Important Formulas

Question 1

Focal length

Answer 1

Focal length = inverse power of the lens in meters

Question 2

Prentice’s rule

Answer 2

Prism power = power of the lens multiplied by amount of decentration distance in centimeters

Prism=F x dcm

If using this equation to solve for decentration distance, remember to convert centimeters to millimeters.

Question 3

How to calculate decentration of the lens before putting it in a lens blocker?

Answer 3

Distance between center of eyes (DBC) = Distance between the lenses (DBL) + the eye size measurement of one lens (A measurement).

DBC = DBL + eye size

After finding DBC use the following formula to find decentration of the lens:

(DBC - PD)/2 = decentration of lens. Where PD is pupillary distance

Question 4

Total set inset

Answer 4

To calculate the centration of a flat-top bifocal, you will need to determine the total seg inset required. The formula is the following:

Total set inset = (eye size + DBL) - near PD. Where DBL is distance between the lenses and near PD is the near pupillary distance.

Question 5

Horizontal decentration

Answer 5

Horizontal decentration = (DBC / 2) - monocular PD. Where DBC is distance between center of the eyes and monocular PD is monocular pupillary distance.

Remember DBC = DBL + eye size. Where eye size is equal to the A measurement of one lens.

Question 6

MRP vertical raise or drop

Answer 6

MRP raise or drop is when you vertically decenter a lens when there is an imbalance between the two eyes. the formula is:

MRP vertical raise or drop = MRP - (B measurement / 2)

Question 7

Seg height change

Answer 7

If the person is wearing a bifocal and has a vertical imbalance, you’ll calculate the amount of vertical seg drop or raise needed. The formula is the following:

Seg height change = seg height - (B measurement / 2)

Question 8

How to calculate compensated or effective power

Answer 8

(D1 x D1) / 1000 = A. Where D1 is the original lens power

A x the change in vertex in millimeters = B

D1 +/- B = compensated or effective power. Where adding or subtracting B depends on whether or not the vertex change would result in a stronger or weaker compensated or effective power

Question 9

Vergence of light leaving the lens is equal to what?

Answer 9

Vergence of light enter a lens + lens power = vergence of light leaving the lens

Question 10

If your client has a prescription for progressive or bifocal lenses and decides to purchase a single vision pair for reading only, how will you arrive at the new prescription?

Answer 10

It’s easy: Simply add the value of the add to the sphere for each eye.

For example, let’s say your client has this original prescription:
O.D. +3.00 -0.50 x 080
O.S. +2.00 -1.00 x 010
Add: +2.00

The single-vision prescription for reading only is:
O.D. +5.00 -0.50 x 080
O.S. +4.00 -1.00 x 010

Question 11

If your client has a prescription for progressive or trifocal lenses and decides to purchase a single vision pair to use while working on the computer, how will you arrive at the new prescription?

Answer 11

Typically, you’ll want to change the prescription to allow the person to see things at the intermediate or midrange distance, since computer monitors are usually about 22 to 24 inches from the eyes instead of the 16 inches used for reading books.

So the add for computer distance is one-half the add for the reading prescription.

what would the new prescription for a single-vision computer or intermediate-distance pair of lenses be?

For example, let’s say your client has this original prescription:
O.D. +3.00 -0.50 x 080
O.S. +2.00 -1.00 x 010
Add: +2.00

What would the new prescription for a single-vision computer or intermediate-distance pair of lenses be?
O.D. +4.00 -0.50 x 080
O.S. +3.00 -1.00 x 010

Question 12

When using a lensometer, how do you calculate the true power of the cylinder?

Answer 12

Follow instructions in lesson 13 for using a lensometer. Remember to focus the lensometer so that the thinly spaced lines come into focus first. Record the value for the sphere power. This is the true power for the sphere. Next, focus the lensometer so that the widely spaced lines come into focus. Record this value for the cylinder. This value is not the true power of the cylinder. To find the true power of the cylinder, remember this formula:

True cylinder power = Sphere power - cylinder power listed on lensometer.

Also note that if the lines for the sphere and cylinder come into focus at the same time, then there is no cylinder in the lens.

Question 13

How to measure add with a lensometer

Answer 13

Measure far distance power of the sphere, then measure the near distance power of the sphere. The add will be equal to the difference between the two values.

Add = Far distance power of lens - Near distance power of the lens.

Question 14

One use of Prentice’s formula is to calculate induced prism. Induce prism is what? How do you solve the following example?

Example:
Brad just got his new glasses, and the optical center is misaligned 2 millimeters nasally (or .2 centimeters) in each eye. How much induced prism does he have in each eye?

Here is Brad’s prescription:
O.D. +5.00 sph
O.S. -2.50 sph

Answer 14

Induced prism is the amount of prismatic effect a patient will experience if the lenses are not cut with the optical center at the patient’s PD.

Example:
Brad just got his new glasses, and the optical center is misaligned 2 millimeters nasally (or .2 centimeters) in each eye. How much induced prism does he have in each eye?

