilary distance of 60 mm but you find the PD in the frame is 64 then by using the formula you can find out how much that 4 mm of distance is inducing prism. .4 cm X 4 (D^) = 1.6^ (prism)total or .8^ in each eye.
 

Prentice's rule
A method of determining prism power at any point on a lens. Prism power equals the product of the dioptric power and the distance, in centimeters, from the optical center.
 
 
 

Now let's try to put this in a real question.
 

Example 1)   The lab ground the following lens at 64mm PD.  The patient has a actual PD of 60mm.  For the following perscription what is the amount of prism induced?

RX:  OD  +5.00 -2.00 X 90
        OS +3.00 -1.50 X 90

First step to this is to obtain the equivalent sphere power, so let's put this on a optical cross.
 

Right Lens: [image]

Left Lens: [image]

Okay now lets look at our decentration of 4mm.  Now since that is a binocular (both eyes) measurment lets divide that in have so we have a monocular decentration of 2mm.  With that lets figure our answer for both eyes:

 

Right Eye Prism:  Prentice Rule is  Prism = Decentration (cm) X Diopters of power In the Right eye we figured our power at +3.00 and our decentration at 2mm or .2cm so Prism = (.2cm) X (+3.00D Power) Prism = 0.6 Now lets figure out where this 0.06 prism is, Base in or Base out? [image] Looking at where the eyes lineup, where is the base facing?  is it facing the nose?  No, then it is facing out away from the nose. So our base is base out.   Our answer is: 0.60D Base Out

Left Eye Prism:  Prentice Rule is  Prism = Decentration (cm) X Diopters of power In the Left eye we figured our power at +1.50 and our decentration at 2mm or .2cm so Prism = (.2cm) X (+1.50D Power) Prism = 0.3 Now lets figure out where this 0.03 prism is, Base in or Base out? [image] Looking at where the eyes lineup, where is the base facing?  is it facing the nose?  No, then it is facing out away from the nose. So our base is base out.   Our answer is: 0.30D Base Out

visible light spectrum

[image][image]

• Wavelengths are measured in Nanometers (nm)

• Wavelengths are measured in Nanometers (nm)

Of the radiation that reaches the earth's surface 60% is infrared, 37% visible, and 3% Ultraviolet (UV). The UV spectrum composes the 200 to 400 nanometers (nm) wavelength band, and is divided into 3 groups:

• UVA is made up of wavelengths 320 to 400 nm
• UVB 280 to 320 nm
• UVC 200 to 280 nm

The earth's atmosphere absorbs UVC rays and only UVA and UVB reach the earth's surface.

• UVB rays are more important in the development of skin cancer than UVA.
• UVA rays cause aging and wrinkling of the skin, but also increase the damaging effects of UVB.

Table 1: The Regions of the Electromagnetic Spectrum

Table 2: Colour, frequency and wavelength of light

Lens Guide Correction and Thickness

[image]

Lens Correction
 

Material Guide
 

Simple decentering edge thickness guide
 

Lens Types
 

Multi-Focal Lens Guide
 

Lens Coatings

A simple fact is that a clear uncoated lens will only allow 92% of the afailable light to pass through while the other 8% is lost in the form of reflected glare.  Not only does this loss of light diminsh visual acuity, it also creates troublesome reflections which tend to mask the wearer's eyes.
Color coatings can add cosmetic versatility to lens.  Color coating can also add reflectance to a glass or plastic surface as in the case of gradient or solid mirror coatings.

The Metric System

The optical industry is a world industry and as such it is gauged by the world standard of measurement which is metric.  Temple lengths, eye sizes, pd's, lens sizes, blank sizes, and segment heights are all measured in millimeters.  To be an effective dispensing technician a good working knowledge of the metric system is necessary.  To be successful on the exam it is often required to convert a correct answer in millimeter form into corresponding meter form.

The Metric Values:
 

CONVERSION:

Meters to Millimeters:

Move the decimal three places to the right by multiplying the Meter Value by 1000.

Example - Convert 0.012 M to Millimeters
                  (0.012M x 1000 = 12 mm)

                  Convert 2 M to Millimeters)
                  (2M x 1000 = 2000 mm)

Millimeters to Meters:

Move the decimal three places to the left by dividing the Millimeter Value by 1000.

