The Smith-Kettlewell Eye Research Institute, 2232 Webster Street, San Francisco, California 94115
Dale Allen
College of Optometry, University of Houston, Houston, Texas 77004
Received June 24, 1991; accepted July 3, 1991
An optical analysis of eccentric photorefraction (photoretinoscopy) of astigmatic eyes is presented. The sizeand the angular tilt of the dark crescent appearing in the subject's pupil are derived as a function of five vari-ables: the ametropia of the eye (Dph, DY, axis), the eccentricity of the flash, e, and the distance of the camerafrom the subject's eye, d,. A simplified solution and a solution of the inverse problem, which enable one to cal-culate the degree of ametropia from the size and the tilt of the crescent, are also presented. If the crescent issmaller than the pupil, both the size and the tilt of the dark crescent are independent of the pupil size. Theangular tilt of the crescent is also independent of the eccentricity. Characteristic changes of the crescent as afunction of the cylinder axis are illustrated for compound and mixed astigmatisms. The validity of the theo-retical predictions was experimentally verified on a model eye.
1. INTRODUCTION
Eccentric photorefraction, also termed photoretinoscopy,is an objective method for estimating ametropia thatbears many similarities to retinoscopy and to the workingprinciple of contemporary automated eye refractors. Inretinoscopy the examiner observes the movements of acrescent-shaped light reflex appearing in the pupil. De-pending on the direction of motion of the reflex, the ex-aminer adds plus or minus lenses to find the neutral point.The automatic eye refractor manufactured by HumphreyInstruments utilizes stationary light sources situated im-mediately adjacent to the optical axis. A four-quadrantdetector positioned at the optical axis determines inwhich part of the pupil a crescent appears. The positionof the crescent is used to control a Badal optometer and aset of variable Stokes cylinder lenses to compensate forthe ametropia.
In contrast to these two compensation (nulling) tech-niques, photoretinoscopes are used to estimate the magni-tude of the ametropia from the absolute size and shape ofthe light reflex in the pupil. Since no attempt is made toneutralize the ametropia, this method may never yield theaccuracy of retinoscopy.' However, it has the unique ad-vantage that the ametropia of both eyes can be deter-mined at the same time, thus guaranteeing equal states ofaccommodation. This advantage is especially importantfor infants or young children without cycloplegia.
After the publication of the research of Howland andIowland 2 and I-owland et al.3 on the isotropic photore-fractor, in which the observation and the illumination
paths are coaxial, several different devices appeared thatincorporated an eccentrically mounted light source.Kaakinen& constructed a photorefractor by placing anelectronic flash unit just outside the entrance pupil of a100-mm objective attached to a 35-mm still camera. Hayet al.7 employed a 1000-mm catadioptric lens and an elec-tronic flash. A similar method was subsequently used byDay and Norcia' and Norcia et al.9 Molteno et al." de-signed a photographic objective in which the first compo-nent of the lens formed the limiting aperture of the system,around which was placed an annular flash. Bobier andBraddick" placed a fiber-optic bundle near the entrancepupil of a f/2.5 105-mm lens attached to a video camera.Schaeffel et al. 2 and Angi and Cocchiglia" presentedinfrared photoretinoscopes that use high-output light-emitting diodes in combination with a video camera. Thistechnique has been fruitfully applied for the study of thedevelopment of myopia (Schaeffel and Howland 4 ).
A number of different names have been suggestedfor these photographic refraction techniques based onFoucault's knife-edge test 5 : static photoskiascopy," ec-centric photorefraction," paraxial photorefraction, 7 andphotoretinoscopy 12,18
Several ray-tracing analyses of knife-edge photorefrac-tors have appeared.""8 However, these analyses are re-stricted to special cases, namely, a spherical ametropia, oran astigmatic error, with its main meridian either parallelor perpendicular to the camera-flash axis. These analy-ses cannot explain the behavior of the crescent in anastigmatic eye with an oblique axis because such a refrac-tive error gives rise to an oblique, tilted crescent. The
Vol. 8, No. 12/December 1991/J. Opt. Soc. Am. A 2039
pupil plane
9 flash
Fig. 1. Photorefractor with a light source mounted at an eccen-tricity e below the entrance pupil (EP) of the camera. The dia-gram denotes the rotating x-y coordinate system used inAppendix A.
analytic derivation presented below is a solution for thegeneral case and enables one to explain the behavior ofthe crescent for all forms of refractive error.
2. BASIC PREMISE
We consider a photorefractor as illustrated in Fig. 1. Theborder of the entrance pupil (EP) of the camera adjacentto the flash is assumed to be a straight horizontal line atan eccentricity e above the flash. We sometimes refer tothe straight horizontal border as knife edge. The flash isassumed to be a point source located on the optical axis.This geometry closely resembles the infrared photo-retinoscope of Schaeffel et al.'2 but differs from the designof other existing photorefractors. These existing photo-refractors use not a straight border but the circular edgeof the camera aperture as the occluding edge.
The derivation makes use of the following approxi-mations:
(1) The optical system of the eye is approximated by athin. lens.
(2) Spherical aberration and other optical errors areneglected.
