When bifocals or trifocals are prescribed in an anisometropic correction, the patient will often perceive the tops of the segments or the dividing line between intermediate and near portions as being at different heights even though they are perfectly aligned in the correction. Although the blur of the bifocal edges and the monocular segment edge diplopia 1 will help mask the misalignment, it is often obvious enough to be a constant irritation to the patient. This is especially annoying for patients who have become accustomed to reading through portions of the segments just below their edges and for patients who frequently use far portions of the spectacles that lie just above the segment edges.
When left and right spectacle lenses magnify unequally, extra-axial image points will be displaced different amounts. Therefore, to fuse off-axis objects, the lines of sight are forced to diverge. The result is that the segment tops in bifocals, or the intermediate line in trifocals, will be seen at different heights in left and right spectacle fields. For example, if one eye is forced to look lower than the other for fusion to occur, the top of the segment in the corresponding lens will appear proportionately higher in the visual field.
Fortunately, the lateral portions of the segment edges, and the bottom portion if it is present, appear so far peripherally that the misalignment is not a problem. However, the top edge of the segment or dividing line between the intermediate and near portion of a trifocal segment lie so close to the immediate and most frequently used portion of the visual field that a misalignment of the segments is noticed even though the edges appear blurred.
The differences in ocular deviations resulting in the perceived misalignment can be expressed in terms of conventional prismatic effects at the spectacle plane. However, because it does not differentiate between object and image eccentricities, Prentice’s rule cannot be used to analyze this problem. Instead, we express the deviations more accurately in terms of dynamic spectacle magnifications 2–5 and the corresponding object and image difference eccentricities. The dynamic spectacle magnification differs from the conventional spectacle magnification in that the ocular rotation centers are used as reference points rather than the entrance pupils and in that it refers to the primary images formed by the spectacles rather than the retinal images. The equation for the dynamic spectacle magnification can be stated as follows 3–5:
where G is the dynamic spectacle magnification, t is the center thickness of the lens, F1 is its front surface power, n is the index of refraction, Fv′ is the back vertex power of the lens, and s is the stop distance, i.e., the distance between the back surface of the lens and the ocular rotation center. G replaces the more cumbersome term Mdyn used in many previous publications. 2–5 When left and right dynamic spectacle magnifications are unequal, we have induced dynamic aniseikonia, which is essentially identical to optical anisophoria, a concept introduced by Friedenwald. 6 In dynamic aniseikonia, the emphasis is on the farpoint images rather than the retinal images.
Fig. 1 a shows a spectacle lens with a flat-top segment on the front. A straight line is drawn from the ocular rotation center to graze the segment top, z, and is produced to the reading plane to an object point, b, in that plane. An image, b′, is formed behind the eye. Object and image distances are not to scale. This image point is projected through c, the ocular rotation center, to b″ in the reading plane. Because of the dynamic spectacle magnification produced by the lens, the projection, b″, is situated below the object point, b. The line of sight has moved from z to z′, and the edge of the segment is seen above the point z′ in the spectacle field. Conversely, in a minus lens, which produces a reduction in the size of the farpoint image, the point z would be seen lower than z′. In general, the distance between the optic axis and z is defined as the object eccentricity, m, and the distance between the optical axis and z′ is defined as the image eccentricity, m′. u is a fixed distance, on the spectacle lens, between the segment top and the optical axis transferred to the back vertex plane, whereas m and m′ vary with the direction of the line of sight. In this situation, u is identical to m; hence u′ is identical to m′. The distance zz′ represents the displacement in the spectacle plane of the line of sight brought about by the dynamic spectacle magnification.
