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Ophthalmic lenses and dispensing Contents of CD-ROM How to navigate the CD-ROM Action of a prism The transverse test Action of a cylinder Use of the focimeter The correction of ametropia Effective power in DV Effective power in NV Vergence impressed in NV Centration of spectacle lenses Effects of centration errors Lens thickness and weight Lens design and performance Iso-V-Prism zones Exit Click on page First time users click here Ophthalmic Lenses & Dispensing

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To advance through the show in the order intended, click anywhere on the screen. Click now to continue. In some topics you will see a small button like this > It enables you to return to the contents page for the topic. To reverse this step, right click and select ‘previous’ from the pop-up menu. You may like to try this now, or just left click to continue. How to navigate the CD-ROM Click here to start Click to skip to next topic Click to return to previous screen Click to return to CD contents Each screen has three action buttons This button returns you to the CONTENTS PAGE. Click on it now to end the navigation instructions. This button enables you to proceed to the next topic. When you click on it you skip the rest of the current topic. This button enables you to go back to the previous screen. Click anywhere to continue Ophthalmic Lenses & Dispensing

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Now the prism has been rotated clockwise through 90° from its original position. The base lies on the left and only the vertical limb is displaced towards the prism apex. The base setting can be marked on the lens when it has been rotated to this position where a continuous appearance of the vertical limb is obtained. This is a crossline chart. On the next mouse click you will place a prism held with its base DOWN in front of the chart… Note how the horizontal limb of the crossline chart appears to be displaced towards the prism apex. When the prism is rotated clockwise before the crossline chart the lines also appear to rotate, always displaced in the direction of the prism apex. Action of a prism Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The lens has now been returned to its original position before the chart, and the limbs move back to their original positions and appear continuous with the limbs viewed outside the chart. The optical centre can be marked on the lens over the intersection of the crosslines. On the next mouse click you will place a plus sphere centrally in front of the chart... Notice that the limbs are in their correct position but they appear magnified through the lens... The lens has now been moved slowly to the right before the chart, and the vertical limb appears to move to the left, AGAINST the movement of the lens. The lens has now been moved slowly downwards before the chart, and the horizontal limb appears to move upwards, again AGAINST the movement of the lens. AGAINST MOVEMENT is obtained from all plus lenses in the transverse test. The transverse test - plus lens. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The lens has now been returned to its original position before the chart, and the limbs move back to their original positions and appear continuous with the limbs viewed outside the chart. The optical centre can be marked on the lens over the intersection of the crosslines. On the next mouse click you will place a minus sphere centrally in front of the chart... Notice that the limbs are in their correct position but they appear minified through the lens... The lens has now been moved slowly to the right before the chart, and the vertical limb also appears to move to the right, WITH the movement of the lens. The lens has now been moved slowly downwards before the chart, and the horizontal limb also appears to move downwards, again WITH the movement of the lens. WITH MOVEMENT is obtained from all minus lenses in the transverse test. The transverse test - minus lens. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The cylinder axis can be marked on the lens when it has been rotated to this position where a continuous appearance of the crosslines is obtained. Action of a plus cylinder This is a crossline chart. On the next mouse click you will place a plus cylinder held with its axis VERTICAL in front of the chart… Notice that the limbs appear to be in their original positions but that just the vertical limb is magnified in the horizontal meridian. On clockwise rotation of the cylinder, the vertical limb rotates anticlockwise and the horizontal limb rotates clockwise towards it in a SCISSORS MOVEMENT. Further clockwise rotation of the cylinder, produces further SCISSORS MOVEMENT. Rotating the cylinder back again causes the limbs to SCISSOR back to their original position. You can emulate the actual movement seen during the rotation by reversing four times through this sequence (right click and select previous). Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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On the next mouse click you will place a minus cylinder held with its axis VERTICAL in front of the chart… Notice that the limbs appear to be in their original positions but that just the vertical limb appears minified in the horizontal meridian. Action of a minus cylinder On clockwise rotation of the cylinder, the vertical limb also rotates clockwise and the horizontal limb rotates anticlockwise towards it in a SCISSORS MOVEMENT. Further clockwise rotation of the cylinder, produces further SCISSORS MOVEMENT. Rotating the cylinder back again causes the limbs to, SCISSOR back to their original position. You can emulate the actual movement seen during the rotation test by reversing three times through this sequence (right click and select previous). The cylinder axis can be marked on the lens when it has been rotated to this position where a continuous appearance of the crosslines is obtained. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Use of the focimeter This is a typical view of the measuring scales seen when you look into a manually operated focimeter. This is the central target area. The circular scales are calibrated in prism dioptres This is the protractor from which cylinder axis and base setting can be read. This is the circular target which is brought into focus by rotation of the dioptre power knob. If the instrument is in correct adjustment, when there is no lens under test, the target should be seen in sharp focus at the centre of the protractor and the power scale should read zero, as is the case here The vertical and horizontal crosslines can be rotated to coincide with the cylinder axis direction, or the base setting of an oblique prism. Here they now lie along the meridians, 150 and The target may be of the linear type which must first be rotated so that the lines coincide with the principal meridians of an astigmatic lens under test. In the following demonstration we will assume the use of a circular target. - + Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Use of the focimeter In order to do this, begin by turning the power adjusting knob right to one end of its reading range. It does not matter whether it is the plus or minus end of the range. You will notice that the green target is so much out of focus that it can no longer be seen. Then rotate the eyepiece slowly back inwards until the scales just come into sharp focus. Stop as soon as they come into focus. You should now find that the dioptric scale reads exactly zero. If it does not now read zero, the instrument needs servicing! Now rotate the adjustable eyepiece ring of the telescope to rack out the eyepiece to its fullest extent. The protractor scale on the graticule will become blurred until it is no longer in focus Before you use a manually operated focimeter you must adjust its focusing eyepiece for your own use Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Use of the focimeter Now that the focimeter is ready for use we can consider how it is used to read the powers of prisms and lenses. We will begin by considering how the focimeter measures prism power You will notice that the green target is exactly centred over the middle of the crosslines. No displacement of the target signifies that there is no prismatic effect at the point on the lens which is being measured If the green target lies in the position indicated here it signifies that the optical element under test incorporates 2 base UP at the measuring point If the green target lies in the position indicated here it signifies that the optical element under test incorporates 4 base IN at the measuring point, assuming that the lens under test is for the right eye. It would be 4 base OUT if it were a left eye The lens under test must be a plano-prism since the dioptre scale is reading zero The target is displaced upwards and its centre is seen to lie over the second ring Finally, if the target lies in the position shown here and a left eye is under test, the reading is 3 base 150 which could equally be expressed as 1.5 base UP and 2.6 base IN The base setting of an oblique prism will be found easier to read if you rotate the crosslines which lie along the 90 & 180 meridians until one limb passes through the centre of the target. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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We will now consider how the focimeter is used to determine the power of spherical lenses Use of the focimeter Continue to refocus until the target appears in sharp focus again. The power of the lens under test can now be read from the power scale and is seen to be Refocus the target by rotation of the power adjusting knob. At some point the target will reappear, blurred and off-centre Adjust the position of the lens to centre the target. (If the lens under test is glazed to a frame, or it incorporates a strong prismatic element, it may not be possible to centre the target.) Here, a spherical lens has been placed on the lens rest and the green target has disappeared. It is too blurred to be seen In order to read the back vertex power of a lens you must ensure that the lens is placed with its concave surface in contact with the lens rest Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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When an astigmatic lens is under test, each dot is drawn out into a line focus which is parallel with one of the principal meridians of the lens Here the best focus is obtained with the vertical lines of the target when the reading on the power scale is Here the best focus is obtained with the horizontal lines of the target when the reading on the power scale is The readings are when the vertical lines are in focus and when the horizontal lines are in focus. Further rotation of the power adjusting knob brings the focal lines in the other principal meridian into sharp focus From these two readings we can deduce the power of the lens under test. Note that in whichever form the prescription is recorded, the first reading is the sphere, the second reading is the sum of the sphere and the cylinder (the power of the cylinder is whatever must be added to the first reading to obtain the second) and the axis direction is the same as the lines in the second reading We will now consider how the focimeter is used to read the power of an astigmatic lens We can record the power of the lens, either as / x or, as / x When a spherical lens is under test the target consists of a circle of dots. - + Use of the focimeter Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Use of the focimeter In the case of an oblique cylinder you should rotate the vertical and horizontal crosslines until they are parallel with the principal meridians of the lens under test Notice that the power scale reads when the lines lie along the 60 meridian In the other principal meridian the lines come into focus when the power adjustment knob is turned to The power of the lens under test is, therefore, / x 150 (or / x 60). Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Try this one by yourself ! Use of the focimeter The power of the lens under test is / x 15 ( or / x 105) Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Here is another one to try yourself ! This time a progressive power lens is under test Use of the focimeter This is one reading at the major reference point in the distance portion This is the second reading at the major reference point in the distance portion This reading is taken at the near reference point in the near portion. What is the power of the lens? Answer given on next mouse click Either / x 45 Add or / x 135 Add Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The correction of ametropia Vertex distance Correction of hypermetropia Correction of myopia Effective power in distance vision Effective power in near vision Vergence impressed in near vision Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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vertex distance BS 3521: Part 1: 1991 Glossary of terms relating to ophthalmic lenses and spectacle frames vertex distance Distance from the visual point of a lens to the corneal apex BS 2738: Part 3: 1991 Method of presentation of prescription orders for spectacle lenses 3.1 On all prescriptions and prescription orders, the power of the sphere (spherical power) shall be stated for each eye or lens. NOTE. If the prescribed power is sufficiently high such that the vertex distance becomes significant, e.g. if the power exceeds 5.00 D, then the distance at which the power was measured should additionally be recorded in the prescription. vertex distance should be indicated by stating the number of millimetres following the prescription, for example: +6.00/-0.50 x 90 at 12 Vertex distance The correction of ametropia Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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k far point distance Correction of Hypermetropia M´ The correction of ametropia In hypermetropia, light from a distant object is focused behind of the retina. This may be due to the refracting surfaces being too weak (purely refractive error), the axial length of the eye being too short (axial error), or, as is usually the case, a combination of these two factors (correlation ametropia). M´ M R There is a point, behind the eye, which is conjugate with the macula. Light converging towards this point would be focused by the eye’s optical system at the macula. This virtual point is called the Far Point, M R. If the eye can make sufficient effort of accommodation, it may increase its power to produce a sharp image of a distant object on the macula, M´. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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f ´ V = k + d The back vertex focal length is made up from the sum of the vertex distance, d, and the far point distance, k. F´ V = K / (1 + dK) The spectacle lens, therefore, lies at its own back vertex focal length from the eye’s far point. f ´ V vertex distance d k far point distance f ‘ V = k + d Correction of Hypermetropia In order for a spectacle lens to correct an eye, its second principal focus, F´, must coincide with the eye’s far point M R. M´ This can be expressed in dioptres......but it is much easier to think of it terms of focal length! The correction of ametropia F´ Ophthalmic Lenses & Dispensing MRMR Light from a distant object is focused by the lens at its second principal focus. Since this is conjugate with the macula, the eye’s own optical system can produce a sharp focus of the distant object at the macula, M´. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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F´ MRMR MRMR k f ´ V d f ´ V = k + d Correction of Myopia The correction of ametropia In myopia, light from a distant object is focused in front of the retina. This may be due to the refracting surfaces being too strong (refractive myopia), the axial length of the eye being too great (axial myopia), or a combination of these two factors (correlation ametropia). The far point, M R, of the myopic eye lies in front of the eye. An object placed at the far point would be in sharp focus at the macula in the unaccommodated eye. M´ Myopia is corrected by minus lenses. Light from a distant object is focused by the lens at its second principal focus, which lies in front of the lens. Since the second principal focus coincides with the eye’s far point, which is conjugate with the macula, the eye can produce a sharp focus of the distant object at the macula, M´. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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f´ V F´ new f ´ V = old f ´ V + x To correct the eye at a greater vertex distance a plus lens must be made weaker. Plus lenses moved away from the eye get stronger. Effective power in distance vision We will now consider what happens when changes are made to the vertex distance. Ophthalmic Lenses & Dispensing x Hypermetropia The condition for a spectacle lens to correct an eye is that the second principal focus of the lens must coincide with the eye’s far point. Thus if a plus lens is moved away from the eye, its focal length must be increased by the change in vertex distance. f´ V Click to skip to next topic Click to return to previous screen Click to return to CD contents

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F´ MRMR x f ´ V Effective power in distance vision new f ´ V = old f ´ V + x To correct the eye at a greater vertex distance a minus lens must be made stronger. Minus lenses moved away from the eye get weaker. Myopia If a minus lens is moved away from the eye, its focal length must be decreased by the change in vertex distance. In myopia, the far point lies in front of the eye. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Effective power in near vision cm B 0 stationary object We will now consider what happens when the vertex distance of a lens is altered when the the lens is being used for near vision. Here, a +3.00D lens is being used for near vision at 33.3 cm in front of the lens The vergence arriving at the front surface of the lens is -3.00D and assuming a thin lens, the vergence leaving the back surface will be zero, the lens collimates the light. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Effective power in near vision B x l < 33.3cm stationary object Suppose that the lens is now moved away from the eye, (i.e., down the nose) towards the object, which, in this case, has remained in its original position. Ophthalmic Lenses & Dispensing The light leaving the lens is now divergent and the eye will need to make an effort of accommodation (about 0.25D) in order to view the near point clearly. Clearly, the new object distance will decrease by the movement of the lens and the vergence arriving at the lens will increase, here to -3.25D. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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+3.00 B x l < 33.3cm It can be shown that, for small movements, x, the change in effective power is given by -xF(2L 1 + F ) where L 1 is the original object distance and F is the power of the lens. x is considered to be negative if the lens moves to the left, away from the eye. A graph of this expression showing how effective power varies with lenses of different powers is given later. Ophthalmic Lenses & Dispensing Effective power in near vision stationary object Click to skip to next topic Click to return to previous screen Click to return to CD contents

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+3.00 We will now consider a second situation where, when the lens is moved, the object also moves through the same distance as the lens, the object distance is, therefore, constant. Effective power in near vision object moves with lens Ophthalmic Lenses & Dispensing B cm x x cm B x Under these circumstances it can be shown that the change in effective power is given by -xF( L + F) 2. The situation shown above is the unique case when the term in the bracket, L + F, is equal to zero, so the change in effective power must also be equal to zero. This result is indicated on the graph of effective power which follows. Here, the lens has moved away from the eye, through a distance, x, and the object has also moved away by the same distance, x. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Effective power in near vision Lens power object remains s tationary Q = -xF(2L 1 + F) object moves with lens Q = -x(L 1 + F) 2 Change in effective power when lenses are used for near vision at 33.3cm and subsequently moved 5 mm away from the eye. Ophthalmic Lenses & Dispensing change in effective power Q This is a plot of the expression, Q = -xF(2L 1 + F) when x = m and L 1 = -3.00D. Note that Q = 0 when F = 0, or when (2L 1 + F) = 0. This is a plot of the expression, Q = -x (L 1 + F) 2 when x = m and L 1 = -3.00D.Note that Q = 0 for the single case when (L 1 + F ) = 0. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Vergence impressed in near vision L 1 = 0 F´ L´ 2 = BVP The back vertex power of a lens represents the vergence leaving the back surface when the incident vergence is zero. L´ 2 = BVP The form of the lens is immaterial. Provided that the BVPs are the same, in distance vision, lenses of different forms are interchangeable. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Clearly, in near vision, lenses of the same back vertex power but made in different forms are not interchangeable. Now the lens form has changed to plano-convex, its other details remaining the same, and, for the same near object position, the vergence leaving the lens will be found to be Again you must find the front curve of the lens before tracing from the near object point. In this case, F 1 is found to be Finally if the lens form is changed to equiconvex, its other details remaining the same, for the same near object position, the vergence leaving the lens is now found to be Again you must find the surface powers of the lens before tracing from the near object point. In this case, F 1 = F 2 = Vergence impressed in near vision B L 1 = -3 B’ L’ 2 = L’ 2 = B’ L´ 2 = B´ In near vision, the vergence impressed by a lens depends not only on its BVP, but also upon its form and thickness. Here, a D lens made with a base curve, in glass, n = 1.5 and an axial thickness of 9mm, is used for near vision at -33.3cm. The vergence leaving the back surface of the lens is +6.58D. Ophthalmic Lenses & Dispensing Note that you must first find the front curve before you can determine L´ 2. F 1 = Click to skip to next topic Click to return to previous screen Click to return to CD contents