(3) For convenience, the eye is considered an air-airsystem, so that the object and the image focal lengthsare equal.
(4) The diameter of the EP of the camera is assumedto be large.
3. SPHERICAL AMETROPIA
The image-forming geometry of a knife-edge photorefrac-tor is plotted in Fig. 2 for a myopic eye. A similar, butslightly incorrect, diagram has already been published.'9
The size of the dark crescent, DCR, in the pupil of arotationally symmetric eye can be derived from Fig. 2by an elementary analysis of similar triangles (see alsoBobier and Braddick"). The size of the dark crescent is
DCR = e/(d0 D5 ph - 1), (1)
where e denotes the eccentricity of the flash. Dh is the
refractive power of the lens that would be required to cor-rect the eye's spherical ametropia (myopia requires anegative lens). The distance of the camera from the sub-ject's eye, d,, is negative because it is measured againstthe conventional direction of light propagation. DCR ispositive when the eye is myopic relative to the camera dis-tance. The positive sign indicates that the dark portionof the crescent appears on the side of the camera lens (inthe upper part of the pupil for the arrangement indicatedin Figs. 1 and 2). DCR is negative in the case of a hyper-opic eye, indicating that the dark crescent appears on theside of the flash.
When the photoretinoscope is used to determine the re-fractive state of an eye having an astigmatism with oragainst the rule (cylinder axis a = 0° or a = 90° in minus-cylinder form), the crescent size can be obtained from amodified Eq. (1) because the horizontally aligned refractoris sensitive to the ametropia only in the vertical mainmeridian of the eye.
In the case of an astigmatism with the rule, the verticalmeridian is the principal meridian with the greater re-fractive power. This meridian is corrected by a lens withpower Dph + D,1, where DC,1 denotes the magnitude of thecorrecting minus cylinder, and the dark crescent sizeDCR1 is given by
DCR1 = e/[dC(D5 ph + D,,i) - 1]. (2)
Given an astigmatism against the rule, the dark crescentsize DCR2 is
DCR2 = e/(d.Dph - 1). (3)
In these special cases the border of the crescent is perpen-dicular to the line connecting the centers of the flash andthe camera lens.
Note that the size of the dark crescent is independent ofthe pupil size. Thus the pupil size can be omitted in thealgebraic formulation of the problem. However, pupil sizeis not irrelevant. In practice, it is normally easier to mea-sure the size of the bright crescent. Furthermore, pupilsize is an important constraint on the sensitivity and theaccuracy of the method (see Section 8).
camera EP pupil / lens retina
Fig. 2. Crescent formation with a myopically defocused eye. Aflash source is positioned on the optical axis at a distance (-)d,in front of the eye. The eye is myopically focused with respect tothe flash. The flash is imaged to a point in front of the retinaand generates a blur circle with diameter AB on the retina. Thelower boundary point of the blur circle B is imaged to a corre-sponding point B' in the far point plane outside the eye. Of allrays passing through B and B', only those inside the dotted areacan enter the entrance pupil of the camera. They create a brightcrescent of size PQ in the lower part of the pupil.
Wesemann et al.
2040 J. Opt. Soc. Am. A/Vol. 8, No. 12/December 1991
(a) 0.0 DS -4.0 DC
750
9'
(b) 1.0 DS -4.0 DC (C) 4.0 DS -4.0 DC
300
(d) 0.0 DS -4.0 DC (e) 1.0 DS -4.0 DC
60,
900
1/DCR2'
30,
(f) 4.0 DS -4.0 DC
Fig. 3. (a)-(c) Polar diagrams of the crescent vector (DCR,-y) for three astigmatic ametropias and parametrically varied cylinder axis.The size of the dark crescent is indicated by the distance of the circumference of the circle from the origin. The numbers on thecircumference of the circle denote the cylinder axis a. The arrows indicate the dark crescent for a cylinder axis of 30°. For the specialcase of an astigmatism with or against the rule (a = 0 or a = 900), a horizontal border of the crescent is obtained (y = 0 or y = 180).In this case the dark crescent size is DCR1 or DCR2 as defined in Eqs. (2) and (3), respectively. The question of whether the crescentwill be visible can be answered by drawing a circle with a radius equal to the pupil diameter around the origin [see the scale in (c)]. Onlythose conditions that generate a crescent vector that lies inside the pupil will create a visible crescent. (d)-(f) Polar diagrams of thetransformed crescent vector (1/DCR,y). Ametropias are as in (a)-(c). The cylinder axis now varies in an orderly manner independentof ametropia.
4. RAY-TRACING APPROACH TOASTIGMATIC EYES: NUMERICAL RESULTSIn the case of an oblique astigmatism the crescent is tilted,and the size of the dark crescent has an intermediatevalue between DCR1 and DCR2.