When a front segment is projected to the back vertex plane, refraction at the back surface results in a very small displacement in image space. However, because the angle of incidence in this situation is usually in the order of a few degrees, the effect is negligible for clinical purposes. For example, for a u distance of 4 mm or approximately 8° with respect to the rotation center, the angle of incidence of the chief ray from the segment top might be in the order of 3° if the pantoscopic angle is taken into account. The corresponding angle of refraction would be 3° × 1.5 = 4.5°, 1.5 being the approximate index of refraction of the glass, and the deviation would be 4.5° − 3° = 1.5°. Projected onto the front surface through, for example, a lens thickness of 3 mm, this would correspond to 3(tan 1.5°) = 3(0.026) = 0.078 mm, or less than one tenth of a millimeter. If we are interested in the difference between left and right segment positions, as in this article, the discrepancy becomes even smaller because the angles of incidence do not differ greatly for the two lenses. Hence, for our analysis, it matters little whether the segment is placed in front or on the back. Therefore, this factor has been ignored.
When a portion of the visual field is displaced downward effect of the lens, any visible object in the spectacle plane will be perceived as being displaced upward by an equal and opposite amount. This is what happens to the visible portions of the segment. If the eye were to first fixate a point in the reading plane without the spectacle lens in place, the projection of that point in the spectacle plane would be the object projection, z, shown in Fig. 1. If a plus lens were now to be inserted with the segment top exactly at z, the eye would turn down to fixate the displaced image of the object point. The observer would be unaware that the eye had turned down and would see the segment top as being above the image being fixated by an amount equal and opposite to the distance between z and the line of sight.
Fig. 1 b shows a projection view of the segment and the locus of the corresponding displacements of the line of sight. If the top edge of the segment is used as reference, the perceived segment outline can be obtained by setting off equal distances from the segment tops the opposite direction to the displacements of the line of sight, along meridians converging in the optical center of the major lens. Because the distances representing the perceived displacement converge toward a point, the perceived segment top will be slightly curved as shown in the diagram. However, because the entire segment top is not seen at the same time, the small curvature distortion is not noticed by the patient and need not concern us here.
When the displacements of the lines of sight are equal for the two eyes, there is no perceived misalignment between left and right segments. However, when the displacements are unequal, one segment top will be seen higher than the other. Fig. 2 a shows a projection view of left and right segments, their displacements, and the resulting perceptions of the displacements. In Fig. 2 b, left and right perceived segments have been superimposed with the original segment top as common reference. The diagram shows the perceived difference, upOU, between the segment tops.
Because the amount of perceived misalignment between the two segments depends on the difference in dynamic spectacle magnification for left and right lenses, an obvious solution to the problem is reduction or elimination of the dynamic spectacle magnification difference between the two lenses. However, sometimes the magnification difference is too large to eliminate completely without producing cosmetically unacceptable lenses. Also, a full correction of the dynamic spectacle magnification may induce an unwanted static aniseikonia in the opposite direction. Thus, in most corrections, it is necessary to accept a certain amount of magnification difference between left and right lenses and, hence, a small amount of deviation between the lines of sight.
Another solution to the problem of perceived segment misalignment is to move the optical centers of the major lenses down. This has the limitation that if the optical centers are moved too far down, an imbalance problem in the upper field may result. Sometimes a partial iseikonic correction can be combined with moving the optical centers down a small amount to reduce the misalignment sufficiently for there to be no discomfort. However, if the misalignment is still significant after introducing these two modifications, it can be reduced or eliminated by placing one segment lower than the other. The example presented in the following will demonstrate how the perceived misalignment can be calculated and eliminated.
EXAMPLE: ANISOMETROPIC PLUS LENS CORRECTION
In the spectacle correction shown in Table 1, the segment tops are tentatively set 5 mm below the optical centers of the major lenses. Is the perceived misalignment large enough to warrant changes in the correction? We may not know the answer until it is quantified at the pertinent observation distances, such as the reading distance. Presumably, a small misalignment can be ignored. Where the line should be drawn between a significant and an insignificant misalignment is determined by the patient’s sensitivity to this problem.
To find the amount of perceived misalignment, we must find the ocular deviations and their difference in the region 5 mm below the optical centers of the major lenses. The first step is to find the dynamic spectacle magnification for left and right lenses:
Let u = 5 mm for each eye if the optical centers coincide with the major reference point. If we select object eccentricities, m = u, the corresponding image eccentricities, m′ and u′ can be found by multiplying m by the dynamic spectacle magnification: u′OD = m′OD = 1.233154(5.0) = 6.165770 mm, and u′OS = m′OS = 1.126327(5.0) = 5.631635 mm.