Here is Brad’s prescription:
O.D. +5.00 sph
O.S. -2.50 sph

For right eye
Prism = F x dcm
Prism = 5.00 x .2 cm
Prism = 5.00 x .2 = 1Δ

For left eye
Prism = F x dcm
Prism = 2.5 x .2 cm
Prism = 2.5 x .2 = .5Δ

You’ll indicate the direction of this effect according to the direction of the base. So it will be base out, base in, base up, or base down.

The right eye (OD) is looking through the part of the minus lens in which the base of the prism points toward his nose. This means that he’s looking through a base-in prism

The left eye (OS) is looking through the part of his plus lens in which the prism base points toward his temple. So on the left side, he’s looking through a base-out prism.

Brad’s right eye is getting a 1 prism diopter base-in effect because his right pupil is now looking through the part of the prism where the base is facing his nose, and his left eye is getting a .5 prism diopter base-out effect since his pupil is now looking through the part of the prism where the base is toward his ear

Question 15

Let’s say you’re cutting down lenses for a client, and there’s a big difference between the client’s PD and the geometric centers of the frame the person chose. (This occurs if the frame they want is really too large, but they insist.) In this case, you may have to figure out how much to decenter the lenses so you won’t induce prism in the new glasses.

How do you figure out how much to decenter the lenses so you won’t induce prism in the new glasses?

Answer 15

First, you’ll need to calculate the difference between your client’s PD and the frame’s geometric center. Assume that the person’s pupils are exactly the same distance apart. When the eyes are symmetrical, you know that the distance from the pupil of one eye to the geometric center of the lens for that eye is the same as for the other eye.

As an example, let’s consider a client named Janet. Let’s say that Janet’s PD is 60, and the distance between the geometric centers of her frame is 70 mm. The total difference between them is 10 mm. But that’s for both eyes. So the amount of decentration for each eye to the geometric center of that lens should be half this number, which is 5 mm.

Question 16

Imagine that you’re looking 5 mm (or .5 cm) above the optical center of a lens with this prescription:

+2.00 -0.50 x 180

How much verticle prism will you experience?

We’ll start with our optical cross. It has +1.50 on the vertical meridian and +2.00 on the horizontal meridian.

Answer 16

Imagine that you’re looking 5 mm (or .5 cm) above the optical center of a lens with this prescription:

+2.00 -0.50 x 180

How much verticle prism will you experience?

We’ll start with our optical cross. It has +1.50 on the vertical meridian and +2.00 on the horizontal meridian.

Prism = Power x decentration in centimeters
Prism = +1.50 x .5 cm
Prism = +0.75Δ BD

Question 17

Vogel’s formula

Answer 17

The optimal base curve for a lens. For a plus lens, the optimal base curve = spherical equivalent + 6 diopters. For a minus lens, the optimal base curve = (spherical equivalent / 2) + 6 diopters. Spherical equivalent = sphere + (cylinder amount / 2)

Question 18

How to calculate the power of the lens cylinder in a meridian that is a certain distance in degrees away from either of the principal axis meridians. There are two ways of doing this: One is the percentage method (quick), and the other is using a formula. What is the percentage method?

Answer 18

The percentage method is as follows:

• Remember that at 30 degrees away from the axis, you will have 25 percent of the cylinder power.
• Remember that at 45 degrees away from the axis, you will have 50 percent of the cylinder power.
• Remember that at 60 degrees away from the axis, you will have 75 percent of the cylinder power.

We are not talking about the power of the lens at axis 30, 45, or 60, but the power of the cylinder 30, 45, or 60 degrees away from either axis. Your answer to this question will be a percentage of the cylinder value added to the total power.

Example: Given a lens -4.00 -1.00 x 150, how would you correct a lens that was not cut correctly, and you need to know the power of the lens at the axis of 180?

Answer:
The difference in the axis between 180 and 150 is 30 degrees. You need to know the power of the cylinder 30 degrees away, which is 25 percent of the cylinder power. So, 25 percent of -1.00 is -0.25. Now add this -0.25 to the spherical power of -4.00, and your answer is -4.25 @ axis 180.

Question 19

Any time the prescribed axis of a lens is at 045 or 135 (called oblique astigmatism), you’ll determine the power of the lens in the 90-degree axis by…

Answer 19

Any time the prescribed axis of a lens is at 045 or 135 (called oblique astigmatism), you’ll determine the power of the lens in the 90-degree axis by taking half the cylinder and adding it to the sphere.

Example: For the prescription -2.00 -1.00 x 045, find the power of the lens in the 90-degree axis.

Answer = -2.50 D.S.

Again, just take half the cylinder, and add it to the sphere. So, the power at the 90-degree meridian is -0.50.

Question 20

What is transposition? List the steps to do it.

Answer 20

Converting minus (negative) cylinder form to plus (positive) cylinder form or vice versa.

The steps involved are:

1. Take the sphere number (including the sign) and the cylinder number (including the sign), and add them together to get your new sphere.
2. Take the cylinder number, and change its sign to the opposite sign to come up with the new cylinder.
3. Take the axis, and change it 90 degrees by either adding 90 or subtracting 90 from it. Remember that the axis cannot be 0 or greater than 180. This will tell you whether to add or subtract.

Example:
Old prescription = OS: -1.50 +3.00 x 005
New transposed prescription = OS: +1.50 -3.00 x 095