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On understanding the optics of intraocular lenses
SURVEY OF OPHTHALMOLOGY
VOLUME 24. NUMBER 1 JULY-AUGUST l
1979
PERSPECTIVES IN REFRACTION MELVIN L. RUBIN, EDITOR
On Understanding Intraocular DAVID MUIR,
the Optics of
Lenses
M.D.
Department of Ophthalmology,
Beth Israel Hospital, Boston, Massachusetts
Abstract. The optical considerations
of intraocular
lenses can be understood, using
only the formulae and principles taught in a typical basic science course in ophthalmology. Some of the more common optical problems associated with intraocular lenses are presented and the methods for their solutions are illustrated and explained. (SW Ophthalmol 24:394, 1979)
Key words. axial length optical resolution optics l
intraocular
lenses
-
keratometry
l
l
A
s one pores over the scholarly formulae and calculator programs concerned with the optics of the intraocular lens,‘-’ one is impressed with the apparent complexity of the area. It almost seems beyond the sophistication of the practitioner armed solely with the simple concepts and formulae expected of a candidate for the American Board of Ophthalmology. I don’t want to sound like those old ads that promised that you could play the piano after one lesson, but it is possible to understand the intraocular lens if you can revive some dormant optics. This paper will attempt to discuss the more common optical problems connected with intraocular lenses, using only the formulae and principles that are taught in a typical basic science course in ophthalmology.
+ 19* refers to the power of the lens (FIoL) surrounded by aqueous (as it is in the eye). n refers to the index of refraction. F,,, =
formula:
where:
napueous r
FIoL = + 19 nIena = 1.49 for plastic (1.62 for glass) = 1.33 n raZZus of curvature of convex surface
calculation: answer..
nIens -
19 =
1.49 - 1.33 r
r = JO84 Meters (or
8.4 mm)
PROBLEMI
*+ 19 diopters was chosen
because
in the
Determine the radius ofcurvature forthe Gullstrand emmetropic schematic eye, the power of the crystalline lens is +19.11 diopters. +19 is convex surface (r) of a +I9 diopter plastic piano-convex intraocular lens. Recall that also the most frequently used power for an IOL. 39
40
Surv Ophthalmol
24 (1) July-August
1979
It is also interesting that the radius of curvature of the convex surface (8.4 mm) is within the distribution of human cornea1 radii of curvature. However, if that is true, why does the intraocular lens have only +19 diopters and a cornea with a similar curvature have +40 diopters? The cornea1 front surface is faced by air, which has an index of refraction of 1, The index of refraction of the cornea is 1.3375. F c0rnea=
1.3375 - 1 = +40.2 diopters .0084M
The intraocular lens convex surface is faced with aqueous, a media with an index of refraction closer to that of plastic. F IOL
=
1.49 - 1.33 = + 19 diopters .0084M
PROBLEM II
Determine the power of a +19 diopter intraocular lens in air. This is a practical
problem, for in order to check the power with most lensometers, or keratometers, one must work in air. Since all the power resides in the convex surface (if the lens is plano-convex), this problem can also be treated with the equation for the power of a surface. F IOL
=
nplastic - nalr r
The radius will be the same as in the other problem, for it does not change with the surrounding media. F,,, air = $$&
= 58.3 diopters
To convert the power of any plastic lens in aqueous to that in air, we can develop a conversion factor. Fro, air
=
FIoL aqueous =
1.49 - 1 r 1.49 - 1.33 r
diopters in air. PROBLEM III
Determine thepower of the intraocular lens needed for emmetropia in a patient with a keratometry of 38 diopters and an ocular axial length of 25mm.