The difference between the right and left misalignments is u′OD − u′OS = 6.165770 − 5.631635 = 0.534135 mm ≈ 0.5 mm. To fuse, the eyes must diverge 0.5 mm at the spectacle plane, the right eye directed lower than the left eye, the right segment edge appearing higher. At a reading distance of 40 cm from the spectacle plane and with a stop distance of 27 mm, this produces a segment top misalignment of 427/27(0.534135) = 8.447246 mm ≈ 8.5 mm, which is about the width of a pencil. This is significant for most patients and should therefore be reduced or eliminated.
The above considerations apply to the distance portion of the correction. When the eyes look down through the segment portion, the back vertex powers are increased by the amount of the adds, and the dynamic spectacle magnifications change. However, because the change is 2.50 D for each lens, the difference between the left and right dynamic spectacle magnifications will not change greatly. Therefore, as long as the adds are the same for both eyes, it is not usually necessary to calculate the perceived misalignment both for the distance field and the segment field. To demonstrate this, let us calculate the misalignment of the segment line when the patient looks through the segment fields just below the segment line.
u′OD = m′OD = 1.341196(5.0) = 6.705980 mm, and u′OS = m′OS = 1.217061(5.0) = 6.085305 mm. The difference between left and right misalignments for the reading field is 6.705980 − 6.085305 = 0.620675 ≈ 0.6 mm. Therefore, the difference between the misalignment in the reading field and the distance field is 0.620675 − 0.534135 = 0.086540 ≈ 0.09 mm, i.e., less than one tenth of a millimeter. At the reading plane, this would be 427/27(0.086540) = 1.368614, or 1.4 mm. In a practical setting, this small difference between the results for the reading field and the distance field is hardly significant and is usually masked by the blur of the segment edges.
EFFECT OF PLACEMENT OF THE OPTICAL CENTERS OF THE MAJOR LENSES
The ophthalmic laboratories sometimes place the optical center of the spectacle lens several millimeters above the major reference point if no instructions to the contrary are supplied. If, for example, the laboratory places the centers 3 mm above the major reference points, the segment top will be 5 + 3 = 8 mm below the optical center. In the present example, to calculate the effect of this larger segment eccentricity, we simply have to multiply the previous value of the perceived misalignment obtained in the above by the factor 8/5; i.e., the resulting spectacle plane misalignment is (8/5)(0.534135) = 0.854616 mm ≈ 0.9 mm. The reading plane misalignment resulting from the higher placement of the optical centers would be 8/5 times that found for the 5-mm distance between the segment top and the major lens optical center, i.e., (8/5) (8.447246) = 13.515594 mm ≈ 13.5 mm.
In some cases, the spectacles have been returned from the laboratory with the optical centers as much as 6 mm above the major reference points. No particular reason was given for this even though it is obviously not a desired feature in an anisometropic bifocal correction. This results in a distance of 5 + 6 = 11 mm between the optical center and the intermediate segment line, producing a spectacle plane misalignment of 11/5(0.534135) = 1.175097 or 1.2 mm. At the reading plane, the misalignment would be 427/27(1.175097) = 18.583941 mm, or nearly 2 cm, almost the width of three pencils.
THE INTERMEDIATE SEGMENT BOTTOM
In trifocals, the dividing line between near and intermediate portions likely causes the greater problem because it is farther from the optical center than the segment top. Using the prescription in Table 1 as an illustration, assume that the top of the segment is fitted at or near the pupil center in the straight ahead viewing position and that the depth of the intermediate segment is 7.0 mm. With the optical center at the major reference point, this would produce a perceived misalignment of (7/5)(0.534135) = 0.747789 ≈ 0.75 mm at the spectacle plane. The corresponding misalignment at the reading plane would be (427/27)(0.747789) = 11.826145 ≈ 11.8 mm. At an intermediate distance of about 1.0 m, the perceived misalignment would be (1027/27(0.747789) = 28.443678 mm ≈ 28.4 mm. Such large magnitudes of the projected misalignments call for a correction.