With the rigorous analytic ray-tracing approach, de-scribed in Appendix A, we derive the size DCR and the tilty of the dark crescent as a function of six variables,namely, (1) the spherical and (2) the cylindrical powers ofthe corrective spectacle lens at vertex distance 0, Dph andD, 1; (3) the axis of the cylinder, a; (4) the eccentricity ofthe flash from the camera, e; (5) the radius of the pupil,Rp; and (6) the distance d, of the camera from the subject'seye:
(DCR,^y) = f(D~ph Dyi, a, e, Rp, dc) . (4)
For each set of input data we obtain a pair of data forminga vector with magnitude DCR and phase angle . A con-venient way to illustrate the behavior of the dark crescentis to draw the vector in a polar diagram for a given set ofinput data. This has been done in Figs. 3(a), 3(b), and3(c) for a fixed camera distance of -1 m, an eccentricityof 10 mm, and ametropias of 0.0 DS, -4.0 DC; 1.0 DS,-4.0 DC; and 4.0 DS, -4.0 DC, respectively. In eachpanel the axis of the cylinder axis was varied parametri-cally from 0 to 180°.
The polar diagrams in Figs. 3(a)-3(c) show that the vec-tor (DCR, y) approaches a perfect circle for a fixed refrac-tive power and varying cylinder axis. The x coordinatesat the right-hand and left-hand x intercepts of the circle,DCR1 and DCR2, define the size of the crescent for anastigmatism with or against the rule. DCR1 and DCR2are numerically equal to the results obtained with thebasic Eqs. (2) and (3). A positive x intercept denotes thatthe respective principal meridian is myopic with respectto the camera; a negative x intercept is generated by ahyperopic principal meridian.
We can answer the question as to whether a crescentappears in the pupil from Figs. 3(a)-3(c) for any givenpupil size simply by drawing a circle with a radius equal tothe diameter of the pupil around the origin. All condi-tions that create a dark crescent vector (DCR, y) that liesfully inside the pupil circle will give rise to a visible cres-cent. All conditions generating a crescent vector thatpoints to a locus outside the pupil circle show a totallydark pupil because the dark crescent size DCR is largerthan the diameter of the pupil.
The circle described by the dark crescent vector is cen-tered at the origin in the case of a symmetric mixed astig-matism [Fig. 3(b)]. This placement means that the sizeof the crescent does not change with varying cylinderaxes. A crescent of constant size appears at all angles aand rotates along the rim of the pupil. [To avoid confu-
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sion, we should note that the ametropia of 1.0 DS, -4.0 DC,assumed in Fig. 3(b), represents a symmetric mixed astig-matism with respect to the camera because the assumeddistance of the camera (-1 m) is equivalent to a myopia of-1 DS. Thus the center of the dead zone of the photo-refractor is located at -1 DS.]
The numbers at the circumference of the circle denotethe values of the cylinder axis a. They enable one to com-pare the pace at which the tilt of the crescent changes. Itis obvious that a simple relationship between the cylinderaxis and the tilt of the crescent exists in symmetric mixedastigmatism [Fig. 3(b)]. In this case the tilt of the cres-cent is always equal to exactly twice the cylinder axis. InFig. 3(a), which was calculated for an asymmetric mixedastigmatism, the tilt of the crescent rotates by a largeamount (900) when the cylinder axis a rotates from 90 to600 and decelerates its rotational speed with decreasingcylinder angle.
In both Figs. 3(a) and 3(b) the crescent rotates throughthe full range of 3600 when the cylinder axis varies from0° to 180°. This means that the dark crescent is locatedin the upper part of the pupil at a = 0 and rotates counter-clockwise along the rim of the pupil with increasingcylinder axis. This behavior is typical for a mixedastigmatism.
In a compound astigmatism [Fig. 3(c)] the tilt of thecrescent y is always smaller than 900 because the circle
180
165
150
135
120
105Ca)
a) 90o
c) 75
o 60
45.
30-
15-
0-0 15 30 45 60 75 90
Cylinder Axis c <deg>Fig. 4. Tilt of the crescent as a function of the cylinder axis.The spherical power of the corrective lens changes parametricallywhile the cylinder power is held constant at -4 D. The cameradistance was assumed to be 1 m. An eye having an ametropia of-1 D would be in focus with respect to the camera plane. Thusan astigmatic ametropia of +1 DS, -4 DC, having a sphericalequivalent of -1 DS, is a symmetric mixed astigmatism with re-spect to the camera distance. The behavior in five distinct re-gions is explained in the text.
lies entirely on one side of the coordinate system. In theexample in Fig. 3(c) the angle of the crescent deviates onlyby less than 450 from the horizontal axis and changes in ascissorlike movement against the rotating cylinder axis.
Figure 4 summarizes further details of the behavior ofthe tilt for a number of astigmatic ametropias. The cylin-der power was kept constant at -4 DC. The sphericalpower was varied parametrically from -6 to +8 DS. Allcurves are plotted as a function of the axis of the correc-tive minus cylinder. The graph is rotationally symmetricwith respect to the point (a = 4 5,y = 90).
Figure 4 shows five distinctly different regions:
(1) In the region of compound myopic astigmatisms,the tilt of the crescent is always smaller than the cylinderaxis. When the cylinder axis is rotated through 90°, thetilt of the crescent initially follows the rotation but even-tually falls back to 00 tilt.
(2) In the case of a simple myopic astigmatism, the tiltis always identical to the axis of the cylinder.
