514 Ophthalmic Lenses PAR T TWO

Where

SM = static spectacle magnificationt = center thickness

F1 =front surface lens powern = refractive indexd = distance from the back surface of the lens to

the entrance pupil of the eye (usually vertexdistance plus 3 mm)

F: = back vertex lens powerS = shape factor

Pstat= static power factor

Remole uses Pst.t to point out that this is the powerfactor for an eye that is not moving. However, whencalculating spectacle magnification when the eye ismoving, Remole maintains that the more logical pointof reference is the center of rotation of the eye. "Becausethe entrance pupils of the eye move as the eye rotates,they can't be used as reference points. Instead the centerof rotation must be used.13"This changes the formulafor the eye when it is no longer in the straight-aheadposition. The formula then becomes a "dynamic specta-cle magnification formula" so that it reads:

G =

[

1

)(

1

)1-~;/1 1-sF;

= (S)(pdyn)

Where

G = dynamic spectacle magnificationt = center thickness

F1 = front surface lens powern = refractive indexs = distance from the back surface of the lens to

the center of rotation of the eyeF: = back vertex lens powerS = shape factor

Pdyn= dynamic power factor

So "in dynamic spectacle magnification, primaryimage sizes are determined with reference to the centersof rotation rather than the entrance pupilsY"

Example 21-12

For a pair of lenses made from Trivex lens material, useRemole's method to find the prismatic effects for right andleft lenses for a reading level 10 mm below the distanceoptical centers. Then determine the vertical imbalancebetween the two eyes. The distance from the back of thelens to the center of rotation of the €ye is 27 mm. Here arethe lens parameters.

TIRight lens:

Power = +2.00 D sphereIndex of refraction = 1.53True base curve = +6.30*

Center thickness = 3.2 mm

Left lens:

Power = +4.00 D sphereIndex of refraction = 1.53True base curve = +8.34

Center thickness = 4.4 mm

Solution

In solving a problem like this, remember that the readinglevel of 10 mm is the level at the back vertex plane of thelens where the object ray strikes the lens. This 10-mm mea-surement is represented by m, the object eccentricity (Figure21-22). Unless the lens has a power of zero, this point isnot the point through which the image ray appears to becoming. The prismatic effect of the lens causes the eye toturn from this point. The distance from the center of the lensto the image point is called the image eccentricity, repre-sented by m'.

To find the prismatic effect using object and imageeccentricities, we first need to find the dynamic spectaclemagnification.

Dynamic spectacle magnification is:

G=

[

~

](

1

)1- ~ Fl 1 - sF;

For the right lens, dynamic spectacle magnification is:

GOD=

[

1

](1- 0.0032 1

/,61 ~ 1:53(+6.30) 1-(0.027)(+2.00)J33.5:' #J;9!iH~~)(1.05708)1..{MJ15"- - / --r- rOI/:1..

A

The object angle (a) is the angle between the optical axisand a line from the object point to the center of rotation ofthe eye (see Figure 21-22). This angle may be calculatedsince m is known and so is s. Remember,

m is the object projection on the back vertex plane of thelens. (In this case m equals the reading level.)

s is the distance from the back vertex plane of the lensto the center of rotation of the eye.

Therefore from the geometry of the figure we know that: TIowe

tan a = ms to

obthiknis

ob.

*Since Trivex has an index of 1.53, the true base curve is also the

refractive power of the front lens surface. If the index of refractionwas any other index than 1.53, it would be necessary to convertthe true base curve to refractive power before entering it in theequation as Fl'

-= ..

CHAPTER 21 Anisometropia 515

Optical axis g'

\\\

9\ri'\\

~\

~\;\ <1--\~

a&09\W~

~'(j.0e~

Figure 2:1-22. In this drawing, the eye is looking through an off-axis point and expe-riencing prismatic effect. The object in the reading plane is denoted by g and the imageof that object by g'. The difference in object and image size resulting from dynamicspectacle magnification is denoted by ~g.

The angle made by the object projection as it goes through the center of rotationof the eye (denoted by c) is abbreviated as a. The projection of the image of that objecthas the abbreviation a'. The letter q is the difference between the object and imageangles, a and a'.

The letters m and m' are the object and image projections of a and a' on the backvertex plane of the lens. m is called the object eccentricity; m' is called the imageeccentricity. The working distance is w; s is the stop distance.

The letter p is the angle formed by the chief ray from the object point and the pro-jection of the chief ray entering the eye. The diagram is not to scale.

(From Remole A: New equations for determining ocular deviations produced byspectacle corrections, Opt Vis Sci 77(10):56, 2000.)

a

a = tan-1ms

For the right lens this is:!tt) 7/1...

m' ~ ~@4G4-§i)(1Q)= J,€h4J15 mm

l'Or')! 1-Knowingm: wecan find a: the imageangle. Usingthe

same method as we did for the object angle, we have:

mAndwe can find a bytaking the inverse tan of -

In other words, S

In our example this angle is:

a =tan-1 mS10= tan-1 -27

=tan-1 0.37037= 20.32314°

((j7 3'?~ 7 0/'= 21:.1-2-4-13°-2,., ~4 tJ.;z I ~

The difference between these two angles is the prismaticeffect caused by the lens. In degrees this is:

m'a' = tan-1 -

S~:-4J~

= tan-1 .27

= tan-1 ~

ft)~ 7 (~

This means that if the lens had a power of zero, the eyewould turn 20.3 degrees to view the object.

