Week 1.06 Spectacle Mag

Question 1

Spectacle mag has what 2 components

Answer 1

Shape factor
Power factor

Question 2

What is the formula for spectacle magnification

Answer 2

Spectacle magnification = shape factor x power factor

Question 3

What is the shape factor formula

Answer 3

1/ 1- (t/n’ x F1)

Question 4

What is the power factor formula

Answer 4

Power factor = 1/ 1- (d x BVP)

Question 5

What is the spectacle magnification if a lens has:
BVP: +4.00DS
F1: +8.00DS
t: 4mm
d:15mm
n: 1.498

Answer 5

Use shape factor formula first so 1/1-(t/n x F1) = 1.022

Then use power factor: 1/1-(d x BVP) = 1.064

1.022 x 1.065 =1.087

This shows this lens will magnify the image as greater than 1

Question 6

What is lens magnification for a lens that BVP is -3.00DS, F1 is +4.00DS, t=2mm, d=15mm n=1.498

Answer 6

Shape factor = 1.005
Power factor = 0.957

Spectacle magnification = 0.962

This will minify the image

Question 7

What happens to the shape and power factor if we increase base curve (F1)

Answer 7

Shape factor provides increased magnification as base curve increases in positive power.
Power factor remains constant

Question 8

What happens to shape and power factor when increasing centre thickness, t

Answer 8

Shape factor provides increasing magnification as centre thickness increases. Power factor remains constant
Spectacle magnification increases as t increases

Question 9

In myopes/hyperopes what is the effect of increasing BVD on BVP

Answer 9

Myopes – increasing BVD = increase negative correcting lens
Hyperopes – increasing BVD = reduction in power of the positive correcting lens

Question 10

What is the formula to work out new BVP of a changed BVD

Answer 10

Reciprocal of BVP
Subtract original BVD
Add new BVD
Do reciprocal to find BVP

Or alternatively use the formula
BVP = F/ 1-xF

F - original lens power
X - difference in BVD in metres (old BVD - new BVD)

Question 11

Px is refracted with a trial frame at a BVD of 10mm. The sphere found is -12.00DS. The px chooses a frame with a BVD of 14mm. What is the BVP required to correct the px?

Answer 11

X= 0.010 - 0.014 = -0.004
BVP = -12.00/ 1- (-0.004 x -12.00)
BVP = -12.61DS

Question 12

Px is refracted with a trial frame at a BVD of 10mm. The sphere found is +8.00DS. Px chooses a frame with BVD of 14mm. What is the BVP required to correct px?

Answer 12

X = 0.010-0.014=-0.004
BVP = +8.00/1-(-0.004x+8.00)
BVP = +7.75DS

Question 13

What’s the effect of increasing BVD in a positive rx and a negative rx

Answer 13

Negative rx - increases BVP, DECREASES spectacle magnification
Positive rx - decreases BVP, INCREASES spectacle magnification