(3) In the region of mixed astigmatisms the crescentalways rotates faster than the cylinder axis. When thecylinder axis is rotated through 90°, the tilt of the crescentrotates through a full 1800. In the case of a symmetricmixed astigmatism the crescent rotates at exactly twicethe rate of the cylinder axis.
(4) The tilt is equal to 90 + a for a simple hyperopicastigmatism.
(5) The tilt is always larger than 900 + a in cases ofcompound hyperopic astigmatisms.
In the case of ametropia close to a simple astigmatismwith respect to the camera distance, the tilt of the crescentcan never be seen near the two discontinuities at (a =0,) = 90) and (a = 90 ,y = 900), where the meridionalrefractive error is zero and the dark crescent size growsbeyond all boundaries. The absence of the crescent couldbe mistaken as evidence for an emmetropic eye, but actu-ally the dark pupil is simply a consequence of the absoluteinsensitivity of knife-edge photorefractors to refractiveerrors in the axis orthogonal to the camera-flash axis.The extent of this angular dead zone depends mainly onthe eccentricity e, but it also depends on the linear dimen-sions of the flash tube and the size of the EP of the camera(see Section 8). It should be evaluated experimentally forthe actual geometry of the photorefractor.
5. SIMPLIFIED FORWARD SOLUTIONFrom the behavior of the size and the tilt of the dark cres-cent described in Section 4, a simplified solution can bederived. As is shown in Figs. 3(a)-3(c), the polar diagramof the crescent approaches a perfect circle, with itsx intercepts, DCR1 and DCR2, being defined by the re-fractive powers in the two respective principal meridians.The starting point for the simplified solution is the factthat the polar diagram can be made symmetric with re-spect to the cylinder axis by the transform
r'= 1/DCR, y' = y. (5)
This transformation replaces the size of the dark crescentby its inverse and leaves the angle of the crescent unal-tered. The effect of this transformation is illustrated
Wesemann et al.
2042 J. Opt. Soc. Am. A/Vol. 8, No. 12/December 1991
in Figs. 3(d)-3(f). The transformed polar diagram(1/DCR, y) is also a perfect circle, but now the cylinderaxis a rotates always at a constant speed around the cen-ter of the circle. After elementary operations, the centerof the transformed circle is found to be situated at
(DCR1 +DCR2) /2 = 2 (D5 Ph + dDYI - d)-
(6)
Thus the center of the transformed circle is located at aposition proportional to the spherical equivalent, Dsph +(1/2)Dyl, corrected for the distance of the camera 1d,.The radius of the circle is
(DCR1 DCR2) 2= 2e * (7)
The radius of the circle in the transformed coordinate sys-tem is independent of the spherical power and propor-tional to the power of the cylinder.
Since the cylinder axis rotates at a constant speed, thetransformed circle is completely described by its paramet-ric form
x' = R'cos(2a) + xm', (8)
y' = R'sin(2a). (9)
Now, it is possible to calculate the absolute value of thedark crescent DCR and the tilt y of the crescent from
DCR = l/(x' 2 + y'2)1/2,
y = arctan(y'/x').
(10)
(11)
After inserting Eqs. (6)-(9) into Eqs. (10) and (11), we find,after elementary operations, the following general expres-sions for the size and the tilt of the dark crescent as afunction of an arbitrary ametropia
where sgn represents the sign function.These more intuitively derived formulas give numerical
results that are in agreement with the rigorous ray-tracingapproach described in Appendix A. In addition, Eq. (12)has the important property of being identical to the basicEq. (1) when the cylinder power is set to 0.
6. THE INVERSE SOLUTION: DEGREEOF AMETROPIA CALCULATED FROMCRESCENT PARAMETERS
Equations (13) and (14) describe the size and the tilt of thecrescent as a function of ametropia. The practitioner isnaturally more interested in the inverse solution, i.e., thecalculation of the degree of ametropia from the measuredcrescent size and tilt. Obviously, the three parametersthat specify the ametropia (Dsph, Dy1, a) cannot be derivedfrom two equations. Thus at least two independent pic-tures of the crescent are required. We can achieve thisrequirement, e.g., by taking two pictures in two orthogonalorientations of the photorefractor.
We assume that the first picture is taken with a hori-zontally aligned photorefractor, as shown in Fig. 1, andthe second with the photorefractor rotated by 90°. In thefirst picture, with normal orientation, the photorefractorsees the actual ametropia of D5ph, Dy1, a. In the secondpicture it sees an ametropia of D8ph, D,1, (a + 900). As-suming that crescents are visible in both pictures, we canspecify the two observed crescents by their sizes, DCRhand DCRV, and tilts, Yh and yv, respectively. By insertingthese four values into Eqs. (12) and (13), we obtain fourequations from which the unknown quantities D,,ph, Dcyl,and a can be determined. After a number of elementarytransformations, we finally obtain
Dc = -(2k, - 4k32 )1"2 , (15)
Dph = k3 + 1 - 1 De (16)d0 2 1 16
a = sgn(DCRh) * sgn(yh) * (1/2)arccos(k2/k4), (17)
with the abbreviations
e2 /1 1\
k, d2 CRh DCR,?2 / 1 1
2= 2 DC-Rh DCRV)
k2[sin(&y0 - Yh)]
I 4[sin(y 0 + Yh)] J
k = 2DCYI(Dsph + -Dey - d)-
One of the problems encountered in the course of thederivation is the fact that the important signs are lostwhen we take the square root. These signs had to be re-covered afterward. For the sign of k3, we adopted thesign rule
A few comments are added concerning the applicabilityof the inverse solution:
(1) The size of the measured tilted crescents is nega-tive when the larger part of the dark crescent lies on theside of the flash.