A simple and relatively accurate solution to this problem will require a number of assumptions: 1. The pre- and post-operative cornea1 curvatures will not be very different. 2. The distance of the implanted IOL from the front cornea1 surface will be 3.5mm (somewhat akin to a standard anterior chamber depth). 3. The keratometry reading is the true power of the entire cornea. This assumption eliminates the influence of the posterior cornea1 surface, and is consistent with the design of most ophthalmometers which also place all the cornea1 refraction on the front surface by introducing the index of tears as the cornea1 index of refraction in the -1 . equation FcOrnea= “aqueoy 4. Ophthalmic ultrasonography is used to determine the axial length of the eye (front of cornea to retina) and ignores compensation for retinal thickness. 5. The index of refraction of all the ocular media (except the IOL) is 1.33. 6. The intraocular lens which in most cases is about 0.5mm thick, will be treated optically, as a thin lens (thus a simple, familiar formula can be used). First trace a ray from infinity through the cornea (Fig. 1). (n divided by infinity always approaches zero.) A. formula: where:
Following this formula, the conversion factor for plastic lens is: $
= 3.06. The conversion
factor for glass lens is: F9 = 2.14. Therefore, a + 17 diopter plastic lens in aqueous equals 17 X 3.06 = 52 diopters in air. A + 17 glass lens in aqueous equals 17 X 2.14 = 36.4
+
naqy
nair = 1.00 naqueous= 1.33 u = 03 (object distance) F = +38 diopters v = image distance
calculation: answer:
+ FcOrnea=
0+38=
+
v = 0.035M (35mm)
Thus, if, the cornea were the only refracting element of the eye, parallel rays would focus 35mm behind the cornea.
FIG I. Parallel rays being brought to a focus by only the cornea1 refracting power.
V= 35mm 0+3a=1.33
n + Fcornea = n’ iG
v
T
q
.035M
V=
v
35mm
-
V = 21.5 mm
FIG. 2. The rays bent by the cornea now traverse the IOL and come to a focus at the retina.
-
n’ +
FIOL= n’
u
u
L
y = 31.5mm
1.33 + FIOL .0315M
z1.33 .0215M
VIOL = 19.8 Diopters
Thus, the image position now becomes the object for the IOL, which we agreed was 3Smm behind the cornea. Therefore, the object distance for the IOL is 35 - 3.5 = 31.5 mm (.0315M). The new image distance (if we want the image to fall on the retina) will be the distance from the IOL to the retina, i.e. axial length - standard Anterior Chamber [25mm - 3.5 mm = 21.5 mm (.0215M)]. The rays are now traced through the IOL (Fig. 2).
naqr B. for,,.,&.
+ FIoL = naqy
naqueoua= 1.33 u = .0315M v = .0215M 1.33 1.33 caiculatiorr .0315M + FIoL = .m where:
amwec
FIoL = 61.9 - 42.2 = +19.7 diopters
42
MILLER
Surv Ophtholmol 24 (1) July-August 1979
PROBLEM
IV
Determine the power of an intraocular lens needed to produce 2.5 diopters of myopia (the IOL may include a reading addition) if keratometry is 46 diopters and axial length is Zlmm.
A. formula: where:
+
+ Fcornea= naqr
n*s*= 1.00 u=al
naqueoua= 1.33 v = image distance calculation: answer:
0 + 46 = -!$
v = .0289M (or 29mm)
To obtain the new object distance, we use the former image distance minus the anterior chamber depth. Anterior chamber depth = .0035M.
%$EE!% + FIoL = *
B. formula: calculation:
+F
IOL
=
1.33 .0289M - .0035M
1.33 .021M - .0035M 1.33 1.33 .0254M + F1oL = .0175M
answer.- For emmetropia, FIoL = 76.2 52.5 = 23.7. For 2.5 myopia, FIo, = 23.7 + 2.5 = 26.2
PROBLEM
V
Why do the manufacturers suggest that the IOL always be implanted with the convex side facing the cornea?
