Optical Ed

Archive for the ‘Problem Solving’ Category

Understanding Prism

Wednesday, November 18th, 2009

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. 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

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Compensated and Effective Power

Wednesday, November 11th, 2009

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.

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Oblique Meridian

Tuesday, November 10th, 2009

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 90^th 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 90^th 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.
$\ sin(40)^2 = 0.413176$
$sin(140)^2 = 0.413176$

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

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

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

$\ F_{\theta} = -10.87\ D$

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Back Vertex Power

Tuesday, November 10th, 2009

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

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

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

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

$\ B_v = 16.03 + (-6.00)$

$\ B_v = +10.03$

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