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Fig. 5. Polar diagram of the crescent vector predicted byEqs. (13) and (14) for an astigmatic ametropia of 0.25 DS,-3.0 DC. Tick marks on the circumference of the circle indicatethe predicted values for the specified cylinder axes. The experi-mental results found on the model eye are depicted by the filledcircles. The experimental results differ from the predictedvalues by less than 5 and 0.4 mm.
(2) The tilt of the crescent has to be measured withrespect to the orientation of the knife edge (the orienta-tion perpendicular to the camera-flash axis). A clock-wise tilt in the range from 0 to 900 is associated with apositive sign. A counterclockwise tilt is associated with anegative sign (range: -0°--90°).
(3) The above form of the inverse solution cannot beapplied to astigmatic ametropias with or against the rule.In these situations both crescents have a tilt of 0, result-ing in undefined 0/0 division problems. In these condi-tions the basic Eqs. (2) and (3) should be used.
(4) Given a symmetric mixed astigmatism, both cres-cents are always of equal size but opposite sign, and bothtilts are identical. In these conditions the cylinderangle a cannot be evaluated from Eq. (17) because both k 2'and k4 are 0. However, in these conditions, a can be sim-ply set to y/2 if DCRh > 0 and to (180 + y)/2 if DCRh <0 (see Figs. 3 and 4).
(5) In all other cases we can avoid division-by-zeroproblems by replacing the denominator of k 3 with a smallnumber.
7. VERIFICATION ON A MODEL EYEIn order to verify the results predicted by the theory, weused a photorefractor similar to Fig. 1. The photorefrac-tor consisted of a 35-mm camera with a 150-mm lens anda halogen lamp mounted at an eccentricity of 11.8 mmbelow the horizontal borderline of an opaque mask thatcovered the lower half of the lens. The model eye con-sisted of an 80-mm achromatic lens and an adjustable,white painted metal plate modeling the retina. The pupildiameter was 16 mm. The model eye was adjusted toemmetropia.2 0 A refractive deficit was introduced by theplacement of trial lenses in front of the model eye. Pic-tures of the reflex from the fundus appearing in the pupilof the model eye were taken from a distance of 1 m. A
series of pictures was taken at a number of mixed andsimple astigmatic errors. For every combination of triallenses, the axis was rotated through 180 in steps of 100.
Figure 5 shows a comparison between experimentallyobtained data points for a trial lens of 0.25 DS, -3.0 DCand the circle described by the crescent vector accordingto the theory. All measured values are in close agree-ment with the predicted values. In general, the tilt of thecrescent differed by less than 30 from the values predictedby Eq. (13); the maximal deviation was 5°. The size of thedark crescent differed by less than 0.5 mm from the val-ues specified by Eq. (12).
8. DISCUSSION
The foregoing analysis showed that the size and the tilt ofthe crescent can be predicted from five of the six param-eters specified in Eq. (4). Equations (12) and (13) indicatethat the dark crescent size and tilt is independent of pupilsize. It has been emphasized in the literature" 8 2 ' thatcare must be taken to measure the pupil size when one isestimating the refractive error with an eccentric photo-refractor. This discrepancy arises from our use of thedark crescent measure, i.e., the size of the pupil not filledwith light. Howland's original Eq. (10) for the dark frac-tion, DF,8 can be rewritten in the present notation as
DF = DCR/2Rp = e/[2RP(dCDSph - 1)].
It can be shown from this formula that the pupil size can-cels out when the dark crescent size, DCR, instead of thedark fraction, DF, is used as the dependent variable andthat it is only necessary to measure the distance betweenthe pupil margin and the crescent border. However, aswe mentioned in Section 3, pupil size is an important con-straint on the sensitivity and accuracy of the method.
For the derivation presented in Appendix A, we assumethat the two intersections, S,1 and S2, between the knifeedge and the blur ellipse lie inside the aperture of thecamera (see Fig. 10 below). This assumption may be vio-lated (1) in the case of a large ametropia (large blur el-lipse), (2) in the case of a simple astigmatism (the blurellipse is constricted to a line that may not hit the EP atall), and (3) when the diameter of the camera aperture issmall. In these cases the crescent size and tilt may differfrom those in the theory, or the crescent border may haveround edges or may be completely invisible. These limi-tations are especially pronounced in photorefractors thatincorporate a small light source or a small observationaperture as, e.g., in the ingenious, simple point-spreadretinoscope2' with which the crescent is observed with thesmall EP of the human eye.
