1. Ophthalmic Lenses & Dispensing
Ophthalmic lenses and dispensing
Contents of CD-ROM
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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
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Ophthalmic Lenses & Dispensing

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Ophthalmic Lenses & Dispensing

3. Action of a prism Now the prism has been rotated clockwise through 90°
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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.
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.
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.
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.
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Ophthalmic Lenses & Dispensing

4. . The transverse test - plus lens
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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 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.
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.
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...
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Ophthalmic Lenses & Dispensing

5. . The transverse test - minus lens
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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.
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 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.
Notice that the limbs are in their correct position
but they appear minified through the lens...
On the next mouse click you will place a
minus sphere centrally in front of the chart...
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Ophthalmic Lenses & Dispensing

6. Action of a plus cylinder
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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).
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.
On clockwise rotation of the cylinder, the vertical limb rotates anticlockwise and
the horizontal limb rotates clockwise towards it in a SCISSORS MOVEMENT.
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.
Further clockwise rotation of the cylinder,
produces further SCISSORS MOVEMENT.
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Ophthalmic Lenses & Dispensing

7. Action of a minus cylinder
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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.
On clockwise rotation of the cylinder, the vertical limb also rotates clockwise and
the horizontal limb rotates anticlockwise towards it in a SCISSORS MOVEMENT.
Notice that the limbs appear to be in their original positions but
that just the vertical limb appears minified in the horizontal meridian.
On the next mouse click you will place a minus cylinder
held with its axis VERTICAL in front of the chart…
Further clockwise rotation of the cylinder,
produces further SCISSORS MOVEMENT.
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Ophthalmic Lenses & Dispensing

8. Use of the focimeter
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180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00
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.
180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00
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 60. 90
100 80
120 60
This is the circular target
which is brought into focus
by rotation of the dioptre
power knob.
140 40
160 20
This is the protractor from
which cylinder axis and
base setting can be read.
180
180 -
0.75
0.50
0.25
0.00 +
This is the central target
area. The circular scales
are calibrated in prism dioptres
This is a typical view of the measuring scales seen
when you look into a manually operated focimeter.
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.
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Ophthalmic Lenses & Dispensing

9. - - Use of the focimeter + +
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17.75
18.00
18.25
18.50
18.75
19.00
19.25
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.
180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00 - + 90
100 80
120 60
140 40
160 20
Then rotate the eyepiece
slowly back inwards until
the scales just come into
sharp focus. Stop as soon
as they come into focus.
0.75
0.50
0.25
0.00
Before you use a manually operated focimeter you
must adjust its focusing eyepiece for your own use. - +
180
180
17.75
18.00
18.25
18.50
18.75
19.00
19.25
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.
You should now find that the dioptric scale reads exactly zero.
If it does not now read zero, the instrument needs servicing!
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Ophthalmic Lenses & Dispensing

10. Use of the focimeter
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180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00
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. 90
100 80
120 60
140 40
The lens under test must
be a plano-prism since the
dioptre scale is reading zero.
0.75
0.50
0.25
0.00
The target is displaced upwards and its
centre is seen to lie over the second ring.
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.
0.75
0.50
0.25
0.00
0.75
0.50
0.25
0.00
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.
160 20
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.
0.75
0.50
0.25
0.00
We will begin by considering how the focimeter measures prism power.
0.75
0.50
0.25
0.00
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.
0.75
0.50
0.25
0.00
180
0.75
0.50
0.25
0.00
180 - +
Now that the focimeter is ready for use we can consider
how it is used to read the powers of prisms and lenses.
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Ophthalmic Lenses & Dispensing

11. - - - Use of the focimeter + + +
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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.
180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00 - +
180 90 80 60 40 20
100
120
140
160
Refocus the target by rotation of the power adjusting knob.
At some point the target will reappear, blurred and off-centre.
4.75
4.50
4.25
4.00
3.75
3.50
3.25
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
180 90 80 60 40 20
100
120
140
160
6.75
6.50
6.25
6.00
5.75
5.50
5.25
Here, a spherical lens has been placed on the lens rest and
the green target has disappeared. It is too blurred to be seen.
180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00 - +
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.)
180 90 80 60 40 20
100
120
140
160
4.75
4.50
4.25
4.00
3.75
3.50
3.25
We will now consider how the focimeter is used
to determine the power of spherical lenses.
180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00 - +
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Ophthalmic Lenses & Dispensing