GENERAL EQUATIONS FOR THE MISALIGNMENT
The process of estimating the perceived misalignment of the left and right segment tops or intermediate lines in trifocals can be shortened by using more general equations. By the above definitions and as shown in Fig. 1, m = u, and u′ = m′. Hence, we can apply the equations relating m and m′ to u and u′. The displacement of the segment edge, u′ − u, = Gu − u = (G − 1)u.
From the above discussion, the perceived displacement is equal and opposite, i.e., up = −(G − 1)u, the displacement being measured from the actual segment edge, along the meridians passing through the major lens optical center. The perceived interocular difference, ΔupOU, between the left and right segments in the spectacle plane can be found as follows:
where ΔGOU is the difference between left and right spectacle magnifications.
For the misalignment in a plane other than the spectacle plane, we simply multiply the expression by the working distance measured to the ocular rotation center, wc, and divide by s, the stop distance. Thus, the perceived misalignment at a given working distance becomes
where Uw/s is the linear misalignment at a plane other than the spectacle plane.
Thus, to estimate the effect of a correction on the perceived segment alignment, we only have to consider the product of the interocular dynamic spectacle magnification difference and the original segment eccentricity at the spectacle plane. To project the misalignment, we multiply the misalignment at the spectacle plane by the ratio between the working distance to the ocular rotation center and the stop distance. Thus, to determine the misalignment at the reading distance, we multiply upOU by 427/27 = 15.814815 ≈ 15.8 mm, assuming a stop distance of 27 mm and a spectacle plane reading distance of 40 cm. To determine the eccentricity difference at a distance of 6 m, we multiply upOU by 6027/27 = 223.222222 ≈ 223.2 mm.
CORRECTING OR ELIMINATING PERCEIVED SEGMENT MISALIGNMENT
There are at least four procedures by which the perceived misalignment can be reduced or eliminated:
- Move both optical centers down.
- Correct for static or dynamic aniseikonia.
- In the lens with the greater spectacle magnification in plus lenses, or lesser minification in minus lenses, move one segment down by the amount determined by Equation 6.
- Similarly, for a trifocal, the intermediate line can be moved down by using a deeper intermediate segment for the lens with the greater magnification or lesser minification.
Any of these methods can be used as partial solutions and in combination with one or more other methods.
Move Both Major Lens Optical Centers Down.
If the major lens optical centers were moved to the level of the segment tops, there would be perfect perceived alignment of the two edges. However, this remedy has the drawback that it increases the interocular prismatic difference in the upper field, and it should therefore be applied with some caution. As a minimum step in this direction, one can ensure that the laboratory does not place the optical centers of the major lenses higher than the major reference point.
Correct for Static or Dynamic Aniseikonia.
In many anisometropic corrections, no attempt has been made to reduce large differences between left and right spectacle magnifications. The rationale is often that the patient has shown no symptoms of aniseikonia; hence, there is presumably no need to reduce the left/right magnification differences. Typically, there has been no check on anisophoria. However, there is often room for partial equalization of the dynamic spectacle magnifications even if no obvious symptoms of aniseikonia are present. The closer the left and right dynamic spectacle magnification, the better will be the perceived alignment of the segment tops. On the other hand, there are many cases where the magnifications cannot or should not be equalized completely or even partially. In such cases, there are other methods of eliminating the perceived misalignment of the segments, such as placing the left and right segment tops at different heights or prescribing different depth intermediate segments.
Move One Segment Down or Make Left and Right Intermediate Heights Unequal.
Assume the two above methods have not been applied. The remaining solution is then to place the left and right segment tops or intermediate lines at different heights. In the above example, the right lens segment top can be moved down 0.5 mm. Had trifocals been prescribed, the segment top would be placed very close to the major reference points, and the segment eccentricity of the intermediate line would likely be about 7 mm. In that case, the solution is to set the intermediate segment portion of the right lens lower than that of the left lens by using an intermediate portion of greater depth. This is not always possible because intermediate segment depths come only in 1-mm increments. In many cases, a best approximation can be selected. For example, in the above case, where the misalignment was found to be 0.75 mm at the spectacle plane when the optical centers coincided with the major reference points, the left intermediate segment could be made 7 mm, whereas the right intermediate segment depth could be made 8 mm, the tops of the intermediate portions located at or near the optical centers of the major lenses. If an 8-mm intermediate portion is unavailable, the 6.0- and 7.0-mm intermediate portions could be chosen for left and right segments as an alternative. This would introduce a small segment misalignment in the opposite direction.
