Vertex Distance, Effective Power, and Compensated Power Flashcards
What is the simplified formula for finding effective power in mm?
EP = (D^2/1000) x mm of change, where D is the dioptric power of the lens, and mm of change is the difference between the vertex distance prescribed and the actual vertex distance.
What is Martin’s Tilt Formula for finding sphere power?
S1 = S{1+[(sin(a))^2/2n]}
Where:
S1 = New sphere power
S = original sphere power
a = Degrees of tilt
n = index of refraction NOTE: Some formulas use 3 instead of 2n, as 2 x index of refraction of CR-39, 1.498, is 3
What is Martin’s Tilt Formula for finding cylinder power?
C1 = S1(tan(a))^2
Where:
C1 = new cylinder power
S1 = new sphere power
a = degrees of tilt
A patient refracted at an 8mm vertex is prescribed a 13.00 sphere. The fitted vertex is 12mm. What should the compensated lens power be, using simplified formula for effective power?
12.37
13^2/1000 x -4 = -0.676
13 - 0.676 = 12.324
As a general rule, round up to the next nearest 1/8th power, so 12.37
Vertex distance compensation for spectacle lenses is most needed for rx’s which begin at what?
+/- 7.00
A patient comes in with an rx requiring a large amount of plus (+12.50) and with a notation of vertex distance 12mm. You find that when fitted with a frame, the lens will sit 14mm from the eyes. What change will be necessary in the lens to be ordered?
Less plus sphere power
If the lenses of a patient wearing -15.00 spheres are brought closer to his eyes, what effect will this have on the effective power of the lens?
Increase minus power effect
Increasing the pantoscopic angle of a cataract lens would most likely do what?
induce plus cyl power and plus sphere power
When a pair of spectacles is tilted so that the bottom of the lenses are are closer to the face than the top, it is called what?
pantoscopic tilt
When a minus lens is moved farther from the eyes, the effective power of the lens becomes what?
Weaker
What is vertex distance?
The straight line distance in mm from the back surface of the lens to the apex of the cornea
When a plus lens is moved closer to the eye, what happens to the effective power?
It decreases
When a minus lens is moved closer to the eye, what happens to the effective power?
It increases
When a minus lens is moved away from the eye, what happens to the effective power?
It decreases
What type of image is produced by a minus lens?
Virtual
Fill in the blanks: If a minus lens is moved closer to the eye, the ___ power increases; therefore, the ____ must _____
Effective, compensated, decrease
Fill in the blanks: If a plus lens is moved closer to the eye, the ___ power decreases; therefore, the ____ must _____
Effective, compensated, increase
When is it necessary to compensate for vertex distance, with regards to prescriptions?
+/- 4.00 D for contact lenses and +/- 7.00 D for spectacles
Which of the following can be used to measure vertex distance? PD Ruler, Distometer, Phoropter
All of the above.
Given an rx of -8.00 sphere refracted at 12mm and the glasses fit at 16mm, what is the effective power of the lens? What is the compensated power of the lens?
D^2/1000 x mm of change = -64/1000 x -4 = 0.256. -8.00 + 0.256 = -7.744, so rounded up it is -7.75 D. The compensated power must be increased by the difference (in this case, 0.25), so it will be -8.25.
Given an rx of -10.00 -2.00 x090 refracted at 9mm and the glasses fit at 12mm, what is the effective power of the lens? What is the compensated power of the lens?
We will need to deal with the sphere power and the cylinder power separately.
First, the sphere power:
D^2/1000 x mm of change = -100/1000 x -3 = 0.3 D. -10.00 + 0.30 = -9.70, so this is our sphere power.
Secondly, the cylinder power. We need to use the total power in the 180th meridian, instead of just the cylinder power:
D^2/1000 x mm of change = -144/1000 x -3 = 0.42 D. -12.00 + 0.42 = -11.58.
So the effective power is -9.70 -1.58 x090. It is important not to round off until the end.
The compensated power of the lens would be -10.30 for the sphere power, since the compensated power must be increased by the difference. For the cylinder power, we must go back to -12.00 and increase it to -12.42, then subtract -9.70 from it to get -2.12.
So the final compensated power is -10.30 -2.12 x090
A +15.00 CR-39 lens has 10 degrees of pantoscopic tilt, and the OC is placed directly in front of the pupil. Using Martin’s Tilt Formulas, what is the EP of this lens?
S1 = S{1+[(sin(a))^2/3]} = 15{1+[(sin(10))^2/3]} = 15.15
C1 = S1(tan(a)^2) = 15.15(tan(10)^2) = 0.47
The axis for tilt is always in the 180, so the EP of the lens is +15.15 +0.47 x180
A -15.00 CR-39 lens has 10 degrees of pantoscopic tilt, and the OC is placed directly in front of the pupil. Using Martin’s Tilt Formulas, what is the EP of this lens?
S1 = S{1+[(sin(a))^2/3]} = -15{1+[(sin(10))^2/3]} = -15.15
C1 = S1(tan(a)^2) = -15.15(tan(10)^2) = -0.47
The axis for tilt is always in the 180, so the EP of the lens is -15.15 -0.47 x180
A -15.00 -4.00 x90 CR-39 lens has 12 degrees of pantoscopic tilt, and the OC is placed directly in front of the pupil. Using Martin’s Tilt Formulas, what is the EP of this lens?
First, we need to find the power in the 180th meridian, which we can do by converting the rx to plus cyl format, which is -19.00 +4.00 x180.
S1 = S{1+[(sin(a))^2/3]} = -19{1+[(sin(12))^2/3]} = -19.274
C1 = S1(tan(a)^2) = -19.274(tan(12)^2) = -0.871
C1 + C = -0.87 + 4.00 = +3.13
The axis for tilt is always in the 180, so the EP of the lens is -19.27 +3.13 x180