12. - - Use of the focimeter + +
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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.
180 90 80 60 40 20
100
120
140
160
0.25
0.50
0.75
1.00
1.25
1.50
1.75
180 90 80 60 40 20
100
120
140
160
1.25
1.50
1.75
2.00
2.25
2.50
2.75
From these two readings we can deduce the power of the lens under test.
180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00
When a spherical lens is under test the target consists of a circle of dots. - +
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.
180 90 80 60 40 20
100
120
140
160
0.25
0.50
0.75
1.00
1.25
1.50
1.75
180 90 80 60 40 20
100
120
140
160
1.25
1.50
1.75
2.00
2.25
2.50
2.75
Here the best focus is obtained with the horizontal lines of
the target when the reading on the power scale is
180 90 80 60 40 20
100
120
140
160
1.25
1.50
1.75
2.00
2.25
2.50
2.75
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.
180 90 80 60 40 20
100
120
140
160
1.25
1.50
1.75
2.00
2.25
2.50
2.75
Here the best focus is obtained with the vertical lines of
the target when the reading on the power scale is
180 90 80 60 40 20
100
120
140
160
0.25
0.50
0.75
1.00
1.25
1.50
1.75
We will now consider how the focimeter is
used to read the power of an astigmatic lens.
180 90 80 60 40 20
100
120
140
160
0.75
0.50
0.25
0.00 - +
180 90 80 60 40 20
100
120
140
160
1.25
1.50
1.75
2.00
2.25
2.50
2.75
We can record the power of the lens, either
as / x
...or, as / x 90.
180 90 80 60 40 20
100
120
140
160
0.25
0.50
0.75
1.00
1.25
1.50
1.75
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Ophthalmic Lenses & Dispensing

13. Ophthalmic Lenses & Dispensing
Use of the focimeter
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180 90 80 60 40 20
100
120
140
160
4.25
4.00
3.75
3.50
3.25
3.00
2.75
In the other principal meridian the lines come into focus
when the power adjustment knob is turned to
180 90 80 60 40 20
100
120
140
160
4.25
4.00
3.75
3.50
3.25
3.00
2.75
The power of the lens under test is, therefore,
-4.25 / x 150 (or / x 60).
180 90 80 60 40 20
100
120
140
160
5.00
4.75
4.50
4.25
4.00
3.75
3.50
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.
180 90 80 60 40 20
100
120
140
160
5.00
4.75
4.50
4.25
4.00
3.75
3.50
Notice that the power scale reads -4.25
when the lines lie along the 60 meridian.
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Ophthalmic Lenses & Dispensing

14. Use of the focimeter Try this one by yourself !
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180 90 80 60 40 20
100
120
140
160
Try this one by yourself !
2.25
2.00
1.75
1.50
1.25
1.00
0.75
The power of the lens under test is / x 15
( or / x 105)
180 90 80 60 40 20
100
120
140
160
0.50
0.75
1.00
1.25
1.50
1.75
2.00
180 90 80 60 40 20
100
120
140
160
2.25
2.00
1.75
1.50
1.25
1.00
0.75
180 90 80 60 40 20
100
120
140
160
0.50
0.75
1.00
1.25
1.50
1.75
2.00
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Ophthalmic Lenses & Dispensing

15. Use of the focimeter What is the power of the lens?
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180 90 80 60 40 20
100
120
140
160
Here is another one to try yourself !
This time a progressive power lens is under test.
1.50
2.00
2.25
2.50
2.75
3.00
3.25
180 90 80 60 40 20
100
120
140
160
1.00
1.25
1.50
1.75
2.00
2.25
2.50
180 90 80 60 40 20
100
120
140
160
4.00
4.25
4.50
4.75
5.00
5.25
5.50
180 90 80 60 40 20
100
120
140
160
1.50
2.00
2.25
2.50
2.75
3.00
3.25
Either / x 45 Add +2.25 or
+1.75 / x 135 Add +2.25
What is the power of the lens?
Answer given on next mouse click
This is the second reading at the major
reference point in the distance portion.
This is one reading at the major
reference point in the distance portion.
This reading is taken at the near
reference point in the near portion.
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Ophthalmic Lenses & Dispensing