To find the prismatic effect caused by the lens we needto find out how much the lens displaces the image of thatobject. Thereforewe need to find the image angle a'. Tofindthis angle, we need to know m'. And to find m: we need toknow how much the image of the object was magnified. Thisis found by multiplying the magnification factor, G, by theobject eccentricity, m. So

q = a' - a

where q is the angle the eye deviates (angle of ocular devia-tion) from where it would be lookingwithout a lens in place.This is prismatic effect in degrees. For the right lens thisis:

?/v&t('D~ tq = ~i.i21'!8° - 20.32314 °

= O.BOe99P

'v:?f7lJ '7 ()

m'= Gm

516 Ophthalmic Lenses PAR T TWO

To convert this to prism diopters, we use the definition of aprism diopter. So

GPO= 100 tan G

= 100 tan~99 f~ ?:>11 15(7

't~ =100(\t.9;l3~) (o'6>;Z~~1)r3.r: =.~ :l, :31' /1

So the pri~' ~ effect for the right eye at the 10-mmreading le~:;r;:~ prism diopters, base up.

For the left eye the procedure is repeated.Dynamic spectacle magnification is:

G=

[

~

]

( 1 I1 - ~ F1 l1 - sF;)

Gas =

[ 1- ~(+8.34) ](1- (0.02~)(+4.00)j

1.53= (1.02457) (1.12108)=1.14862

The object angle (a) for the left eye is the same as for theright. But the image eccentricity (m) will change as will theimage angle (a').

m'= Gm

For the left lens, image eccentricity is:

m' = (1.14862)(10)= 11.4862 mm

And the image angle for the left lens is:

m'a' = tan-1 -

s1 11.4862= tan-

27= tan-1 0.38635= 23.04562°

For the left eye, the prismatic effect in degrees is:

G= 23.04562° - 20.323140= 2.72248°

In prism diopters this is equal to:

GPO = 100 tan G

= 100tan2.72248= 100(0.04755)= 4.76Li

So the prismatic effect for the left eye at the 10-mmreading level is 4.76 prism diopters, base up.

The vertical imbal'Wce between left ??d right eyes is:~f ;/0 (} j 1,f.1 '/ti L>4.76Li base up -~ base up =.~ base up left eye

...

Shorteningthe Procedure. It is possible to shortenthe procedure some by finding the image angles (a') forboth right and left eyes. But instead of converting theseangles to prism diopters, subtract the two angles first.Then convert to prism diopters. Here is how it wouldwork for the above example. We find this relative pris-matic effect (relative q) as the difference between right andleft image angles in degrees.

relative q = (eye with greater a') - (eye with lesser a')

= (23.04562°) - (::;11""'"1:~11i3°).J. ~~&/-102,I Q

"'.O2~49c°- 1,+f0$ tf/ Q

This relative difference in degrees is then convertedinto prism diopters.

qpo = 100tan (relative q)= 100 tanh921z1o91,¥!(J!:Ju<:{I

= 100 EO.0335'5) COJ.. (j /),<15:3 '. = 3.36~ Z'tS-t1 /J

J..'I f{Since the grea~ plus power is on the left eye, the

imbalance is ~~ base up left eye. This agrees eKl'letl}li"fl,with what was found earlier.

Note that this amount of vertical imbalance is ~.JL ~GMWy greater than would be found using Prentice's

rule. Prentice's rule wouldhavegiven2~ baseup left eyeas the imbalance.

For a summary of how to find vertical imbalanceusing Remole'smethod, see Box21-6.

Remole has also presented a method for determiningprismatic effect for any point on a lens in the presenceof oblique cylinder using magnification ellipses. Hestates that "For determining differential prismatic effectsin cylindrical lenses, the methods based on the dynamicspectacle magnification and magnification ellipses arefar easier than the multiple vector methods. . . . Theycan be applied to all types of spectacle lenses as well asto iseikonic corrections.!4" (For more on this topic seeRemole A: A new method for determining prismaticeffects in cylindrical spectacle corrections, Optom VisSci 77:4, 2000.)

Comparing Results: Remole's Method versusPrentice's Rule

Results obtained using Remole's method will not yieldthe same results as found using Prentice's rule. BecauseRemole's method takes both lens thickness and base

curve into consideration, it would be expected to givemore accurate results. So how do the two compare?

Fortunately, "The conventional application of Pren-tice's rule can be applied to most low and moderateanisometropic minus lens corrections without producingclinically significant errors.!3" However, "The differen-tial prismatic effects found with minus lenses by usingthe exact formula are smaller than the estimate made byPrentice's rule, whereas with plus lenses, the prismatic