Example - Convert 3000 mm to Meters.
                  (3000 divided by 1000 = 3M)

                  Convert 1200 mm to Meters.
                  (1200 divided by 1000 = 1.2M)

Meters to Centimeters:

Move the decimal two places to the right by multiplying the Meter Value by 100.

Example - Convert 0.012 M to Centimeters
                  (0.012M x 1000 = 1.2 cm)

                  Convert 4 M to Centimeters
                  (4M x 100 = 400 cm)

Centimeters to Inches:

There are 2.54 centimeters per inch, so in order to convert you must divide the number of centimeters to be converted by 2.54

Example - Convert 36 cm to Inches
                  (36 divided by 2.54 = 14.17 Inches)

                  Convert 300 cm to Inches
                  (300 divided by 2.54 = 118.11 Inches)

Millimeters to Inches:

There are 25.4 Millimeters per inch, so in order to convert you must divide the number of Millimeters to be converted by 25.4

Example - Convert 500 Millimeters to Inches.
                  (500 divided by 25.4 = 19.68 Inches)

                  Convert 72 Millimeters to Inches.
                  (72 divided by 25.4 = 2.83 Inches)

Inches to Millimeters:

In order to go from Inches to Millimeters it is necessary to multiply the number of inches to be converted by 25.4

Example - Convert 17 Inches to Millimeters.
                  (17 x 25.4 = 431.8 Millimeters)

                  Convert 2 Inches to Millimeters.
                  (2 x 25.4 = 50.8 Millimeters)
 

** Remember -The Metric System works on multiples of 10, Based on the standard Meter.

[image]

Cornea

The clear front window to the eye. It is the most important part of the focusing system of the eye and brings objects into focus on the retina.

Anterior Chamber

The space between the cornea and the iris which is filled with fluid aqueous.

Chamber Angle

The structure formed by the junction of the iris with the corneo-sclera. Normal fluid drainage from the eye occurs through the trabecular meshwork which lies in the chamber angle. Malfunction of the chamber angle is the cause of glaucoma (high pressure in the eye).

Iris

The colored tissue behind the cornea - color varies from pale blue to dark brown.

Pupil

The round hole in the iris. Its size is automatically varied to regulate the amount of light entering the eye.

Lens

A clear biconvex structure behind the iris which works in conjunction with the cornea to focus light onto the retina. When the lens becomes opaque or cloudy, it is called a cataract.

Ciliary Body

A ring of tissue lying at the base of the iris. Its muscle fibers serve to change the curvature of the lens and thereby provide fine focusing of light onto the retina.

Zonules

Threadlike filamentous attachments which hold the lens in place in the eye.

Vitreous

The clear jelly which fills the space between the lens and the retina. Eighty percent of the eye is filled with vitreous.

Sclera

The white, tough protective outer wall of the eye.

Choroid

A spongy layer filled with blood vessels. It lies between the sclera and the retina. The choroid nourishes the outer layers of the retina.

Retina

The vital thin layer of tissue composed of millions of visual cells which lines the inside back two-thirds of the eye. The retina is analogous to a film in a camera. It receives light and sends tiny electrical impulses to the brain to give sight.

Optic Nerve

A cable-like structure composed of thousands of nerve fibers which carry impulses from the retina to the brain where visual perception occurs.

Macula

The highly developed central zone of the retina. The macula gives critical vision for reading and discrimination small objects.

[image][image]

Muscles of the Eye

The Eye has the ability to move in a variety of direction.  It achieves this great versatility by vitrue of the
6 Musclesthat are attached to the sclerotic coating of the globe.  Working together the following muscles achieve rotation:

Six Muscles of the Eye

Superior Rectus	Turns the eye upwards and inwards
Superior Oblique	Turns the eye downwards and outwards
Internal Rectus	Turns the eye inwards and toward the nose
Inferior Oblique	Turns the eye upwards and outwards
Inferior Rectus	Turns the eye downards and inwards
External Rectus	Turns the eye outwards

When these muscles are working correctly in conjunction with one another the eyes can achieve great range for distance vision as well as Convergence necessary to achieve Binocular Vision.  Binocular vision refers to the two eyes Fusing their two images to form a single three dimensional image that the brain can readily interpret

Light Wave Theory

A fundamental understanding of light is important to the optical technician since it is throught the vehicle of light that the whole system of vision functions.  There are two separate theories of light and while neither one totally explains the mysterious properties of light both are accepted since each approach explains some feature of light which the other cannot.  The two theories which we will briefly outline are known as the Corpuscular Theory and the Electromagnetic Wave Theory.