16. The correction of ametropia
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Vertex distance
Correction of hypermetropia
Correction of myopia
Effective power in distance vision
Effective power in near vision
Vergence impressed in near vision
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17. The correction of ametropia
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Vertex distance
vertex distance
BS 2738: Part 3: 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.
BS 3521: Part 1: Glossary of terms relating to ophthalmic lenses and spectacle frames
vertex distance Distance from the visual point of a lens to the corneal apex
vertex distance should be indicated by stating the number of millimetres
following the prescription, for example:
+6.00/-0.50 x 90 at 12
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18. The correction of ametropia
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Correction of Hypermetropia k
far point distance M´ M´ MR
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, MR.
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).
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´.
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Ophthalmic Lenses & Dispensing

19. F´V = K / (1 + dK) f ‘V = k + d f ´V = k + d
The correction of ametropia
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Correction of Hypermetropia
vertex distance d k
far point distance M´ F´ MR
f ´V
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´.
In order for a spectacle lens to correct an eye, its second
principal focus, F´, must coincide with the eye’s far point MR.
The back vertex focal length is made up from the sum
of the vertex distance, d, and the far point distance, k.
The spectacle lens, therefore, lies at its own
back vertex focal length from the eye’s far point.
F´V = K / (1 + dK)
f ‘V = k + d
f ´V = k + d
...but it is much easier to think of it terms of focal length!
This can be expressed in dioptres...
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20. The correction of ametropia
Click to return
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Correction of Myopia k M´ MR
f ´V d
The far point, MR, 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.
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´.
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).
f ´V = k + d
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21. Effective power in distance vision
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Hypermetropia x F´
f´V
f´V
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.
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.
We will now consider what happens when changes are made to the vertex distance.
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22. Effective power in distance visionx
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Myopia F´
f ´V MR
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.
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.
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23. Effective power in near vision
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stationary object
-3.00
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. B
-33.3 cm
+3.00
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.
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24. Effective power in near vision
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stationary object x
-3.25
-0.25 B
l < 33.3cm
+3.00
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.
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.
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.
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25. Effective power in near vision
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stationary object x B
-0.25
-3.25
l < 33.3cm
+3.00
It can be shown that, for small movements, x, the change in effective
power is given by -xF(2L1 + F ) where L1 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.
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26. Ophthalmic Lenses & Dispensing
Effective power in near vision
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object moves with lens x
-33.3 cm B
+3.00
-33.3 cm B
+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.
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.
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27. Effective power in near vision
object moves with lens
Q = -x(L1+ F)2
Effective power in near vision
Change in effective power when lenses are used for near vision
at 33.3cm and subsequently moved 5 mm away from the eye.
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+1.00
+0.80
+0.60
+0.40
+0.20
0.00
-0.20
-0.40
object remains stationary
Q = -xF(2L1+ F)
change in effective power Q
Lens power
This is a plot of the expression, Q = -xF(2L1 + F) when x = m and L1 = -3.00D.
This is a plot of the expression, Q = -x (L1 + F)2 when x = m and L1 = -3.00D.
Note that Q = 0 for the single case when (L1 + F ) = 0.
Note that Q = 0 when F = 0, or when (2L1 + F) = 0.
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28. Vergence impressed in near vision
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L´2 = BVP
L1 = 0
L´ 2 = BVP F´
The form of the lens is immaterial. Provided that the BVPs are the
same, in distance vision, lenses of different forms are interchangeable.
The back vertex power of a lens represents the vergence
leaving the back surface when the incident vergence is zero.
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29. Vergence impressed in near vision
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Note that you must first find the front
curve before you can determine L´2.
F1 =
L’2 = B’
L´ 2 = B´ B
L1 = -3 B’
L’2 = +6.58
+10.00
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 -3.00
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.
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, F1 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, F1 = F2 =
Clearly, in near vision, lenses of the same back vertex
power but made in different forms are not interchangeable.
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30. Ophthalmic Lenses & Dispensing
Vergence impressed in near vision
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Near vision effectivity error
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).
L1 = -3
L’2 = +6.58
For example, in the case of the lens made as a base
meniscus the vergence leaving the lens was found to be +6.58D.
The anticipated vergence (found from L´ = L + F ) for the lens
forms which have just been considered is = D.
For this form the error due to near vision effectivity is D.