The effect of a limited EP was also observed in our modelexperiments. For example, we were not able to observe thecrescent for some cylinder angles in the case of a simpleastigmatism. With an ametropia of -1.0 DS, -3.0 DC anda camera distance of 1 m, the tilt of the crescent showed anorderly behavior for a cylinder axis of less than 600: Thetilt was always equal to the cylinder axis. At an axis of600, however, a discontinuous transition from an easilyvisible crescent to a totally dark pupil occurred.
The influence of a limited EP of the camera is illus-trated in Fig. 6(a). An ametropia that creates a narrow
Wesemann et al.
2044 J. Opt. Soc. Am. A/Vol. 8, No. 12/December 1991
(a)
P3
ExtendedLight Source
(b)Fig. 6. Dimensions of the EP and the light source have an im-portant influence on the detectability and the orderly behavior ofthe crescent (see text).
ellipse in the camera plane as shown in Fig. 6(a) will gen-erate a crescent according to the theory when the ellipse istilted as depicted in position P1. The same ametropiawill not create a crescent at all when the ellipse is tilted asdepicted in position P2 because the blur ellipse lies out-side the camera EP.
The effect of a limited camera EP may be overcome inpart with an extended linear light source orientated par-allel to the knife edge instead of a point source. An ex-tended flash tube consists of a large number of laterallydisplaced light sources that create laterally displaced blurellipses. This setup enhances the ability to observe thecrescent in the pupil. As illustrated in Fig. 6(b), the blurellipse created by the centered light source, P3, does nothit the camera EP. However, a laterally displaced portionof the light source creates a displaced ellipse, P4, that willgenerate a regular crescent according to the theory pre-sented above.
From the foregoing analysis it is clear that a single pic-ture of the crescent cannot effectively be used to screenfor astigmatic errors. In particular, astigmatisms withor against the rule give rise to a crescent with a horizontalborder when a horizontally aligned camera is used, thusmaking the crescent indistinguishable from a crescentcreated by a purely spherical ametropia. More impor-tantly, simple astigmatisms at an axis orthogonal to theorientation of the photorefractor are undetectable. How-ever the foregoing analysis shows that two pictures aresufficient for determination of the spherical and cylindri-cal power of the ametropia, provided that both picturesshow a measurable crescent.
The model experiment has shown that the theory canpredict the size and the tilt of the crescent from the opticalproperties of the model eye. However, one should keep inmind that our model eye had a higher optical quality thana normal human eye (i.e., a high-quality lens, a large pupil,and a highly reflecting image plane). In a real screeningsituation for the human eye, one will encounter additionalproblems that are not covered by the geometric-opticsapproach described here. Among these problems are(1) ill-defined crescent borders owing to the shallow con-tinuous transition at the edge of the crescent, (2) coloredcrescent borders owing to chromatic aberration, and (3) ascissor effect similar to streak retinoscopy, i.e., a visiblecrescent appears on opposite sides of the pupil in a singlepicture. However, the theoretical framework of photo-retinoscopy presented here may serve as a guideline thathelps to improve the accuracy of the existing designs.
APPENDIX A
In this appendix the size and the tilt of the dark crescentwill be determined for an astigmatic eye with arbitrarycylinder axis. It will be shown that the size and the tiltof the crescent can be completely described in terms ofthe spherical and the cylindrical power, Dsph and D,,, thecylinder axis, a, the eccentricity of the flash, e, andthe distance of the camera, d,. Further variables in thecourse of the derivation are the pupil radius, Rp, and thedistance of the retina from the nodal point, dr. In orderto simplify the derivation, we will use a coordinate systemthat rotates with the cylinder axis a (Fig. 1).
Step I: Dimensions of the Blur Ellipse on the RetinaAn ametropic eye creates a blurred retinal image of theflash. First we determine the radii of the blur ellipse Rrxand Rry.
Assuming an astigmatic ametropia of Dsph combinedwith Dcy axis a at vertex distance 0, we can obtain therefractive power of the eye's optical system in the twoprincipal meridians by subtracting the refractive errorfrom the refractive power of an emmetropic eye, Deye, thathas the same length dr as the eye under consideration(Deye = 1/dr). The resulting expressions for the refractivepower in the two principal meridians D. and Dy and theassociated focal lengths f, and fly are
D,= Deye - Dph 1/f,
Dy =Deye - Dsph -Dcy = l/fy (Al)
The actual value of Deye is unimportant. Dee is set to 60Din the computations described below. Throughout thederivations f., and fly will be used instead of Dph and D,,.
From the meridional focal length f., or fy, the posi-tions of the meridional flash images b5 and by (Fig. 7) can
lens retina
Fig. 7. Marginal rays define the radii of the blur ellipse on theretina, R,y.
Wesemann et al.
Vol. 8, No. 12/December 1991/J. Opt. Soc. Am. A 2045
border of the camera EP
P, (x,, y,)
Fig. 8. Retina: the extreme point Pr(x,, y,) on the tilted retinalblur ellipse is specified by a horizontal tangent.
be found:
b = (lidc + 1/fx ,)', (A2)
where the abbreviation bx y stands for b and by when fxyrepresents the meridional focal length fx or fy, respectively.This short-form notation, which enables one to write asingle equation instead of two, will be used throughoutthis appendix.