This is done so that a standard anterior chamber depth can be used to calculate the needed power of the lens. From an optical point of view, the anterior chamber depth is measured from the cornea1 surface to the major refracting surface of the IOL. In our calculations, we assume a figure of 3.5mm. If the lens were implanted with the plano surface facing the cornea, the anterior chamber
depth would be increased by the thickness of the lens. One would have to recalculate lens power using an anterior chamber depth of 3.5 mm + the lens thickness; new anterior chamber depth is about 4-4.1 mm for plastic lenses, and 3.8 mm for glass lenses. PROBLEM
VI
If during the removal of the IOL from the package (in the operating room). the convexpiano orientation is lost, how does one determine which side is convex?
If the surfaces of the lens are viewed as reflecting surfaces, the convex surface will produce a smaller image than the plano surface. In Fig. 3, the ceiling fluorescent light is then considered the object, y (its distance is d from the lens), the virtual image y1 will be located midway between the convex surface and the center of curvature C (r being the radius of curvature). Thus, applying the concept of similar triangles 2y/d+r
= +,
the
image size is proportional to radius of curvature. Since the radius of curvature of a plano surface is almost infinity, the reflected image from the plano side is much larger than from the convex side.
PROBLEM
VII
The quality of the optics of an IOL is discussed in terms of resolvingpower, i.e., the number of lines per mm that can be resolved by the lens. For example, a good lens can resolve between 100 and 200 lines per mm. Can lines per mm be reinterpreted into Snellen notation?
A 20/20 E subtends 5 minutes of arc. Thus, each bar or space in the E subtends 1 minute of arc. One would say the resolving power of the eye (the ability to see a space between two bars) was 1 minute of arc. We must now convert 1 minute of arc into mm on the retina (X) (Fig. 4). From the trigonometry table, Tan 1’ = .0003. Since the nodal point of the eye is 17mm from the retina, Tanl’=.O&3=
17zm
.
’
,
43 fluorescent light t y
IY=II: d+r
if2yzK d-r
d
I 2
2Y -=d+r
Y’ r
r(K)= Y’
FIG. 3. Consider the convex surface of an IOL to be a reflecting surface. The object (y) is a ceiling fluorescent light, and its distance from the lens (d). The image of the light (yl) is midway between the lens surface and the center of curvature of the lens surface.
Tan 1’ =X 17 mm
17 mm
C
x=.005
FIG. 4. Converting angular subtense into a linear retinal dimension, X. Note Tan’ =- X where 17mm is the distance from the nodal point of the eye to the retina in a 17mm’ schematic eye.
To determine how many of these 1 minute spaces are in lmm: 1.OOmm = 200 lines/spaces or lOO/lines mm. .005mm Therefore lOO/lines is equivalent to 20/20 and 20/ 10 is the equivalent of 2OO/lines/mm.
CONCLUSION
We trust that this brief review will show the reader that the optical considerations of intraocular lenses are not difficult to understand. However, I cannot promise that the optics of the self-focusing IOL, which is sure to appear by the 21st century, will be as simple to follow.
44
Surv Ophthalmol 24 (1) July-August 1979
References 1. Binkhorst RD: The optical design of intraocular lens implants. Ophthalmic Surg 6:17-31, 1975 2. Binkhorst RD, Loones CH: Intraocular lens Trans Am Acad Ophthalmol power. Otolaryngol 81:OP 70-79, 1976 3. Colenbrander NC: Calculation of the power of an iris clip lens for distant vision. Br J Ophthalmol 57:735-740,_ 1973 . _ . 4. Drews RC: A practical approach to lens lmplant power. Am Intraoc Implant Sot J 3:170-176, 1977 5. Drews RC: Programs for the HP 25/c calculators for lens implant power. Am Intraoc
Implant Sot J 3:48-49, 1977 Fyodorov SN, Galin MA, Linksz A: Calcula” tion of the optical powers of intraocular lenses. Invest Ophthalmol 14:625-628, 1975 7. Thijssen JM: The emmetropic and the iseikonic implant lens: Computer calculation of the its accuracy. power and refractive Ophthalmologica 171:467-486, 1975
Reprint requests should be addressed to David Miller, M.D., Department of Ophthalmology, Beth Israel Hospital, 330 Brookline Avenue, Boston, Massachusetts 022 15.