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31. Vergence impressed in near vision
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Near vision effectivity error
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!
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.
final lens
L´ 2 = +6.58
NVEE =
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.
trial lens of
common form
L´ 2 = +7.02
NVEE =
symmetrical
trial lens
L´ 2 = +6.87
NVEE =
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32. Ophthalmic Lenses & Dispensing
Vergence impressed in near vision
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Lenses of different forms are not interchangeable In near vision.
In practice, tables of correction factors are available giving compensation for NVEE.
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
Correction Factors
Near addition (D)
Lens
Power
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.
-10.0
-8.0
-6.0
-4.0
-2.0
0.0
+2.0
+4.0
+5.0
+6.0
+7.0
+8.0
+10.0
+12.0
+14.0
-0.25
0.0
+0.25
+0.50
+0.75 37 38
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33. Centration of spectacle lenses
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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
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34. Horizontal centration of lenses
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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.
Furthermore, if progressive power lenses were dispensed without specifying monocular centration distances, their corridors of clear vision would be offset.
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 MR and ML.
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.
The asymmetry becomes very
obvious if we bisect the face!
Do you notice how asymmetric it is?
Look very carefully at this face. MR ML PD OC
These distances are equal.
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35. Ophthalmic Lenses & Dispensing
Horizontal centration
Centration distance
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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.
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.
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.
What we are trying to achieve is demonstrated here.
Optimum position
for optical centre
monocular CD
spectacle plane
mid-point of
frame bridge . R
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36. Horizontal centration
Near centration distance
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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.
NCD = CD. l
l + s
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.
Optimum position
for near optical centre
NCD
spectacle plane
centres of
converging pupils CD
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37. . . Horizontal centration Obtaining the measurements Click to return
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Obtaining the measurements
The following routine will be found to provide
accurate and consistent results. . .
Fit the frame
Mark centre of bridge
Attach tape
- if empty frame
Dot pupil centres
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38. . . Horizontal centration Obtaining the measurements
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The monocular centration distances can now be measured and recorded. . .
Left eye value
Right eye value
Monoc CDs are recorded as follows: 32 / 35.
The right eye value is always written first.
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39. Horizontal centration
Bifocal inset
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seen only by L eye
seen only by R eye
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.
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.
If the near fields do not coincide
there will be areas which fall
only within the field of one eye.
binocular field
To make the fields coincident
they should overlap exactly
in the near point plane.
segment
aperture
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40. Bifocal segment insetting
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The following method ensures that the segment
insetting brings the near fields into coincidence. B
mid-line
spectacle plane
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.
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.
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.
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.
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.
It has now been placed in position
in the spectacle plane...
... and inset so its centre lies
on the converging visual axis. OD OD
This is the bifocal segment
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41. Geometrical insetting to bring the near fields into coincidence
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monoc CD
= p
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: R s
The following table of geometrical insetting
has been prepared from this expression OS
Table of geometrical insetting
monocular centration distances
g = p.L / (L + F - S)
g = 3.p / (40 - F)
distance OC g l
Visual
axis
Tables such as this may be provided in
practice in order to assist bifocal fitting.
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42. Specification of segment top position
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segment
top position
HCL 22 23
If the final frame is fitted, the height can be measured with a simple ruler.
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.
For example, if the segment heights are measured as 23 mm,
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.
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.
Remember to measure the other eye also.
and the vertical box dimension of the frame is 44 mm,
then the segment top position = 1 above HCL.
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43. Horizontal centration
Prismatic aspheric lenses
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An aspherical surface such as the
ellipsoid shown here has a pole.
A1 = pole of aspherical surface
s = centre of rotation distance
P = prismatic effect of lens A1 x s P
When dispensing aspheric lenses which incorporate prism, there is a further consideration.
The amount of decentration
can be found as follows.
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.
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 pole of the aspherical surface no longer lies on the visual axis.
The amount of decentration x = P.s / 100 = 0.3 P
So when the prism power is 3, the necessary decentration is about 1mm.
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44. Vertical centration of spectacle lenses
