Vertex Distance, Effective Power, and Compensated Power Flashcards

1
Q

What is the simplified formula for finding effective power in mm?

A

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.

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2
Q

What is Martin’s Tilt Formula for finding sphere power?

A

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

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3
Q

What is Martin’s Tilt Formula for finding cylinder power?

A

C1 = S1(tan(a))^2
Where:
C1 = new cylinder power
S1 = new sphere power
a = degrees of tilt

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4
Q

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?

A

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

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5
Q

Vertex distance compensation for spectacle lenses is most needed for rx’s which begin at what?

A

+/- 7.00

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6
Q

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?

A

Less plus sphere power

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7
Q

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?

A

Increase minus power effect

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8
Q

Increasing the pantoscopic angle of a cataract lens would most likely do what?

A

induce plus cyl power and plus sphere power

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9
Q

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?

A

pantoscopic tilt

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10
Q

When a minus lens is moved farther from the eyes, the effective power of the lens becomes what?

A

Weaker

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11
Q

What is vertex distance?

A

The straight line distance in mm from the back surface of the lens to the apex of the cornea

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12
Q

When a plus lens is moved closer to the eye, what happens to the effective power?

A

It decreases

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13
Q

When a minus lens is moved closer to the eye, what happens to the effective power?

A

It increases

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14
Q

When a minus lens is moved away from the eye, what happens to the effective power?

A

It decreases

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15
Q

What type of image is produced by a minus lens?

A

Virtual

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16
Q

Fill in the blanks: If a minus lens is moved closer to the eye, the ___ power increases; therefore, the ____ must _____

A

Effective, compensated, decrease

17
Q

Fill in the blanks: If a plus lens is moved closer to the eye, the ___ power decreases; therefore, the ____ must _____

A

Effective, compensated, increase

18
Q

When is it necessary to compensate for vertex distance, with regards to prescriptions?

A

+/- 4.00 D for contact lenses and +/- 7.00 D for spectacles

19
Q

Which of the following can be used to measure vertex distance? PD Ruler, Distometer, Phoropter

A

All of the above.

20
Q

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?

A

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.

21
Q

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?

A

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

22
Q

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?

A

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

23
Q

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?

A

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

24
Q

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?

A

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