Unit I - Problem Solving

Question 1

If the right temple is angled in, this adjustment will:

Answer 1

Push the right side of the front away from the right eye.

Question 2

A ____ prism may be temporarily applied to a lens to evaluate patient acceptance of a correction for vertical imbalance.

Answer 2

Fresnel

Question 3

As the pantoscopic tilt is increased, the distance OC should be:

a. lowered
b. raised
c. left alone
d. rotated

Answer 3

a. lowered

Question 4

Spectacle lens powers over ____ diopters require compensation for vertical imbalance.

Answer 4

7

Question 5

As a plus lens moves closer to the eye, it:

a. becomes more effective
b. becomes less effective
c. does not change
d. requires an axis change

Answer 5

b. becomes less effective

Question 6

A new Rx with with a large increase in plus power will require the patient to converge ____ when reading.

a. more
b. less
c. the same
d. not at all

Answer 6

a. more

Question 7

A patient complains that she could see better with her old glasses (Rx is the same). What should you, the optician, do first?

Answer 7

Check the vertex distance and base curve.

Question 8

In regards to troubleshooting problems patients have, what is S.O.A.P.?

Answer 8

Subjective
Objective
Assess
Plan

Question 9

What might uneven vertex distance feel like on a patient’s head?

Answer 9

It feels like there’s more pressure on one side of their face.

Question 10

What are the top 3 main uses of aspheric curves?

Answer 10

1. reduce aberrations in Rxs over +7.00 and -23.00.
2. cosmetic
3. power changes (used in PALs often)

Question 11

What is anisometropia and what are the parameters?

Answer 11

Uneven vision.

Exists when the spherical equivalent of the refraction in both eyes differs by 1.00 D or more.

Question 12

What is the formula for vertex distance?

Answer 12

D2d/1000 = power change
(D2 = D squared)

Question 13

OD: -1.00-1.25x090
OS: +1.75-1.00x120
Add +1.75 OU
Reading level = 10 mm
What's the amount of vertical imbalance?

Answer 13

OD: -1.00 @ 90
OS: +1.50 @ 90
algebraic difference = 2.50
2.50 prism base up

Question 14

Rx +16.00
vertical distance = 13 mm
Fit at 8 mm
What is the compensated Rx?

Answer 14

13 - 8 = 5 mm difference
Glasses are dispensed 5 mm closer than where fit.
CAP = closer add plus
D2d/1000
(+16)(+16)(5mm)/1000 = 1.28 D
16+1.28 = +17.25 D

Question 15

Rx +16.00
vertical distance at exam = 13 mm
Fit at 8 mm
What is the compensated Rx?

Answer 15

13 - 8 = 5 mm difference
Glasses are fit 5 mm closer than during her exam.
CAP = closer add plus
D(squared)(d)/1000
(16x16)(5)/1000
(256)(5)/1000 = 1.28 D
We need to add 1.28 D to their current Rx.
Answer: order +17.25