The illuminating light rays diverge behind the meridio-nal focal points (Fig. 7) and form a blur ellipse on theretina. The principal radii Rrx and Rry of the ellipse arefound from similar triangles as
Rrxy = (bxy - dr)Rp/bx,y. (A3)
Step II: Position of the Point on the Retinal Blur EllipseThat Appears Lowest on the RetinaIn this important step of the derivation, we calculate thecoordinates of the retinal point Pr(Xr, Yr), which appearslowest on the retina. This point is solely responsible forthe border of the crescent because it will be imagedhighest in the plane of the camera (see P in Figs. 9 and 10below). As is shown in Fig. 8, P is specified by
(1)(2)
90' <
a slope of the tangent of -a andxr,Y < 0 if 0 < a < 900 or x < 0 and Yr > 0 if
a < 1800.
The general form of the slope of the tangent to an ellipsein Pr is
Step III: Position and Radii of the Blur Ellipse Generatedin the Camera Plane by the Retinal Point Pr(xr, Yr)The light reflected off the fundus at Pr forms a blur ellipsein the camera plane. We determine the equation of thisblur ellipse.
Each point inside the retinal blur ellipse gives rise to aconoid of Sturm in object space. The meridional focallines are situated in the meridional far-point planes at dis-tances of ax and ay from the eye (Fig. 9):
axy = (1/dr - l/fy)-' = (Deye -Dxy)l- (A10)
In the plane of the camera, the conoid of Sturm forms ablur ellipse with principal radii Rcx and RCY R and Rcycan be easily obtained from similar triangles (Fig. 9):
RCXY = (ax y - dc)Rplax. (All)
The blur ellipse generated by the extreme retinal point Prwill be centered at an off-axis point Pc(xc, yc) that is de-fined by the intersection between the chief ray through Pr(Fig. 8) and the camera plane. P has the coordinates
x = dcXr/dr, Yc = dcYr/dr. (A12)
Thus the blur ellipse in the camera plane centered at P isdescribed by
(x - xc)2/Rcx2 + (y - yc)2/RCY2 = 1. (A13)
Step IV: Intersection between the Blur Ellipse [Eq. (A13)]and the Border of the Entrance Pupil of the CameraThe blur ellipse in the camera plane centered at Pc mayhave intersections with the knife edge of the camera EP.The light rays passing above this borderline (hatched areain Fig. 10) enter the camera and form the bright crescent.Here we derive the coordinates of the intersections S 1and Sc2.
Given an ametropia with cylinder axis a, the border ofthe EP is described in the rotated coordinate system bythe basic formula
x cos(,B) + y sin(3) - e = 0, (A14)
tan(-a) = - xrRry2/yrRrx2. (A4)
After inserting Xr and Yr from the parametric form ofthe ellipse,
Xr = Rrx cOs(tr),
Yr = Rry sin(tr),
(A5)
(A6)
and elementary operations, we obtain the polar angle trof Pr as
where , = 900 - a is the angle between the x axis and theperpendicular to the border of the EP (Fig. 9). After atransformation to the normal form, we find that
y = mx + n, (A15)
with
m = -cos(,0)/sin(13),
n = e/sin(3). (A16)
tr = arctan{Rrx/[Rry tan(a)]}. (A7)
Observing the sign rule described above, we find that thecoordinates of the extreme point Pr are given by
Xr = sgn xiIRrx cos(tr)l,
Yr = sgn yitRry sin(tr)l,
(A8)
(A9)
with sgn x = -1, sgn y = -1 if 00 < a < 900 or sgn y =1 if 900 < a < 1800.
Fig. 9. Each point inside the retinal blur ellipse generates a blurellipse with main radii R,y in the camera plane. The blur el-lipse in the camera plane generated by Pr is centered at P.
P, (,, Y)
Wesemann et al.
2046 J. Opt. Soc. Am. A/Vol. 8, No. 12/December 1991
Fig. 10. Camera plane: the light rays of the blur ellipse passinginside the EP (hatched area) create the bright crescent. Thedark crescent is greater than the pupil radius when P, lies outsidethe EP.
lipse depicted in Fig. 10. These light rays originate in theretinal point Pr (Fig. 9), pass through the circular pupil ofthe eye, and form a conoid of Sturm outside the eye. Inevery cross section the conoid of Sturm has an ellipsoidshape (including the two focal lines, the circular pupil, andthe circle of least confusion as special cases of the ellipse).
From the laws of projective geometry, it can easily beshown that two given points on the boundary of the conoidof Sturm (e.g., S,1 and SC2 of Fig. 10) are associated withtwo conjugated points at the margin of the pupil (i.e., P,1and P,2 of Fig. 11).
The coordinates of these two points at the margin of thepupil can be obtained mathematically by the transforma-tion of the ellipse [Eq. (A13)] into a circle with pupil radiusR, centered at the origin.
Furthermore, one has to identify a possible sign change.A sign change occurs, e.g., in a meridian that is myopicwith respect to the camera because the light rays arefocused and cross one another in front of the camera (seeFig. 9):
sgn x2 = -1 if (Deye - 1/d) < D.,
sgny2 = -1 if (Deye - 1/d) < Dy.