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For example, in the case of the lens made as a base meniscus the vergence leaving the lens was found to be +6.58D. Vergence impressed in near vision Near vision effectivity error L 1 = -3L’ 2 = The anticipated vergence (found from L´ = L + F ) for the lens forms which have just been considered is = +7.00D. For this form the error due to near vision effectivity is D. Ophthalmic Lenses & Dispensing The difference between the actual vergence leaving the lens, L´ 2, and the anticipated vergence on the basis of thin lens theory, has been called the error due to near vision effectivity (or near vision effectivity error, NVEE). Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Vergence impressed in near vision symmetrical trial lens L´ 2 = NVEE = trial lens of common form L´ 2 = NVEE = Ophthalmic Lenses & Dispensing Near vision effectivity error final lens L´ 2 = NVEE = Errors due to near vision effectivity are a problem with medium to high-power plus lenses. Suppose that the trial lens used to determine the near vision prescription was equiconvex in form. We have seen that the NVEE of the final lens form which is likely to be dispensed is -0.42D. Its NVEE is seen to be only -0.12D. Changing from this form to the final lens form without adjusting the power of the lens would mean that there is a loss in power of about 0.3 D. If the trial lens was of the usual plano-convex form, with the curved surface designed to face the eye, the “error” would be even worse, almost 0.50D! Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Vergence impressed in near vision Correction Factors 3738 When lenses are prescribed for near vision, the form of the final lens is usually different from that of the trial lens. In such cases the back vertex power of the final lens must be increased by the amount shown in the Table opposite so that the effect of the final lens is the same as that of the trial lens. The lens is then said to be compensated for errors due to near vision effectivity. Typically, compensation is required for both single vision lenses and for the near addition of bifocal lenses. Note that in the case of bifocal additions, the compensation is also required for the near addition when the DP precription is minus. This correction factor is valid when the seg is on the back surface of the lens. Near addition (D) Lens Power Lenses of different forms are not interchangeable In near vision. Ophthalmic Lenses & Dispensing In practice, tables of correction factors are available giving compensation for NVEE. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Centration of spectacle lenses Horizontal centration of lenses Near centration distance Obtaining the measurements Geometrical insetting of bifocal segments Specification of segment top position Prismatic aspheric lenses Vertical centration - the centre of rotation condition The centre of rotation condition for near vision Centration errors - dispersion - off-axis blur Ghost images due to prism in a lens Graphical construction to find prismatic effect Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Furthermore, if progressive power lenses were dispensed without specifying monocular centration distances, their corridors of clear vision would be offset. MRMR MLML Horizontal centration of lenses PD Ophthalmic Lenses & Dispensing Look very carefully at this face.Do you notice how asymmetric it is? The asymmetry becomes very obvious if we bisect the face! This distance is called the interpupillary distance or PD. You will see that, really, the PD is only useful as a check for our measurement of the horizontal centration distance. OC The type of centration errors we could make if we just specified a binocular PD is illustrated here. It is assumed that the horizontal centre distance of the frame exactly matches the PD, so that no horizontal decentration needs to be specified. The horizontal centration specifications which are actually required are the monocular centration distances, measured from the centre of the bridge of the frame. They are shown here as M R and M L. These distances are equal. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Under these circumstances the centration points would lie where the visual axes intersect the spectacle plane. In the absence of prescribed prism, the optical centre of the lens would be positioned at the centration point. We have already noted that we need to know the distance from each centration point to the mid-point of the bridge of the frame which the subject is to wear. This distance, of course, is the same as the horizontal distance between the mid-point of the bridge of the frame and the eye’s centre of rotation. Optimum position for optical centre mid-point of frame bridge monocular CD spectacle plane R. Horizontal centration Centration distance Ophthalmic Lenses & Dispensing What we are trying to achieve is demonstrated here. First, note that we assume that the visual axes pass through the centres of the eyes’ pupils and through the eyes’ centres of rotation and that these axes are parallel in distance vision and are shown directed towards infinity. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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spectacle plane Optimum position for near optical centre centres of converging pupils NCD CD s l NCD = CD. l l + s Horizontal centration Near centration distance Ophthalmic Lenses & Dispensing In near vision, we are interested in the monocular near centration distances measured in the spectacle plane. Notice that this distance is not the same as the distance between the pupil centres. The near centration distance is seen to be a function of the distance CD. If we know the reading distance, l, and the centre of rotation distance, s, then the NCD can be expressed in terms of the CD. Remember, that it is the monocular NCDs which should be recorded. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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.. Dot pupil centres Horizontal centration Obtaining the measurements Ophthalmic Lenses & Dispensing The following routine will be found to provide accurate and consistent results. Fit the frame Mark centre of bridge Attach tape - if empty frame Click to skip to next topic Click to return to previous screen Click to return to CD contents

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.. Monoc CDs are recorded as follows: 32 / 35. The right eye value is always written first. Horizontal centration Obtaining the measurements Ophthalmic Lenses & Dispensing The monocular centration distances can now be measured and recorded. Right eye value Left eye value Click to skip to next topic Click to return to previous screen Click to return to CD contents

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segment aperture seen only by L eye seen only by R eye binocular field Horizontal centration Bifocal inset Ophthalmic Lenses & Dispensing Bifocal segments are usually inset to bring the near fields into coincidence. Suppose in the diagram below that the bifocal segments are simply apertures in otherwise, opaque, occluders. The fields of view obtained through the right and left apertures would have the same shape as the apertures, here supposed to be D-shape, flat-top segments. To make the fields coincident they should overlap exactly in the near point plane. If the near fields do not coincide there will be areas which fall only within the field of one eye. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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ODOD B mid-line spectacle plane Bifocal segment insetting Ophthalmic Lenses & Dispensing The following method ensures that the segment insetting brings the near fields into coincidence. This diagram shows the distance portion of a plus bifocal lens which has been correctly centred for distance vision in front of the eye. If there were no spectacle lens in front of the eye, it would rotate into this direction in order to view the near object point, B, on the midline. This is the bifocal segment It has now been placed in position in the spectacle plane and inset so its centre lies on the converging visual axis. It goes without saying, that we want the centre of the segment aperture to lie on the visual axis in order for the eye to obtain the maximum field of view. ODOD Now, a plus lens which is correctly centred for distance vision has been placed in front of the eye and it is seen that, owing to the base out prism exerted by the lens, the eye must converge more to view the near object. In order to obtain the maximum field of view. the centre of the bifocal segment should be placed at the point where the visual axis intersects the spectacle plane. Clearly, when the distance prescription is positive the segment must be inset more than we would decentre single vision lenses for near vision. Now, we will place a bifocal lens whose distance prescription is negative in front of the other eye. Note that the minus lens exerts prism base in at the near visual point, relieving the effort of convergence,so minus bifocal lenses should be inset less than single vision lenses. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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monoc CD = p Visual axis distance OC R Geometrical insetting to bring the near fields into coincidence l s g g = p.L / (L + F - S)g = 3.p / (40 - F) Table of geometrical insetting monocular centration distances OSOS Ophthalmic Lenses & Dispensing It can be shown that the geometrical inset, g, for a distance lens of power, F, mounted S (D) in front of the eye’s centre of rotation and L (D) from the near point is given by: Substituting -3.00D for L and D for S (so the lens-eye separation is 27mm) we obtain the useful average rule for geometrical inset: The following table of geometrical insetting has been prepared from this expression Tables such as this may be provided in practice in order to assist bifocal fitting. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Specification of segment top position For example, if the segment heights are measured as 23 mm, segment top position and the vertical box dimension of the frame is 44 mm, Ophthalmic Lenses & Dispensing Bifocals for general purpose use are usually fitted so that the segment top is tangential with the lower edge of the iris. With most subjects, this coincides with the line of the lower lid. If the final frame is fitted, the height can be measured with a simple ruler. Note, however, that segment height is defined as the distance from the segment top to the lower horizontal tangent to the lens periphery and should be measured as illustrated for the right eye. HCL Although the segment heights could be specified as measured, it is better to give the segment top position, which is the vertical distance of the segment top, above or below the horizontal centre line. Remember to measure the other eye also then the segment top position = 1 above HCL. 22 Click to skip to next topic Click to return to previous screen Click to return to CD contents