VOLUME 24. NUMBER 1 JULY-AUGUST l
1979
PERSPECTIVES IN REFRACTION MELVIN L. RUBIN, EDITOR
On Understanding Intraocular DAVID MUIR,
the Optics of
Lenses
M.D.
Department of Ophthalmology,
Beth Israel Hospital, Boston, Massachusetts
Abstract. The optical considerations
of intraocular
lenses can be understood, using
only the formulae and principles taught in a typical basic science course in ophthalmology. Some of the more common optical problems associated with intraocular lenses are presented and the methods for their solutions are illustrated and explained. (SW Ophthalmol 24:394, 1979)
Key words. axial length optical resolution optics l
intraocular
lenses
-
keratometry
l
l
A
s one pores over the scholarly formulae and calculator programs concerned with the optics of the intraocular lens,‘-’ one is impressed with the apparent complexity of the area. It almost seems beyond the sophistication of the practitioner armed solely with the simple concepts and formulae expected of a candidate for the American Board of Ophthalmology. I don’t want to sound like those old ads that promised that you could play the piano after one lesson, but it is possible to understand the intraocular lens if you can revive some dormant optics. This paper will attempt to discuss the more common optical problems connected with intraocular lenses, using only the formulae and principles that are taught in a typical basic science course in ophthalmology.
+ 19* refers to the power of the lens (FIoL) surrounded by aqueous (as it is in the eye). n refers to the index of refraction. F,,, =
formula:
where:
napueous r
FIoL = + 19 nIena = 1.49 for plastic (1.62 for glass) = 1.33 n raZZus of curvature of convex surface
calculation: answer..
nIens -
19 =
1.49 - 1.33 r
r = JO84 Meters (or
8.4 mm)
PROBLEMI
*+ 19 diopters was chosen
because
in the
Determine the radius ofcurvature forthe Gullstrand emmetropic schematic eye, the power of the crystalline lens is +19.11 diopters. +19 is convex surface (r) of a +I9 diopter plastic piano-convex intraocular lens. Recall that also the most frequently used power for an IOL. 39
40
Surv Ophthalmol
24 (1) July-August
1979
It is also interesting that the radius of curvature of the convex surface (8.4 mm) is within the distribution of human cornea1 radii of curvature. However, if that is true, why does the intraocular lens have only +19 diopters and a cornea with a similar curvature have +40 diopters? The cornea1 front surface is faced by air, which has an index of refraction of 1, The index of refraction of the cornea is 1.3375. F c0rnea=
1.3375 - 1 = +40.2 diopters .0084M
The intraocular lens convex surface is faced with aqueous, a media with an index of refraction closer to that of plastic. F IOL
=
1.49 - 1.33 = + 19 diopters .0084M
PROBLEM II
Determine the power of a +19 diopter intraocular lens in air. This is a practical
problem, for in order to check the power with most lensometers, or keratometers, one must work in air. Since all the power resides in the convex surface (if the lens is plano-convex), this problem can also be treated with the equation for the power of a surface. F IOL
=
nplastic - nalr r
The radius will be the same as in the other problem, for it does not change with the surrounding media. F,,, air = $$&
= 58.3 diopters
To convert the power of any plastic lens in aqueous to that in air, we can develop a conversion factor. Fro, air
=
FIoL aqueous =
1.49 - 1 r 1.49 - 1.33 r
diopters in air. PROBLEM III
Determine thepower of the intraocular lens needed for emmetropia in a patient with a keratometry of 38 diopters and an ocular axial length of 25mm.