(A19)
(A20)
Hence the coordinates in the pupil that define theboundary of the crescent are obtained as
Fig. 11. Pupil plane: the two points PI and P2 at the rim of thepupil that are associated with the intersections S and S in thecamera plane (Fig. 10) define the border of the crescent.
The x and y coordinates of the two intersections S,1 andS,2 between the border of the EP [Eq. (A15)] and the blurellipse centered at P, [Eq. (A13)] are obtained after ele-mentary calculations as
Xsl2 = - - h)1 (A17)
Ys1,2= mx + n, (A18)
with
g = R~,2 + m2 R'X2 ,
k = 2m(n - y,)R,. - 2xcRcy2,
h = x,2RC 2 + (n - y)RCX2 - RCX2RCY
2.
No crescent appears when the square root from Eq. (A17)is negative (no intersection).
Step V: Identification of the Marginal Rays Generatingthe CrescentThe intersections Sc1 and Sc2 between the blur ellipse cen-tered at Pc and the border of the EP [Eqs. (A17) and (A18)]define two marginal rays. These rays emerge in the reti-nal point P. (Fig. 9) and pass the pupil at the two points P 1and Pp2 (Fig. 11). These two points at the rim of the pupildefine the boundary of the crescent.
In order to understand how the coordinates of these twopoints can be determined, the reader should try to visualizethe optical path of the light rays that create the blur el-
Xpl,2 = sgn X21(Xsl,2 - xc)RplXl,
Ypl,2 = sgn y21(ysl,2 - Y)Rp,/Ry,:l-
(A21)
(A22)
Step VI: Size and Tilt of the CrescentFinally, the tilt of the crescent with respect to the rotatedcoordinate system can be calculated from the slope of theline connecting the two boundary points P,1,2, resulting in
4, = arctan[(y, 2 - y,1)/(X, 2 - Xp0)]. (A23)
The final value for the magnitude of the tilt in the normalhorizontal coordinate system can be obtained by the addi-tion of the cylinder axis a (see Fig. 11):
y = 0 + a. (A24)
In order to determine the size of the dark crescent, weinitially calculate the distance d, of the secant P,1P, 2 fromthe origin:
d, = [(Xpl + Xp2)2/4 + (Ypi + Yp2)2/4]1/2 (A25)
The remaining problem to be solved is to determinewhether dp has to be added to or subtracted from the pupilradius so that we can obtain the size of the dark crescent(i.e., to decide whether the dark crescent is larger orsmaller than the pupil radius). We can answer this ques-tion by considering the locus of the central point insidethe blur ellipse, P,, as illustrated in Fig. 10. When P, liesoutside the EP of the camera, the dark crescent, formedby the portion of the ellipse outside the EP, is larger thanthe portion of the ellipse inside the EP. This indicatesthat the dark crescent is larger than the pupil radius, anddp has to be added to Rp. Mathematically, this problemcan be solved by determining whether the point Pcm on theborder of the EP in the direction of P, has a larger dis-tance from the origin than does P (Fig. 10). The distance
Wesemann et al.
Vol. 8, No. 12/December 1991/J. Opt. Soc. Am. A 2047
di of P, [Eq. (A12)] from the origin is
d = (x,2 + Yc2) 12 . (A26)
We can derive the coordinates of Pcm by calculating theintersection between the line that describes the border ofthe EP [Eq. (A14)] and the line connecting the origin andP., i.e., y = (yc/xc)x, resulting in
Xcm = e/[cos(,3) + (yc/xc)sin(3)], (A27)
Ycm = e/[sin(3) + (x:/yc)cos(f3)]. (A28)
From Eqs. (A27) and (A28), the distance d2 of theboundary point Pcm from the origin in the direction of P isfound as
d2= (Xcm2 + Ycm2 )1/2. (A29)
Finally, the size of the dark crescent can be written as
DCR = R + sgn(d2 - d)dp. (A30)
The distance d of the secant P1P2 from the origin hasto be subtracted when the sign function sgn(d2 - d) isnegative.
Equations (A24) and (A30) completely describe the sizeand the tilt of the dark crescent. In order to calculatethese values, we need Eqs. (A1)-(A3), (A7)-(A12), and(A16)-(A29).
Numerical EvaluationThe results presented in Section 4 of this paper have beencalculated on a microcomputer from analytical expres-sions with the use of double-precision variables. Problemsarise in some special cases, e.g., when the cylinder anglesare 1800 or 900 and when f or are equal to dr. In thelatter case the blur ellipse is contracted into a focal line.These problems occur because arguments in the denomi-nator of some equations are 0 [see, e.g., Eq. (A10)]. Inthose cases the 0 in the denominator was replaced by avery small number (10-6), a technique that yields a suffi-ciently accurate result.
ACKNOWLEDGMENTS
This research was carried out at the Smith-KettlewellEye Research Institute while W Wesemann was supportedby grant 1-FO5-TW033830-01 from the Fogarty Interna-tional Center, National Institutes of Health, Bethesda,Maryland. Further support was provided by grantsEY 06883, EY 06579, and G 008635115. We thank thethree anonymous reviewers for many valuable suggestions.
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