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A 1 = pole of aspherical surface s = centre of rotation distance P = prismatic effect of lens A1A1 x s P Horizontal centration Prismatic aspheric lenses An aspherical surface such as the ellipsoid shown here has a pole. Ophthalmic Lenses & Dispensing When dispensing aspheric lenses which incorporate prism, there is a further consideration. To obtain the best performance from the aspheric design, the pole of the aspherical surface should lie on the visual axis. When prism is incorporated in the lens, the visual axis is deviated towards the apex of the prism. The pole of the aspherical surface no longer lies on the visual axis. In order to coincide with the the visual axis, the pole of the aspherical surface must be decentred towards the prism apex, that is, in the opposite direction to the prism base. The amount of decentration x = P.s / 100 = 0.3 P So when the prism power is 3 , the necessary decentration is about 1mm. The amount of decentration can be found as follows. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Vertical centration of spectacle lenses optical axis optimum position for optical centre visual axis optimum position for optical centre optical axis pantoscopic angle Typically the pantoscopic angle is It is easily shown, from the geometry of the figure, that for each 1 0 pantoscopic angle, the optical centre should be lowered by 0.5 mm Ophthalmic Lenses & Dispensing If the spectacle frame is fitted like this, the optimum position for the optical centre would be directly in front of the centre of the pupil. However, spectacle frames are normally fitted so that the front is parallel with the line joining the supra -orbital ridge and the chin, i.e.,the front is tilted before the eyes. The degree of tilt is called the pantoscopic angle. In order to compensate for the pantoscopic tilt the optical centre of the lens must be lowered from its position before the pupil centre. The amount depends upon the pantoscopic angle. The optical axis of the lens then continues to pass through the eye’s centre of rotation. The centre of rotation condition Click to skip to next topic Click to return to previous screen Click to return to CD contents

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.. Vertical centration of spectacle lenses Taking note of the pantoscopic angle, the required height of the optical centre can be obtained by measuring the pupil centre height from the lower horizontal tangent to the lens periphery and simply subtracting 0.5mm from this measured value for each 1º of pantoscopic tilt. or, better still, give the required heights of the OC from the HCL. HCL Ophthalmic Lenses & Dispensing The centre of rotation condition for vertical centration is easily satisfied by fitting the frame and marking the position of the pupil centres when the head is held in the primary position. Required position of optical centre.. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The centre of rotation condition for near vision O Optical axis R R = eye’s centre of rotation O = optical centre of lens Vertical centration of spectacle lenses We see that in satisfying the centre of rotation condition for distance vision, the requirement for near vision is satisfied at the same time! N Near visual axis N = near visual point D = distance visual point D Primary visual axis for DV Ophthalmic Lenses & Dispensing This figure indicates that the centre of rotation condition has been satisfied for distance vision. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Centration errors Ophthalmic Lenses & Dispensing The most obvious problem which arises from incorrectly centred lenses is that unprescribed prismatic effect is introduced before the eyes. For example if plus lenses are not decentred inwards for near vision there will be base out prism exerted at the near visual points... Errors in the vertical meridian of the lenses may give rise to intolerable vertical differential prismatic effects. In addition to problems with binocular vision, the prism which is introduced by poorly centred lenses may produce other noticeable effects which are a source of complaint. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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dispersion by a prism incident white light is dispersed by a prism into its monochromatic constituents. The angle between the emergent red and blue ends of the spectrum represents the angular dispersion or chromatic aberration of the prism. chromatic aberration = P vdvd Centration errors Ophthalmic Lenses & Dispensing The chromatic aberration exhibited by a plano prism of power, P, made in a material whose V-value or Abbe Number is denoted by v d, is given by: Click to skip to next topic Click to return to previous screen Click to return to CD contents

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incident white light is dispersed by a lens into its monochromatic constituents transverse chromatic aberration TCA = just as in the case of a plano-prism. P vdvd c OC lens power = F TCA = c.F vd vd dispersion by a lens Centration errors Ophthalmic Lenses & Dispensing The same effect occurs with lenses, the dispersion being due to the prismatic effect of the lens at the point of incidence. In the case of a lens, the effect is known as transverse chromatic aberration, TCA. P represents the prismatic effect exerted by the lens at the point of incidence and is given by Prentice’s rule, P = c.F. So, in the case of a lens, where, c.F is the prismatic effect. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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high-contrast target The effect of transverse chromatism upon the wearer depends upon the object being viewed. Now consider the low-contrast target illustrated on the right. When viewed through a prism whose base-setting coincides with the lines, no effect can be seen on the target. When the prism is turned through 90º, however, the edges of the bars of the target appear blurred, reducing acuity. low-contrast target effects of dispersion Centration errors Ophthalmic Lenses & Dispensing Consider first a high contrast target such as the one shown on the left. When viewed through a prism whose base setting coincides with the lines, no effect can be seen on the target. When the prism is turned through 90º, however, coloured fringes may be seen on the edges of the bars of the target. This effect is sometimes described as off-axis blur and is due to the dispersion caused by the prism along its base-setting. It is minimized, by using lens materials with the highest possible V-value, or Abbé Number. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Notice that the spot appears displaced towards the prism apex and that sitting above the spot you can see a ghost image which is in focus, like the spot. Rotate the prism and notice how the ghost image also rotates. It always appears displaced towards the prism apex. spot room A H E D X R C E G S K B Y T Q P K L N V D X A effects of poorly centred lenses Centration errors This experiment demonstrates another effect of poorly centred lenses. Now click here to >> turn off the room light. A H E D X R C E G S K B Y T Q P K L N V D X A spot room On the next mouse click you will introduce a low-power prism (say ½ ) base DOWN in front of the spotlight... The optics of this ghost image are considered in the next slide. Ophthalmic Lenses & Dispensing Click here to turn on >> the muscle spotlight. Once you have located this ghost in a darkened consulting room, switch the room light on again and see that the ghost is still visible, although somewhat more difficult to discern. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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If the refractive index of the prism material is 1.5 the intensity of this ghost image is only 0.15%. Despite its dimness, you will have seen in the experiment that it is quite noticeable, even when the lights are switched back on in the consulting room. effect of poorly centred lenses Centration errors This shows the formation of the ghost image, which is produced by total internal reflection at the lens surfaces. The deviation of this catoptric image is (3n - 1) a Dioptric image Ghost image The dioptric image produced by the prism is seen in this direction. The deviation of this image, d = (n - 1)a In order for a ghost image to be troublesome, three conditions must be satisfied. 1) The ghost should be bright enough to be noticeable. 2) The vergence of the ghost should be similar to that of the lens. The vergence of this ghost can be shown to be (3n-1).F / (n-1). When n = 1.5, this turns out to be 7F. For a plano-prism its vergence is zero and it is in sharp focus. 3) The ghost must lie close to the fixation line (but not superimposed). These requirements are all met by prisms of low power! This ghost image is often complained of by wearers of low-power prisms, by subjects who are wearing multifocals which have been prism-thinned and those who are wearing low-power, poorly centred lenses. A multi-layer anti-reflection coating reduces the intensity of the image, almost to zero. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Graphical construction to find prismatic effect on an astigmatic lens Example: find the vertical and horizontal prismatic effects at a point which lies 10mm down and 4mm inwards from the optical centre of the lens R / x 30. Begin by finding the prism due to the sphere.. OC This is the optical centre of the spherical component. This is point at which we are finding the prismatic effect.. R Notice the cross-sectional shape of the lens in the region of R. In the vertical meridian the lens shape resembles a prism base UP. In the horizontal meridian, the cross-sectional shape resembles a prism with its base OUT. The prismatic effect due to the sphere is found from Prentice’s law, P = cF. In the vertical meridian, c = 1cm In the horizontal meridian c = 0.4cm. Hence: P V = 1 x 4 = 4 base UP P H = 0.4 x 4 = 1.6 base OUT Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Graphical construction to find prismatic effect on an astigmatic lens Now we must consider the prism due to the cylinder x 30 There is no power along the axis meridian of a plano-cylinder, hence the cylinder can exert no prismatic effect along its axis meridian. All the power of a cylinder lies at right angles to its axis, i.e., along its power meridian, so a cylinder exerts prismatic effect only along its power meridian x 30 Example: find the vertical and horizontal prismatic effects at a point which lies 10mm down and 4mm inwards from the optical centre of the lens R / x 30. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Graphical construction to find prismatic effect on an astigmatic lens Example: find the vertical and horizontal prismatic effects at a point which lies 10mm down and 4mm inwards from the optical centre of the lens R / x x 30. R We will first consider how to find the base direction of the prism due to the cylinder. The cylinder can only produce a prismatic effect at right angles to its axis, i.e., along the 120 meridian of the lens. This is the point at which we are finding prismatic effect. Now notice where the thickest part of the cylinder lies with respect to point, R. It is up and out along 120 with respect to R. In order to determine the prismatic effect at R, we must resolve the vertical and horizontal decentration along the power meridian of the cylinder. P We must find the perpendicular distance, PR, of the point R from the cylinder axis. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Graphical construction to find prismatic effect on an astigmatic lens Example: find the vertical and horizontal prismatic effects at a point which lies 10mm down and 4mm inwards from the optical centre of the lens R / x 30. The simplest method is to use a graphical construction, details of which follow. N First, construct an origin and mark the nasal side of the lens 30º Next draw in the cylinder axis along its prescribed axis direction, here, 30º. 30º Then mark the position of the point, R, to scale on the diagram, here 10mm down and 4mm in from the origin. Choose a large scale, say, 10:1.. R P Now drop a perpendicular from R to the cylinder axis meeting it at P. Does P lie above R or below it? Does P lie on the nasal side of R or does it lie on the temporal side of R? Next, determine the base direction of the prismatic effect at R. Ask yourself, “where does P lie with respect to R?” Evidently, P lies above R and on its temporal side. When the cylinder is positive in sign, as in this example, P represents the position of the prism base, so here, the base is UP & OUT. If the cylinder had been negative in sign, P would represent the position of the prism apex. The base would have been DOWN & IN Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Graphical construction to find prismatic effect on an astigmatic lens Example: find the vertical and horizontal prismatic effects at a point which lies 10mm down and 4mm inwards from the optical centre of the lens R / x 30. N. R P The distance, PR, represents the effective decentration of the plano- cylinder along its power meridian. Since we require the vertical and horizontal prismatic effects we can now resolve PR into vertical and horizontal components, PQ and QR. Q PQ is the effective decentration of the plano-cylinder in the vertical meridian. The vertical prismatic effect at R due to the cylinder is given by: PQ (cm) x Fcyl. QR is the effective decentration of the plano-cylinder in the horizontal meridian. The horizontal prismatic effect at R due to the cylinder is given by: QR (cm) x Fcyl. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Graphical construction to find prismatic effect on an astigmatic lens Example: find the vertical and horizontal prismatic effects at a point which lies 10mm down and 4mm inwards from the optical centre of the lens R / x 30. N. R P Q In this example: by measurement: PQ = 9.2mm QR = 5.3mm Hence the prism due to the cylinder is: P V = 0.92 x 2.00 = 1.84 base UP P H = 0.53 x 2.00 = 1.06 base OUT Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Graphical construction to find prismatic effect on an astigmatic lens Example: find the vertical and horizontal prismatic effects at a point which lies 10mm down and 4mm inwards from the optical centre of the lens R / x 30. Prism due to cylinder = 1.84 base UP & 1.06 base OUT OC.. R So we have found: Prism due to sphere = 4 base UP & 1.6 base OUT Prism due to sphere N. R P Q Prism due to cylinder Total prismatic effect = 5.84 base UP & 2.66 base OUT Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens thickness and weight Edge thickness of a minus lens Thickness of plus lens Variation in thickness of aspheric lens Astigmatic prismatic lenses Graphical construction for thinnest point on edge Lens weight Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens thickness Edge thickness of minus lens e t s This is a cross-sectional view of a plano-concave lens e is the edge thickness t is the centre thickness s is the sag of the concave surface From the geometry of the figure e = t + s Ophthalmic Lenses & Dispensing The quantity, s, is calculated from the sag formula, s = y2y2 r + r 2 - y 2 r y Centre of curvature of surface Suppose we wanted to find the sag of a 10.00D curve at 60 diameter, the refractive index of the lens material being Sag at 60, n = 1.56 First, find the radius of the surface from: r = 1000(n - 1) / Curve = 1000( ) / 10 = 56mm The semi-diameter of the lens is 60 / 2 = 30mm We have: y = 30, r = 56 s = y2y2 r + r 2 - y 2 = = 8.71mm. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens thickness Edge thickness of minus lens s2s2 This is a cross-sectional view of a curved minus lens e is the edge thickness t is the centre thickness s 1 is the sag of the convex surface From the geometry of the figure: t + s 2 = e + s 1 so e = t + s 2 - s 1 e t s1s1 s 2 is the sag of the concave surface Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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-6.00 at 60 edge subs = 7.0mm At diameter 60mm: s 1 = 3.5 s 2 = 9.5 e = t +s 2 - s 1 = 7.0mm Lens thickness Edge thickness of minus lens Consider a lens, made in glass, n = 1.523, with base and centre thickness of 1.0mm. 7.0mm at 50 edge subs = 5.0mm If the diameter is reduced to 50mm: s 1 = 2.4 s 2 = 6.4 e = t +s 2 - s 1 = 5.0mm 5.0mm If the diameter is reduced to 40mm: s 1 = 1.5 s 2 = 4.0 e = t +s 2 - s 1 = 3.5mm at 40 edge subs = 3.5 mm 3.5mm The cross-sectional shape of a minus lens reminds us that the smaller the lens diameter, the thinner will become the lens. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Note: If the overall diameter of the lens is 50mm but it has been decentred 4mm inwards, the edge thickness would change as follows: Lens thickness Edge thickness of minus lens Consider a lens, made in glass, n = 1.523, with base and centre thickness of 1.0mm. 50. This is the geometric centre of the lens This is the optical centre of the lens The edge thickness on the temporal side of the lens, e T, is that of a 58 diameter lens. The edge thickness on the nasal side of the lens, e N, is that of a 42 diameter lens. e T = 6.2mm e N = 3.7mm Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens thickness Thickness of plus lens t e This is a cross-sectional view of a plano-convex lens e is the edge thickness t is the centre thickness s is the sag of the convex surface From the geometry of the figure t = s + e s Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens thickness Thickness of plus lens This is a cross-sectional view of a curved plus lens e is the edge thickness t is the centre thickness s1s1 s2s2 e s 1 is the sag of the convex surface From the geometry of the figure: t + s 2 = e + s 1 so t = e + s 1 - s 2 s 2 is the sag of the concave surface t Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens thickness Thickness of plus lens At diameter 65mm: F 1 = t = 7.5mm s 1 = 11.3 s 2 = 4.3 e = t - (s 1 - s 2 ) = 0.5mm at 65 edge subs = 0.5mm 7.5mm At diameter 50mm: F 1 = t = 7.5mm s 1 = 6.4 s 2 = 2.5 e = t - (s 1 - s 2 ) = 3.6mm at 50 edge subs = 3.6mm Consider a uncut, made in CR 39, n = 1.498, with base and edge thickness of 0.5mm. When this uncut is edged down to a diameter of 50mm the edge thickness increases. 7.5mm At diameter 50mm: F 1 = t = 7.5mm s 1 = 4.0 s 2 = 1.6 e = t - (s 1 - s 2 ) = 5.1mm at 40 edge subs = 5.1mm 7.5mm If edged down to the smaller diameter of 40mm the edge thickness increases even more. This edge thickness is unacceptable and it is quite clear that plus lenses over about +2.00D should not be edged down to much smaller sizes. In order to obtain a reasonable edge thickness, plus lenses should be surfaced down to a suitable diameter for the shape when all the prescription details are known. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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1.5 mm We will custom-design the lens to give 1.5mm at this point on the edge. It will be realised that the edge substance of this lens varies from only 1.5mm to 1.6mm round the edge of the lens. 1.5 mm 1.6mm Lens thickness Consider the specification / x 180 which is to be edged to a 48 x 40 oval shape The centre thickness will then be 3.5 mm. Thickness of plus lens The problems are even greater with plus astigmatic lenses The horizontal diameter of the lens is 48mm The vertical diameter of the lens is 40mm The horizontal power of the lens is The vertical power of the lens is e thin The thinnest points on the edge of the lens lie at the extremities of the vertical meridian. e thick 1.6 mm The thickest points on the edge of the lens lie at each end of the horizontal meridian. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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e thin The thinnest points on the edge of the lens now lie at the extremities of the horizontal meridian. It will be realised that the edge substance of this lens now varies from 1.5mm to 3.3mm round the edge of the lens. 3.3 mm 1.5mm We will again custom-design the lens to give 1.5mm at this point on the edge. 1.5mm Not only is the lens 1.0mm thicker in the centre, the edge now varies from 1.5mm at the thin edge, to 3.3mm at the thick edge simply because the axis lies at 90. e thick 3.3mm The thickest points on the edge of the lens now lie at each end of the vertical meridian. Lens thickness Now consider the specification / x 90 which is to be edged to the same oval shape The centre thickness will then be 4.5 mm. Thickness of plus lens The problems are even greater with plus astigmatic lenses The horizontal diameter of the lens is 48mm The vertical diameter of the lens is 40mm The horizontal power of the lens is The vertical power of the lens is Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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At diameter 65mm: Spherical lens F 1 = t = 7.5mm Aspheric lens F 1 = t = 6.4mm Spherical Aspheric uncuts at 65 edge thickness = Variation in thickness of aspheric lens Lens thickness At diameter 50mm: Spherical lens F 1 = t = 7.5mm Aspheric lens F 1 = t = 6.4mm Spherical Aspheric uncuts at 50 edge thickness = Aspheric lenses have the advantage that they are flatter and thinner than spherical lenses so that when edged down the edge thickness is not so great Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Astigmatic prismatic lenses Lens thickness We will now consider the thickness of circular plus lenses which incorporate both cylindrical and prismatic power. Variation in edge thickness of circular lenses In the general case, a plus lens may have spherical, cylindrical and prismatic power. spherical element This is the spherical component of the lens. It may be considered to incorporate the bending of the lens. Note that if the element is centred, its edge thickness will be the same all round the periphery. cylindrical element This is a plano-cylindrical component with its axis at 90. Note that this component has zero edge thickness at the extremities of the 180 meridian. prismatic element This is the prismatic component with its base DOWN (at 270). Note that this component has zero edge thickness at its apex (at 90). Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Astigmatic prismatic lenses Lens thickness spherical element cylindrical element prismatic element The thickness of a plus sphero-cylindrical lens which incorporates prism (or decentration) is seen to be made up from the thickness of each of these three individual components. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Astigmatic prismatic lenses Lens thickness The spherical element is The cylindrical element is x 90 Consider the prescription R / x 90 with 4 base OUT which is to be produced with an uncut diameter of 60mm and zero edge thickness. The prismatic element is 4 base OUT The centre thickness of this component is 1.9mm The centre thickness of this component is 2.4mm The centre thickness of this component is 3.6mm Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The spherical element +2.00, t C = 1.9 Astigmatic prismatic lenses Lens thickness Consider the prescription R / x 90 with 4 base OUT which is to be produced with an uncut diameter of 60mm and zero edge thickness. The thickness of the specification is made up as follows: The prismatic element 4 base OUT, t C = 2.4 The cylindrical element x 90, t C = 3.6 The total centre thickness of the lens is seen to be 7.9mm. Note that the edge thickness of this lens is zero at the nasal edge. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Astigmatic prismatic lenses Lens thickness The spherical element is The cylindrical element is x 90 Now consider the prescription R / x 90 with 4 base DOWN also to be produced with an uncut diameter of 60mm and zero edge thickness. The centre thickness of this component is 1.9mm The centre thickness of this component is 2.4mm The centre thickness of this component is 3.6mm The prismatic element is 4 base DOWN Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The spherical element +2.00, t C = 1.9 Astigmatic prismatic lenses Lens thickness R / x 90 with 4 base DOWN 60mm and e thin = 0. The thickness of the specification is made up as follows: If this lens is to have zero edge thickness it is seen that the centre thickness must be reduced. The prismatic element 4 base DOWN, t C = 2.4 The cylindrical element x 90, t C = 3.6 Note that the minimum edge thickness of the lens is no longer zero. Both prism and cylinder contribute some thickness to the thinnest point on the edge. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The spherical element +2.00, t C = 1.9 Astigmatic prismatic lenses Lens thickness R / x 90 with 4 base DOWN 60mm and e thin = 0. To obtain zero edge thickness, the centre thickness can be reduced by 2.0mm. To obtain zero edge thickness the centre thickness of the lens need be only 5.9mm. The prismatic element 4 base DOWN, t C = 0.4 The cylindrical element x 90, t C = 3.6 The thickness here can be reduced by 2.0mm Now the edge thickness of the lens is zero. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Astigmatic prismatic lenses Lens thickness R / x 90 with 4 base DOWN 60mm and e thin = 0. Note the position on the edge of the lens where the edge thickness is zero. e min In fact, there are two points on the edge where the edge thickness is a minimum. 180 meridian Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Astigmatic prismatic lenses Lens thickness The position on the edge of an astigmatic, prismatic lens where the thickness is a minimum can be determined by means of a graphical construction. e min minus cyl axis The cylindrical element Note first, that the thinnest point on the edge of the cylinder lies at the extremities of the minus cyl axis. The prismatic element Also note that the thinnest point on the edge of a plano prism lies at the prism apex. prism apex When these two components are combined, the thinnest point on the edge of the combination must lie somewhere between the prism apex and the minus cylinder axis. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The angle between the edge of the ruler (line EFD) and the base line AB, (angle FEB in the figure) gives the angle, , between the minus cylinder axis direction and the point on the edge of the astigmatic, prismatic lens where the thickness is a minimum. Note the angle is always measured in the acute angle between the minus cylinder axis direction and the prism base setting, in the direction towards the prism apex. Astigmatic prismatic lenses Lens thickness The position on the edge of an astigmatic, prismatic lens where the thickness is a minimum can be determined as follows. First, calculate the quantity: z = 200P dF where P is the prism power d is the lens diameter and F is the power of the cylinder Next, draw a horizontal line, AB, 10 units long to represent the minus cylinder axis direction. A B C Then, construct a perpendicular, BC, from B Next, calculate the acute angle, , between the base setting of the prism and the minus cylinder axis direction and draw a line from B, of length, z, inclined at angle, , to AB, (line BD in the figure below). D z Finally place a ruler on the construction and adjust its position until the zero lies on line AB, the edge passes through point D and the distance from zero to the perpendicular, BC, is exactly 10 units. (EF = 10 units in the diagram below.) E F Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Then, construct a perpendicular, BC, from B Finally place a ruler on the construction and adjust its position until the zero lies on line AB, the edge passes through point D and the distance from zero to the perpendicular, BC, is exactly 10 units. (EF = 10 units in the diagram below.) where P is the prism power, 4. d is the lens diameter, 60 and F is the power of the cylinder, 4 By measurement angle = 20º. Remember that when = 90 there are two minima on the edge of the lens, each symmetrical with the minus cylinder axis on the side of the prism apex. Next, calculate the acute angle, , between the base setting of the prism, which lies at 90, and the minus cylinder axis direction, which is at 180. The angle, , is 90º in this example, so we must draw a line from B, of length, 3.33 units, inclined at 90º to AB (line BD in the figure below). Astigmatic prismatic lenses Lens thickness First, calculate the quantity: z = 200P dF Next, draw a horizontal line, AB, 10 units long to represent the minus cylinder axis direction. A B C = 90 D z = 3.33 E F By way of example, take the prescription R / x 90 with 4 base DOWN which is to be produced with an uncut diameter of 60mm and zero edge thickness. z = 200x4 60x4 = 3.33 Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Ophthalmic Lenses & Dispensing Density One of the physical properties which is published by lens material manufacturers is the density of the material. Material n d Density CR mid-index Polycarbonate high index high index high index White White White White White Glasses Plastics The density is expressed in grams per cubic cm (g/cm 3 ) Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Ophthalmic Lenses & Dispensing Density Density expresses the weight in grams of one cubic centimetre of the material. 10mm CR 39 material has a density of This means that 1cm 3 of CR 39 weighs 1.32g. Water has a density of exactly 1.0 so 1cm 3 of water weighs 1gram. 1cm 3 of CR 39 weighs 1.32 grams so CR 39 is 1.32x heavier than water. 1cm 3 of crown glass weighs 2.54g so this glass is 2.54x heavier than water and about twice as heavy as CR 39! Note that the weight is found by multiplying the volume by the density of the material. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Ophthalmic Lenses & Dispensing Density Naturally, we want spectacle lenses to be as light as possible. This means that the density of the material must be as low as possible. The density of plastics materials is about half that of glass so that plastics lenses are about half the weight of glass lenses. But, in reality we cannot just compare the densities of the materials since higher refractive index materials will have less volume, owing to the fact that the same curve will produce a higher surface power on a higher refractive index material. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Weight of plus lens Ophthalmic Lenses & Dispensing The weight of a circular spectacle lens is found by multiplying the volume of the lens by the density of the lens material. A plano-convex lens can be considered to be made up from two separate components. a spherical cap… Consider, first, the cylindrical plate y e The volume of this plate is .y 2.e …combined with a cylindrical plate. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Weight of plus lens Ophthalmic Lenses & Dispensing Now consider the spherical cap y The volume of the cap is .s 2 (3r - s) / 3 y s r Click to skip to next topic Click to return to previous screen Click to return to CD contents

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The volume of a plano-convex lens is the sum of the volumes of the cylindrical plate and the spherical cap i.e., volume = . (s 2 (3r - s) / 3 + y 2.e). It can be seen that if the plus lens is of the usual curved form it is necessary to subtract the spherical cap which forms the concave surface from the sum of the convex cap and the plate which represents the edge thickness. So the volume of a plus meniscus lens is .(s 1 2 (3r 1 - s 1 ) / 3 - s 2 2 (3r 2 - s 2 ) / 3 + y 2.e) where s 1 and r 1 relate to the convex surface and s 2 and r 2 relate to the concave surface. Lens weight Weight of plus lens Ophthalmic Lenses & Dispensing The volume of the plate is .y 2.e spherical cap cylindrical plate The volume of the cap is .s 2 (3r - s) / 3 The volume of a plano-convex lens is the sum of the volumes of the cylindrical plate and the spherical cap. Note that the density of the lens material is normally quoted in grams per cubic centimeter so the volume of the lens must be calculated in cubic centimeters, by substituting s, r and y in cm into the above formula. Also, the formula relates only to the volume of a round lens. The weight of the lens is found by multiplying its volume by the density of the lens material. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Weight of minus lens Ophthalmic Lenses & Dispensing The volume of the plate is .y 2.e and the volume of the cap is .s 2 (3r - s) / 3 A plano-concave lens can be considered to consist of a thick cylindrical plate of thickness, e, from which a spherical cap has been removed. So the volume of a plano-concave lens is the volume of the plate which represents the edge thickness minus the volume of the spherical cap. e hence, the volume of the lens is .(y 2.e - s 2 (3r - s) / 3) Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Weight of minus lens Ophthalmic Lenses & Dispensing In the case of a curved form minus lens, the volume of the convex spherical cap which represents the front surface must be added... … so the volume of a minus meniscus lens is .(s 1 2 (3r 1 - s 1 ) / 3 - s 2 2 (3r 2 - s 2 ) / 3 + y 2.e) where s 1 and r 1 relate to the convex surface and s 2 and r 2 relate to the concave surface. This is exactly the same result as that obtained for plus meniscus lenses. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Weight of minus lens Ophthalmic Lenses & Dispensing …the volume of a flat cylindrical plate which represents the edge thickness... The volume of meniscus lenses is .