MINUS LENS SPECTACLE CORRECTION
With a minus lens correction, one would expect the vertical misalignment of the segment tops to be smaller. This is because the base curves are flatter and the center thickness is relatively small. Table 2 presents an example of a minus lens correction with back vertex anisometropia of 2.50 D, the same as in the plus lens correction in the previous example.
Will the left/right spectacle magnification difference produce a significant perceived segment top misalignment? The first step in answering this question is to find the dynamic spectacle magnification for left and right lenses:
ΔGOU = 0.919091 − 0.864844 = 0.054247, where ΔGOU is the difference between left and right dynamic spectacle magnifications.
MISALIGNMENT PRODUCED BY BIFOCALS
Let the segment top of a bifocal be 5 mm below the major reference point. Assume that the laboratory has placed the optical center 3 mm above the major reference point because no specific instructions have been given. Find the misalignment at (a) the spectacle plane when the stop distance is 27 mm, (b) the reading distance (40 cm), and (c) the 6-m distance.
- (a) The misalignment at the spectacle plane can be found as follows: u, the segment edge eccentricity = 5.0 + 3.0 = 8.0 mm; ΔupOU = ΔGOUu = (0.054247)(8.0) = 0.433976 mm ≈ 0.4 mm. (2)
- (b) The perceived misalignment at 40 cm: U427/27 = (wc/s)ΔGOUu = (427/27)(0.054247)(8.0) = (15.814815)(0.054247)(8.0) = 6.863250 mm ≈ 6.9 mm, where U427/27 is the perceived misalignment at the spectacle plane reading distance of 40 cm. (3)
- (c) The perceived misalignment at 6 m: U6027/27 = (wc/s)ΔGOUu = 6027/27(0.054247)(8.0) = 96.873087 ≈ 96.9 mm, where U6027/27 is the perceived misalignment at 6.0 m from the spectacle plane. (3)
MISALIGNMENTS PRODUCED BY TRIFOCALS
Assume that the intermediate line of the trifocal is 7 mm below the major reference point. Again, assume that the laboratory has placed the major optical centers 3 mm above the major reference points. The only factor that changes is u, which becomes 10 mm instead of 7 mm. Therefore, multiply the above values for the perceived misalignment by 10/8. As was demonstrated in the previous example, when the adds are identical for both eyes, it is not necessary to change to the dynamic spectacle magnifications produced by the intermediate field or the reading field of the spectacles.
- (a) The perceived misalignment at the spectacle plane becomes (10/8)(0.433976) = 0.542470 ≈ 0.5 mm.
- (b) The perceived misalignment for trifocals at the reading distance becomes (10/8)(6.863250) = 8.579063 ≈ 8.6 mm.
- (c) The perceived misalignment at 1.0 m becomes (1027/27)(0.433976) = 16.507161 ≈ 16.5 mm.
Thus, even with a minus lens of moderate power specifications, the vertical segment misalignment can be substantial enough to warrant correction. The methods are identical to those outlined for the earlier example.
CORRECTION FOR PERCEIVED BIFOCAL MISALIGNMENT
First, consider correcting for all or part of the dynamic spectacle magnification difference. Also, consider the effect of moving the optical centers down, but not to the extent that undue prismatic demand is introduced in the upper field. A combination of the two procedures may reduce the misalignment sufficiently so that the segments will not have to be placed at different heights. If no changes are made in the dynamic spectacle magnification or the location of the optical centers, the right segment would have to be moved down 0.4 mm. If the optical centers are moved down 3 mm to the major reference points, the right lens segment would have to be moved down (0.05247)(5) = 0.262450 ≈ 0.3 mm. At the reading distance, the misalignment is 427/27(0.262450) = 4.150598 ≈ 4.1 mm.