CORPUSCULAR THEORY

ELECTROMAGNETIC THEORY

1 inch is = 25.4 mm so
6 inches X 25.4mm = 152.4mm
Round that to the closest correct answer and your answer would be D) 150mm

Focal length is equal to 1/diopter value and your answer is in meters
1/2 = 0.5M
0.5M = 50cm
0.5M = 500mm
your answer would be C) 500mm

Wavelengths are measured in nanometers. your answer would be D) Nanometers

• UVA is made up of wavelengths 320 to 400 nm
• UVB 280 to 320 nm
• UVC 200 to 280 nm
• Infrared 1000nm - 3000nm
  Your answer would be A) UVC 280 to 320 nm

Light waves out of phase would either Destroy one another or Reduce one another.
Light waves in phase would Compound one another.
Your answer would be C) Compound one another

ANSI standard for cylinder axis tolerance is Plus or Minus 3 degrees
Your answer is C) Plus or Minus 3 degrees

Since the thickest part of a plus lens is in the center, the closer the center is to the edge of the frame
the thicker that part of the edge will be.  So if a plus lens is decentered out twords the temples the finished lens would be thicker temporally.
Your answer is B) Thicker temporally

Since the thickest part of a minus lens is on the edge, the closer the center is to the edge of the frame
the thinner that part of the edge will be.  So if a minus lens is decentered out twords the temples the finished lens would be thicker nasally.
Your answer is C) Thicker nasally

A clear uncoated lens will only allow 92% of the available light to pass through while the other 8% is lost in the form of reflected glare.
Your answer is C) 8%

• A Meniscus lens has one convex side and one concave side
• A Biconvex and two convex sides
• A Biconcave has two concave sides
  Your answer in this case would be C) Biconcave
   

The tonometer is used in order to ensure a person's optic nerves are healthy. Optometrists check the pressure placed on them by the fluid in the eyes. This pressure is called intraocular pressure and should measure between 10 mmHg and 21 mmHg. Measurements that are higher than normal can be a sign of early glaucoma or retinal detachment.

The tool used to measure intraocular pressure is called a tonometer. Is used to measure the production of aqueous humor, the liquid found inside the eye, and the rate at which it drains into the tissue surrounding the cornea. Usually, the tests performed are simple and require only a short span of time to complete.

• Converge, in front of

• 1 meter / 1 millimeter

• 2.5 mm

• Reflection

• 3D base out each eye

• 1(FORMULE TO YOU power X dec /10)

Once it is found that either wanted or unwanted prism is present in any given lens and the direction of the base is determined, the power of the prism can then be calculated utilizing a simple formula called Prentice's Rule. It is important to become familiar with Prentice's Rule since it is used many times each day as a routine matter in the optical dispensary.

Prentice's Rule states: The power of the prism is equal to the power of the lens in diopters times the amount of decentration in millimeters divided by 10.

Stated Algebraically:

where:

[image] = prism diopters
F = power of the lens in diopters
dec = decentration or distance in mm away from the optical center of the lens.

Example 1: How much prism is induced 4 mm away from the optical center of a +3.00 D spherical lens?

Applying Prentice's Rule, F = +3.00 and dec. = 4.0 mm
so:

[image] = F x dec/10
    = 3.00 x 4/10
    = 1.2 [image]

Example 2: A patient's PD was mistakenly noted to be 62 mm when the actual measurement is 66 mm. If the power of the lenses are ­2.00 diopters in each eye, how much prism will be induced in the finished spectacles and in which direction is the base?

How do we know the direction of the prism is base out? The optical centers were positioned a total of 4 mm narrower than the actual PD. By referring to the figure D on page 13 which illustrates this situation with a minus lens. It can be observed that the direction of the base of the prism through which the patient is looking is