(s 1 2 (3r 1 - s 1 ) / 3 - s 2 2 (3r 2 - s 2 ) / 3 + y 2.e) Remember that the volume must be calculated in cubic centimetres by substituting s, r and y in cm into the above formula. Also, the formula relates only to the volume of a round lens. Once again, the weight of the lens is found by multiplying its volume by the density of the material. The total volume of a curved lens is made up from the volume of the convex spherical cap... This statement is true for both plus and minus lenses. …minus the volume of the concave spherical cap. .s 1 2 (3r 1 - s 1 ) / 3+ .y 2.e - .s 2 2 (3r 2 - s 2 ) / 3 Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens weight Weight of minus lens Ophthalmic Lenses & Dispensing The volume of the lens is .(s 1 2 (3r 1 - s 1 ) / 3 - s 2 2 (3r 2 - s 2 ) / 3 + y 2.e) Weight = volume x density = x 2.54 = 27.56g. As an example, consider a -6.00D lens made in spectacle crown glass, n = 1.523, D = 2.54, with a +4.00D base curve, a centre thickness of 1.0mm and edged to a 60mm diameter. The convex spherical cap has a radius of mm and a sag at ø 60 of 3.49mm. r 1 = cm s 1 = 0.349cm r 2 = 5.23cm s 2 = 0.946cm The edge thickness of the lens is 6.97mm e = cm y = 3.0cm The concave spherical cap has a radius of 52.3mm and a sag at ø 60 of 9.46mm. = .( (3x ) / (3x ) / x0.697) = 10.85cm 3 Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Off-axis performance of spectacle lenses Field diagrams Best form lenses - point focal lenses Best form lenses - Percival lenses Best-form lenses - Minimum T-Error lenses Aspheric spectacle lenses Tscherning’s Ellipses Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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+4.00 Lens design and performance The back vertex power of a lens is the vergence leaving the back surface when the incident vergence is zero, the light emanating from a distant object. It expresses the effect of the lens when the eye looks along the optical axis. The eye is not stationary but rotates to view through off-axis portions of the lens. The effect of the lens under these conditions varies with the ocular rotation and the form of the lens. When the eye rotates 20º the effect of the lens is +4.00/ The refracted pencil is afflicted with aberrational astigmatism. When the eye rotates 40º the effect of the lens is +4.00/ / º / º The form of this lens is quite shallow, the back surface power is only -1.50D. Off-axis performance of spectacle lenses Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance vertex sphere Field diagrams / º / º In practice, the off-axis powers are measured from the vertex sphere, an imaginary spherical surface concentric with the eye’s centre of rotation. For a 40º rotation of the eye, the sagittal oblique vertex sphere power is and the tangential oblique vertex sphere power is Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance vertex sphere FIELD DIAGRAM S S = plot of sagittal powers T T = plot of tangential powers Field diagrams / º / º.... The off-axis performance is usually shown in the form of a field diagram where oblique vertex sphere powers are plotted against ocular rotation. oblique vertex sphere powers ocular rotation For 0º rotation the effect is For 20º rotation the effect is +4.00/ For 40º rotation the effect is +4.00/ In practice, field diagrams are plotted by the computer. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Best form lenses FIELD DIAGRAM T & S Point focal lens The first best-form lenses were quite steeply curved and were designed to be free from oblique astigmatism. A +4.00D lens would need an inside curve of -6.00D. The field diagram for a point focal lens would look like this. Point focal lenses are free from oblique astigmatism but suffer from mean oblique error; the mean oblique power of the lens decreases as the eye rotates away from the axis. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Best form lenses Percival lens A second type of best-form lens was introduced which was flatter in form and designed to be free from mean oblique error. A small amount of oblique astigmatism was tolerated providing that the disk of least confusion of the astigmatic pencil fell on the retina. This followed a suggestion made by the ophthalmologist, Dr A. Percival and the design was subsequently named after him. A Percival form lens would need an inside curve of -4.00D FIELD DIAGRAM ST MOP = BVP The field diagram for a Percival form lens would look like this. The mean oblique power, (MOP) is seen to be the same as the back vertex power of the lens, the mean oblique error is zero. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Best form lenses Minimum T-Error lens Modern best-form lenses are designed with a compromise bending so that the tangential power is the same as the back vertex power. They are free from tangential error and known as Minimum T-Error forms. They exhibit a small amount of oblique astigmatism, but only about half that of a Percival design. A +4.00D lens would need an inside curve of -5.00D. The field diagram for a Minimum T-Error form would look like this FIELD DIAGRAM S T Minimum T-Error forms have the advantage that they perform well over a wide range of vertex distances. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Minimum T-error at design vertex distance S T Point focal at long vertex distance T & S Percival at short vertex distance ST Best form lenses Effects of changes in vertex distance with Minimum T-Error forms Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Aspheric spectacle lenses Point focal lens made in CR 39. The flatter a spectacle lens is made, the thinner it may become. Best-form lenses made with spherical surfaces are usually quite steeply curved and therefore thicker than if made in flatter forms. For example, a point focal +4.00Dlens made in CR 39 material, would have back curve of -5.25, and if the uncut diameter is 70mm and the edge thickness is assumed to be 1.0mm, then the centre thickness of the lens would be 6.6mm and the weight of the uncut lens would be 20.3 grams. Ophthalmic Lenses & Dispensing 6.6mm 1.0mm 70 Weight = 20.3g Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Aspheric spectacle lenses FIELD DIAGRAM T & S Point focal lens The field diagram for this point-focal design would look like this. Ophthalmic Lenses & Dispensing 6.6mm 1.0mm 70 Weight = 20.3g Click to skip to next topic Click to return to previous screen Click to return to CD contents

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FIELD DIAGRAM base curve The field diagram for this flatter form +4.00Dlens would look like this. The off-axis performance of this flatter curved form is unacceptable! Ophthalmic Lenses & Dispensing 6.0mm 1.0mm 70 Weight = 18.1g S T Lens design and performance Aspheric spectacle lenses If the +4.00D lens is now flattened so that its back curve is only -1.50D, the centre thickness will reduce to 6.0mm and the weight of the lens will reduce by some 2 grams. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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FIELD DIAGRAM for aspheric form with convex hyperboloidal surface, p = base curve The field diagram for this aspheric +4.00D lens would look like this. The off-axis performance is the same as the steeper curved lens with spherical surfaces! Ophthalmic Lenses & Dispensing 5.4mm 1.0mm 70 Weight = 16.0g Lens design and performance Aspheric spectacle lenses If the front, spherical curve of this flatter form lens is now replaced by a suitable aspherical surface, the negative surface astigmatism which is inherent in the aspherical surface can neutralise the astigmatism of oblique incidence. Furthermore, since the sag of the aspherical surface is less than the sag of a spherical surface of the same power, the centre thickness of the lens will reduce even further to 5.4mm. The weight of the aspheric lens will reduce to just 16 grams. T & S Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Ophthalmic Lenses & Dispensing Lens power (F) Back curve (F 2 ) Tscherning’s Ellipses - point focal lenses Before the advent of the computer, accurate trigonometric ray-tracing had to be performed by hand, using six or seven-figure logarithm tables. The procedure was lengthy and to save time, approximate equations were employed to provide a starting point for the design. These so-called, third-order equations for spectacle lenses were first investigated by G B Airy (1825), and later by F Ostwalt and M Tscherning towards the end the 19th century. The latter published comprehensive details of the forms in The equations are quadratic in type and plot as ellipses. These were plotted and described by A Whitwell earlier this century who called them the Tscherning’s Ellipses. According to third-order theory, lenses of power F made in a material of refractive index 1.5 and mounted 26.67mm in front of the eye’s centre of rotation, will be free from astigmatism for distance vision when the inside curve, F 2 is given by: F 2 = {14F -375± ( F-140F 2 )} / 28 Ostwalt DV Wollaston DV Ostwalt point-focal form Wollaston point-focal form D lenses It can be seen from the graph that there is a certain range of powers which can be made free from oblique astigmatism and that within this range there are two forms for each power. The shallower form, which is the one used in practice, is named after Ostwalt and the steeper form after W H Wollaston who had proposed such forms early in the 19th C. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Ophthalmic Lenses & Dispensing Lens power (F) Back curve (F 2 ) Similar ellipses can be constructed for point-focal, near vision lenses at a specified working distance, and also for Percival lens forms or minimum tangential error forms again, for either distance or near vision. Naturally, the ellipses differ in size and position in each of these different circumstances. For example, lenses made in a material of refractive index 1.5 and mounted mm in front of the eye’s centre of rotation, will be free from astigmatism for near vision at 25cm when the inside curve, F 2 is given by: F 2 = {14F -335± ( F-140F 2 )} / 28 Ostwalt DV Wollaston DV It is seen from the graph that Ostwalt near vision forms are somewhat shallower than the forms required for distance vision.The Wollaston NV forms are virtually the same. The Ostwalt near vision design for a D lens required for near vision at -25cm would be biconcave in form. In practice, the eye does not demand optimum acuity in near vision and best form lenses are invariably designed for distance vision. Ostwalt NV Wollaston NV Tscherning’s Ellipses - point focal lenses Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Ophthalmic Lenses & Dispensing Lens power (F) Back curve (F 2 ) If we assume that the lenses are mounted 26.67mm in front of the eye’s centre of rotation, we obtain the following ranges of powers for lenses which can be made free from oblique astigmatism, according to third-order theory, when made in materials of different refractive indices: n F max F min Although based upon approximate equations, these limits do give a good indication of the ranges of lenses which can be made free from astigmatism for different media The ellipses illustrated so far have assumed that the lenses are made in a