CORRECTING MISALIGNMENT PRODUCED BY TRIFOCALS
In trifocals, the top of the segment is often fitted so closely to the major lens optical center that it does not produce a significant perceived misalignment. To ensure perfect alignment of the segment tops, the laboratory can be instructed to place the optical centers at the level of the top of the intermediate portion. However, the dividing line between intermediate and near portions is sufficiently far from the optical centers to produce a significant perceived misalignment. In the minus lens correction above, this dividing line may need to be placed at different levels in left and right lenses. This can be done only if the perceived misalignment is large enough to fit approximately the available intermediate segment depths.
In the example with the minus lens, instruct the laboratory to place the optical centers of the major lenses in coincidence with the tops of the segments. Assume that the intermediate segments are 7-mm deep. Initially assume that the base curves and thickness are to remain unchanged. The misalignment of the intermediate line would then be upfOU = Gdiff(m) = (0.054247)(7.0) = 0.379729 ≈ 0.4 mm. Because intermediate segments do not come in half millimeter steps, this cannot be corrected by making the right lens intermediate deeper than the other. However, a moderate change in base curves to make them more equal would reduce the misalignment to an insignificant value.
If base curves are not changed, one compromise solution to consider in some cases is to lower the optical centers of the major lenses to the middle of the intermediate bands. Both the segment top or the intermediate line would then be misaligned to a lesser amount, in opposite directions. For an intermediate segment portion height, the spectacle plane misalignment would be (0.054247)(3.5) = 0.189865 mm. At the reading distance, this represents a misalignment of (427/27)(0.189865) = 3.002680 mm ≈ 3.0 mm, which would be acceptable to many patients. Any induced effects due to the increased prismatic difference in the upper field should be checked.
The perceived misalignment can be reduced further by making the left and right dynamic spectacle magnifications approach each other in magnitude. From the general equation, Uw/s = (w/s)ΔGOUu, it is obvious that the misalignment is proportional to the interocular dynamic spectacle magnification difference. Thus, a reduction of this difference to half its present value would make the misalignment at the reading plane about 1.5 mm for the top and intermediate lines of the trifocal, in opposite directions. At the same time, this would reduce any adverse effects of having moved the optical centers down. Theoretically, one could reduce the interocular dynamic spectacle magnification difference to zero, but this is usually impractical for technical and cosmetic reasons. Furthermore, it is not recommended because it might induce an unwanted static aniseikonia.
PRISMATIC EFFECTS PRODUCED BY MOVING ONE SEGMENT DOWN
Moving one segment down will produce a change in the prismatic effect produced by the segment and hence a change in the total interocular deviation difference. However, this change is quite small and can usually be ignored. To estimate the effect of moving a segment down, we can use Prentice’s rule. This rule is valid for a thin lens considered by itself and not in combination with the lens before the other eye. 7–9 The add is considered thin, and we are moving only one segment. Prentice’s rule states that the prismatic effect = cF, where F is the equivalent power and coincides with the back vertex power and c is the eccentricity in centimeters. Although this is not exactly the way Prentice 10 formulated the rule, it is the most common interpretation. Thus, if we move a +2.50 add down 1 mm, the object eccentricity with respect to the add changes by 1 mm, and the prism induced by the segment is altered by 0.1(+2.50) = 0.25 pd, usually not a significant amount.
With a flat-top segment whose optical center is above the reading area, the prismatic effect produced by the add decreases by that amount, but with a round add whose optical center is below the reading area, the prismatic effect increases by that amount. In the former case, the total interocular prismatic difference increases, whereas in the latter case, it decreases.
Checking, for illustrative purposes, by using more accurate equations based on the dynamic spectacle magnification, we have the following: when a lens becomes theoretically thin, the center thickness of the lens is zero, which results in a shape factor of 1.0 that can be left out of the equation. Hence, the formula for the dynamic spectacle magnification applied to the add becomes
where Gadd represents the dynamic spectacle magnification of the add.