A simple and relatively accurate solution to this problem will require a number of assumptions: 1. The pre- and post-operative cornea1 curvatures will not be very different. 2. The distance of the implanted IOL from the front cornea1 surface will be 3.5mm (somewhat akin to a standard anterior chamber depth). 3. The keratometry reading is the true power of the entire cornea. This assumption eliminates the influence of the posterior cornea1 surface, and is consistent with the design of most ophthalmometers which also place all the cornea1 refraction on the front surface by introducing the index of tears as the cornea1 index of refraction in the -1 . equation FcOrnea= “aqueoy 4. Ophthalmic ultrasonography is used to determine the axial length of the eye (front of cornea to retina) and ignores compensation for retinal thickness. 5. The index of refraction of all the ocular media (except the IOL) is 1.33. 6. The intraocular lens which in most cases is about 0.5mm thick, will be treated optically, as a thin lens (thus a simple, familiar formula can be used). First trace a ray from infinity through the cornea (Fig. 1). (n divided by infinity always approaches zero.) A. formula: where:
Following this formula, the conversion factor for plastic lens is: $
= 3.06. The conversion
factor for glass lens is: F9 = 2.14. Therefore, a + 17 diopter plastic lens in aqueous equals 17 X 3.06 = 52 diopters in air. A + 17 glass lens in aqueous equals 17 X 2.14 = 36.4
+
naqy
nair = 1.00 naqueous= 1.33 u = 03 (object distance) F = +38 diopters v = image distance
calculation: answer:
+ FcOrnea=
0+38=
+
v = 0.035M (35mm)
Thus, if, the cornea were the only refracting element of the eye, parallel rays would focus 35mm behind the cornea.
FIG I. Parallel rays being brought to a focus by only the cornea1 refracting power.
V= 35mm 0+3a=1.33
n + Fcornea = n’ iG
v
T
q
.035M
V=
v
35mm
-
V = 21.5 mm
FIG. 2. The rays bent by the cornea now traverse the IOL and come to a focus at the retina.
-
n’ +
FIOL= n’
u
u
L
y = 31.5mm
1.33 + FIOL .0315M
z1.33 .0215M
VIOL = 19.8 Diopters
Thus, the image position now becomes the object for the IOL, which we agreed was 3Smm behind the cornea. Therefore, the object distance for the IOL is 35 - 3.5 = 31.5 mm (.0315M). The new image distance (if we want the image to fall on the retina) will be the distance from the IOL to the retina, i.e. axial length - standard Anterior Chamber [25mm - 3.5 mm = 21.5 mm (.0215M)]. The rays are now traced through the IOL (Fig. 2).
naqr B. for,,.,&.
+ FIoL = naqy
naqueoua= 1.33 u = .0315M v = .0215M 1.33 1.33 caiculatiorr .0315M + FIoL = .m where:
amwec
FIoL = 61.9 - 42.2 = +19.7 diopters
42
MILLER
Surv Ophtholmol 24 (1) July-August 1979
PROBLEM
IV
Determine the power of an intraocular lens needed to produce 2.5 diopters of myopia (the IOL may include a reading addition) if keratometry is 46 diopters and axial length is Zlmm.
A. formula: where:
+
+ Fcornea= naqr
n*s*= 1.00 u=al
naqueoua= 1.33 v = image distance calculation: answer:
0 + 46 = -!$
v = .0289M (or 29mm)
To obtain the new object distance, we use the former image distance minus the anterior chamber depth. Anterior chamber depth = .0035M.
%$EE!% + FIoL = *
B. formula: calculation:
+F
IOL
=
1.33 .0289M - .0035M
1.33 .021M - .0035M 1.33 1.33 .0254M + F1oL = .0175M
answer.- For emmetropia, FIoL = 76.2 52.5 = 23.7. For 2.5 myopia, FIo, = 23.7 + 2.5 = 26.2
PROBLEM
V
Why do the manufacturers suggest that the IOL always be implanted with the convex side facing the cornea?
This is done so that a standard anterior chamber depth can be used to calculate the needed power of the lens. From an optical point of view, the anterior chamber depth is measured from the cornea1 surface to the major refracting surface of the IOL. In our calculations, we assume a figure of 3.5mm. If the lens were implanted with the plano surface facing the cornea, the anterior chamber
depth would be increased by the thickness of the lens. One would have to recalculate lens power using an anterior chamber depth of 3.5 mm + the lens thickness; new anterior chamber depth is about 4-4.1 mm for plastic lenses, and 3.8 mm for glass lenses. PROBLEM
VI
If during the removal of the IOL from the package (in the operating room). the convexpiano orientation is lost, how does one determine which side is convex?