material of refractive index 1.5. When the refractive index of the material increases, the range of lenses which can be made free from oblique astigmatism also increases, but curiously, only in the minus range. Whatever the refractive index, plus lenses over about cannot be made free from astigmatism when restricted to spherical surfaces. It can be shown that the limits of Tscherning’s ellipses, for lenses made in a material of refractive index, n, and mounted L´ 2 dioptres in front of the eye’s centre of rotation are given by: F max = F min = 2L´ 2 (n - 1) 2-3n (n -1) - n3n3 n + 2 2L´ 2 (n - 1) 2-3n (n -1) + n3n3 n Tscherning’s Ellipses - point focal lenses Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Lens design and performance Ophthalmic Lenses & Dispensing Lens power (F) Back curve (F 2 ) Similar equations to those given for point-focal lenses can be derived for third-order, Minimum Tangential Error forms. Assuming the same details as those used for the point-focal designs, namely, n = 1.5 and the lenses mounted 26.67mm in front of the centres of rotation, we find that lenses will exhibit minimum tangential error when their inside curves are given by: F 2 = {49F ± ( F-1295F 2 )} / 88. This equation also plots as an ellipse with limits and Ostwalt DV Wollaston DV Ostwalt Min T-error Wollaston Min T-error D lenses Tscherning’s Ellipses - minimum tangential error forms Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Iso -V-prism theory Spherical lenses Plano-cylindrical lenses Sphero-cylindrical lenses Iso-V-differential prism zones Graphical construction for iso-V-prism zones Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Consider the prescription R L -2.00/-2.00 x 180 In the right eye, the eye can roam 1cm from the optical centre before it meets 2 of prismatic effect (from c = P / F). In the left eye, the vertical power of the lens is -4.00, so the eye can roam 0.5cm from the optical centre before it meets 2 of prismatic effect. Iso-V-prism zone theory An iso-V-prism zone is a zone on a lens where vertical prismatic effect does not exceed a stipulated amount. All horizontal prismatic effect is ignored. R 20 mm wide L 10 mm wide In the case of spherical lenses the iso-V-prism zones are simply horizontal bands, their width can be calculated from Prentice’s law P = c.F. Hence the 2 iso-V-prism zones are horizontal bands. Spherical lenses Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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From the decentration relationship, the eye can roam, 2 / 2cm, or 10mm, along the 150 meridian before it meets 2 base 150 or 1 of vertical prism. Hence the 1 iso-V-prism zone is a band 20mm wide along the 60º meridian. The result could be depicted as follows: Power meridian Iso-V-prism zone theory In the case of plano-cylindrical lenses the iso-V-prism zones are still bands but they lie parallel to the axis meridian of the cylinder. Their dimensions can be determined as follows. The power of a cylinder lies at right angles to its axis meridian and it is necessary to determine how far along the power meridian the eye can roam before it encounters the vertical prismatic effect which is stipulated. 11 22 1 iso-V-prism zone for x 60 is a band, 20mm wide, lying along mm J J´ K K´ Note that along JJ´ the prism base is DOWN whereas along KK´ the prism base is UP. Plano-cylindrical lenses A A´ Along the axis meridian, AA´, the prismatic effect is zero. Ophthalmic Lenses & Dispensing 1 base UP is produced by 2 base 150 If the vertical prismatic effect is denoted by P V then the oblique prism, P R, which gives rise to P V, along the power meridian of the cylinder, is found from: P R = P V / sin (90 + ) Where is the axis direction of the cylinder. For example, suppose it is required to find the 1 iso-V-prism zone for the cylinder x 60. The power meridian lies along 150. The prism along 150 which gives rise to 1 vertically is: P R = P V / sin (90 + ) = 1 / sin 150 = 2 base 150. This could have been determined by a graphical construction as follows. Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Iso-V-prism zone theory In the case of sphero-cylindrical lenses the iso-V-prism zone will be parallel to the axis only when the cylinder axis lies at 90 or 180. In all other cases the band will not be parallel with the cylinder axis but will lie along some other meridian. These two points are plotted, one on each principal meridian of the lens, using the same method which was just described for plano-cylinders. It is essential to check the base direction along each principal meridian to ensure that the vertical prism components are both base UP or both base DOWN. Failure to do this will result in the band being 90º off! Since the boundary of the zone is a straight line, it is necessary to be able to locate two points in order to be able to construct the line. We must now determine how far the eye can roam along the 30 meridian before it meets 2 base UP at 30. The power along 30 is The prism along 30 is 2 . Sphero-cylindrical lenses Consider the 1 iso-V-prism zone for the lens /+3.50 x 30. The principal powers of the lens are -2.00D along 30 and +1.50D along Consider first the 30 meridian of the lens. The prism along 30 which will give rise to 1 base UP, is 1 / sin 30 = 2 base UP at base UP From the decentration relationship, the eye can roam 2 /2 cm or 10mm UP along the 30 meridian before it meets 2 base 30, (or 1 of vertical prism). 2 base UP at 30. J Choosing a suitable scale, point, J, can be plotted on the 30 meridian of the lens 10mm up along 30 from the optical centre O O 10 Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Iso-V-prism zone theory We must now determine how far the eye can roam along the 120 meridian before it meets 1.15 base UP at 120. The power along 120 is The prism along 30 is 1.15 . Sphero-cylindrical lenses Consider now the 120 meridian of the lens. The prism along 120 which will give rise to 1 base UP, is 1 / sin 120 = 1.15 base UP at base UP From the decentration relationship, the eye can roam 1.15 / 1.5 cm = 7.7mm DOWN along the 120 meridian before it meets 1.15 base 120, (or 1 of vertical prism) base UP at 120. J Choosing a suitable scale, point, J´, can be plotted on the 120 meridian of the lens, 7.7mm down from the optical centre, O.. J´ O 7.7 Note that the eye must rotate DOWN along 120 in order to encounter prism base UP. We can now construct the 1 base UP iso-V-prism line through the points JJ´. 1 base UP Note that the vertical prismatic effect at any point on this line is base UP. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Iso-V-prism zone theory 1 iso-V-prism zone for / x mm By measurement, the band is 12.2mm wide and lies along the meridian, 67.6º. Sphero-cylindrical lenses J. J´ O We can also construct the 1 base DOWN iso-V-prism line through the points KK´. Note that OK = OJ, and OK´ = OJ´, so that JJ ´ and KK´ are equidistant from HH´. 1 base UP Note that the vertical prismatic effect at any point on the line KK´ is base DOWN. It will be realised that when the eye views through any point on this line it will encounter only horizontal prismatic effect. H H´ We can now draw a line, HH´, through O parallel to JJ´. Assuming a right eye, the base direction of this prism will be IN at any point on the line segment, OH and OUT at any point on the line segment OH´. K K´ 1 base DOWN Finally, we measure the bandwidth, which is the perpendicular distance between JJ´ and KK ´, and the orientation of the band, i.e., the direction of HH ´. 67.6º K J J´ K´ The result could be depicted as follows. 12.2mm 67.6º If the eye remains within the band it will not encounter more than 1 of vertical prismatic effect. Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Iso-V-prism zone theory Perhaps the most important application of iso-V-prism zone theory is to enable us to locate the areas on a pair of spectacles within which the vertical differential prism does not exceed a stipulated amount. These iso-V-differential-prism zones represent the areas on a pair of spectacles within which comfortable binocular vision is most likely to occur. The iso-V-differential-prism zones are simply the iso-V-prism zones plotted for the differential prescription. The differential prescription is simply, how much stronger one lens is than the other. For example, consider the prescription R L-2.00 / x 180. In this case, where the powers of the lenses are known in the vertical meridians (we are only interested in vertical prismatic effects), the power of the left lens in the vertical meridian is -4.00, so the left eye is stronger than the right eye by -2.00D. The 1 iso-V-prism zone for a -2.00D lens is simply a horizontal band 10mm wide. 10mm The eye must remain within this zone so as not to encounter more than 1 of vertical differential prism. The zone is not very wide. In the vertical meridian the eyes can only roam 5mm from the optical centres before they meet 1 of vertical differential prismatic effect! If the subject had been dispensed a pair of bifocal lenses, the segment zone which incorporates the reading prescription may not even lie within the band! Iso-V-differential-prism zones Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Iso-V-prism zone theory It was stated earlier that the differential prescription is simply, how much stronger one lens is than the other. In the general case, with astigmatic lenses whose axes may not be parallel, the difference in power between the eyes may not be obvious. Iso-V-differential-prism zones In the previous example, R L-2.00 / x 180, changing the sign of the sphere in the right eye and adding the result to the left eye prescription results in x 180, which is the differential prescription. The differential power, is therefore, 2.00D along 90. In these circumstances, the differential prescription can be obtained by changing the signs of the sphere and the cylinder in one eye (usually, choosing the weaker lens) and adding the two prescriptions together, if necessary, using astigmatic decomposition or Stoke’s construction x 180 Differential prescription Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Consider the prescription: R /+2.00 x 135 L / x 105 Iso-V-prism zone theory In the general case, the direction of the iso-V-differential prism band may not coincide with either cylinder axis direction. Iso-V-differential-prism zones The differential prescription is found, as before, by changing the sign of the power in one eye and then adding the two prescriptions together. We find: / x / x 105 = x x 105 = / x 75 So the differential Rx = / x 75 Using the graphical construction, described earlier, for locating iso-V-prism zones we find in this instance that the 2 zones are bands 32 mm wide lying along 24 º. 32mm 24 º Ophthalmic Lenses & Dispensing Click to skip to next topic Click to return to previous screen Click to return to CD contents

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Ophthalmic lenses and dispensing Ophthalmic Lenses & Dispensing This presentation was created by Professor Mo Jalie © Reed Educational and Professional Publishing Ltd 1999 © Original articles, Optician Click to exit show Click to return to CD contents

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