If a flat-top add is moved down 1 mm, the object eccentricity with respect to the add is reduced, and the displacement of the line of sight becomes less than before. In general, the ocular deviation produced by a lens and projected on the spectacle plane is (G − 1)m, where m is the object eccentricity. This is identical to Equation 6 except that u has been replaced with m, the general term for object eccentricity. madd, the eccentricity with reference to the segment, has now been reduced by 1 mm, and the corresponding reduction in prismatic effect is 100(1.0)(madd)/s = 100(1.0)(0.072386)/27 = 0.268096 ≈ 0.27 pd, practically identical to what was found with Prentice’s rule. As pointed out previously, a round add would produce a change in the prismatic effect in the opposite direction.
A segment seldom needs to be moved down by a very large amount. However, although the amount of additional interocular prismatic difference is very small, it must be remembered that in a flat-top add, the increase is in the direction of the existing interocular prismatic difference. For this reason, it is best first to explore the other remedies as much as possible, and then move one segment down only after these have been taken advantage of. In the above example, the increase in interocular prismatic difference of 0.25 pd is a lesser problem than the original misalignment. However, should the downward placement of the segment be so large as to create a significant increase in the interocular prismatic difference, this could be remedied by ordering segments so that their optical centers remain at the same height. The latter procedure is likely to be required so seldom that its practical implementation will not be discussed in this article.
SUMMARY OF METHODS
In general, the procedures for ensuring that the segments appear aligned are quite simple. First, find the misalignment, upOU, at the spectacle plane. Translate this into the perceived misalignment, Uw/s, at the pertinent observation planes. The misalignment at the reading plane, the 1-m distance, and the distance plane are for communication with the patient and for visualizing the magnitude of perceived misalignment. The amount of misalignment that is significant is determined by the patient’s response. The various methods of reducing the misalignment are best considered together and used in conjunction with each other.
One method that has not been emphasized in this paper is used by many patients often without the knowledge of the practitioner, that is, bending the frame so that one lens is slightly higher than the other. However, this cannot be recommended as a professional solution because many patients will notice even a small tilt or asymmetry of the spectacle frame. The method will also introduce a misalignment between the optical centers of the major lenses. To resort to this remedy should be unnecessary if the problem is analyzed in the terms set out above.
Unlike the reaction to an acuity chart, the response to perceived misalignment of the segment tops is not clear cut. The discomfort will depend to some extent on which areas of the spectacle field are most frequently used by the patient. The problem is not necessarily always associated with reading but is just as likely to occur with any other activity. For example, when descending a stairway, some patients wearing anisometropic corrections will see the segment lines pop into view at different times for left and right eyes. It is particularly noticeable in this situation because the stairway steps offer a powerful fusion lock and force the eyes to diverge vertically for fusion to occur. Another example of the occurrence of perceived segment top misalignment is when the patient is seated in a car and is glancing from the road to the dashboard. In this case, an added factor is the high level of illumination in the daytime, which will make the pupils smaller and, hence, the segment line sharper and thinner in the field of view. Typically, one segment edge will appear to lag behind the other as the patient looks up and down. For example, if the left segment edge persistently appears above the right segment edge, immediate relief can be demonstrated with a suitable size lens held before the right eye.
In the above discussions, we selected two intermediate-power anisometropic corrections as examples. With higher-power corrections and larger differences between left and right lens back vertex powers, the importance of aligning the segments in the patient’s field of view becomes increasingly more important.
It is somewhat surprising that the problem of perceived misalignment of segments has not been discussed in standard textbooks. One reason is that the misalignment projected to the spectacle plane is very “small” in physical terms. However, this very small vertical deviation between the lines of sight at the spectacle plane can produce quite large misalignments between the segment tops as they appear at the working distance. Another reason for the lack of information on the problem is perhaps that the concept of dynamic spectacle magnification was not introduced until the latter part of the 20th century. 2–4 Without this concept, and without the application of the eccentricities and eccentricity differences calculated by means of the dynamic spectacle magnifications, determining the perceived misalignment is awkward and difficult. However, once the problem is analyzed in these terms, it becomes relatively simple and can always be eliminated or reduced.