If the surfaces of the lens are viewed as reflecting surfaces, the convex surface will produce a smaller image than the plano surface. In Fig. 3, the ceiling fluorescent light is then considered the object, y (its distance is d from the lens), the virtual image y1 will be located midway between the convex surface and the center of curvature C (r being the radius of curvature). Thus, applying the concept of similar triangles 2y/d+r
= +,
the
image size is proportional to radius of curvature. Since the radius of curvature of a plano surface is almost infinity, the reflected image from the plano side is much larger than from the convex side.
PROBLEM
VII
The quality of the optics of an IOL is discussed in terms of resolvingpower, i.e., the number of lines per mm that can be resolved by the lens. For example, a good lens can resolve between 100 and 200 lines per mm. Can lines per mm be reinterpreted into Snellen notation?
A 20/20 E subtends 5 minutes of arc. Thus, each bar or space in the E subtends 1 minute of arc. One would say the resolving power of the eye (the ability to see a space between two bars) was 1 minute of arc. We must now convert 1 minute of arc into mm on the retina (X) (Fig. 4). From the trigonometry table, Tan 1’ = .0003. Since the nodal point of the eye is 17mm from the retina, Tanl’=.O&3=
17zm
.
’
,
43 fluorescent light t y
IY=II: d+r
if2yzK d-r
d
I 2
2Y -=d+r
Y’ r
r(K)= Y’
FIG. 3. Consider the convex surface of an IOL to be a reflecting surface. The object (y) is a ceiling fluorescent light, and its distance from the lens (d). The image of the light (yl) is midway between the lens surface and the center of curvature of the lens surface.
Tan 1’ =X 17 mm
17 mm
C
x=.005
FIG. 4. Converting angular subtense into a linear retinal dimension, X. Note Tan’ =- X where 17mm is the distance from the nodal point of the eye to the retina in a 17mm’ schematic eye.
To determine how many of these 1 minute spaces are in lmm: 1.OOmm = 200 lines/spaces or lOO/lines mm. .005mm Therefore lOO/lines is equivalent to 20/20 and 20/ 10 is the equivalent of 2OO/lines/mm.
CONCLUSION
We trust that this brief review will show the reader that the optical considerations of intraocular lenses are not difficult to understand. However, I cannot promise that the optics of the self-focusing IOL, which is sure to appear by the 21st century, will be as simple to follow.
44
Surv Ophthalmol 24 (1) July-August 1979
References 1. Binkhorst RD: The optical design of intraocular lens implants. Ophthalmic Surg 6:17-31, 1975 2. Binkhorst RD, Loones CH: Intraocular lens Trans Am Acad Ophthalmol power. Otolaryngol 81:OP 70-79, 1976 3. Colenbrander NC: Calculation of the power of an iris clip lens for distant vision. Br J Ophthalmol 57:735-740,_ 1973 . _ . 4. Drews RC: A practical approach to lens lmplant power. Am Intraoc Implant Sot J 3:170-176, 1977 5. Drews RC: Programs for the HP 25/c calculators for lens implant power. Am Intraoc
Implant Sot J 3:48-49, 1977 Fyodorov SN, Galin MA, Linksz A: Calcula” tion of the optical powers of intraocular lenses. Invest Ophthalmol 14:625-628, 1975 7. Thijssen JM: The emmetropic and the iseikonic implant lens: Computer calculation of the its accuracy. power and refractive Ophthalmologica 171:467-486, 1975
Reprint requests should be addressed to David Miller, M.D., Department of Ophthalmology, Beth Israel Hospital, 330 Brookline Avenue, Boston, Massachusetts 022 15.