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The centre of rotation condition
pantoscopic angle
optical axis
optimum position
for optical centre
visual axis
Typically the pantoscopic angle is 100.
optimum position
for optical centre
optical axis
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.
It is easily shown, from the geometry of the
figure, that for each 10 pantoscopic angle, the
optical centre should be lowered by 0.5 mm
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.
The optical axis of the lens then continues
to pass through the eye’s centre of rotation.
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45. Ophthalmic Lenses & Dispensing
Vertical centration of spectacle lenses
Click to return
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Required position
of optical centre. .
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
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.
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46. Vertical centration of spectacle lenses
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The centre of rotation condition for near vision
D = distance visual point D
Primary visual
axis for DV
R = eye’s centre of rotation
O = optical centre of lens N
Near visual axis
N = near visual point R O
Optical axis
This figure indicates that the centre of rotation
condition has been satisfied for distance vision.
We see that in satisfying the centre of rotation condition for distance
vision, the requirement for near vision is satisfied at the same time!
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47. Centration errors
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Errors in the vertical meridian of the lenses may give
rise to intolerable vertical differential prismatic effects.
For example if plus lenses are not
decentred inwards for near vision
there will be base out prism
exerted at the near visual points.
The most obvious problem which
arises from incorrectly centred
lenses is that unprescribed prismatic
effect is introduced before the eyes.
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.
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48. Ophthalmic Lenses & Dispensing
Centration errors
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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.
The chromatic aberration exhibited by a plano prism of power, P, made in
a material whose V-value or Abbe Number is denoted by vd , is given by:
chromatic aberration = P vd
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49. TCA = just as in the case of a plano-prism.
Centration errors
dispersion by a lens
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incident white light
is dispersed by a lens into its
monochromatic constituents c OC
transverse chromatic aberration
lens power = F
TCA =
c.F vd
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.
TCA = just as in the case of a plano-prism. P vd
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50. Centration errors effects of dispersion low-contrast target
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effects of dispersion
low-contrast target
When the prism is turned through 90º,
however, coloured fringes may be seen
on the edges of the bars of the target.
When the prism is turned through 90º,
however, the edges of the bars of the
target appear blurred, reducing acuity.
When viewed through a prism whose
base-setting coincides with the lines,
no effect can be seen on the target.
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.
Consider first a high contrast target
such as the one shown on the left.
The effect of transverse chromatism upon the wearer depends upon the object being viewed.
high-contrast 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.
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51. A A Centration errors H E H E effects of poorly centred lenses
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effects of poorly centred lenses
This experiment demonstrates another effect of poorly centred lenses. 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
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.
On the next mouse click you
will introduce a low-power
prism (say ½  ) base DOWN
in front of the spotlight... A
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.
H E
D X R C
spot
room
E G S K B Y
Click here to turn on >>
the muscle spotlight.
T Q P K L N V D X A
Now click here to >>
turn off the room light.
The optics of this ghost image
are considered in the next slide.
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52. Centration errors effect of poorly centred lenses Ghost image Dioptric
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effect of poorly centred lenses
Ghost
image
Dioptric
image
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.
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.
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).
The dioptric image produced by the prism is seen in this direction.
The deviation of this image, d = (n - 1)a
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
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53. . . Graphical construction
to find prismatic effect on an astigmatic lens
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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.
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.
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: PV = 1 x 4 = 4 base UP
PH = 0.4 x 4 = 1.6  base OUT
This is the optical centre of the
spherical component. . OC
In the horizontal meridian, the cross-sectional
shape resembles a prism with its base OUT.
This is point at which we are
finding the prismatic effect. . R
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54. Graphical construction
to find prismatic effect on an astigmatic lens
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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.
There is no power along the axis
meridian of a plano-cylinder, hence
the cylinder can exert no prismatic
effect along its axis meridian.
+2.00 x 30
Now we must consider the prism due to the cylinder x 30
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.
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55. . Graphical construction
to find prismatic effect on an astigmatic lens
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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.
+2.00 x 30 P
The cylinder can only produce a prismatic
effect at right angles to its axis, i.e., along
the 120 meridian of the lens.
We will first consider how to find the base direction of the prism due to the cylinder.
In order to determine the prismatic effect
at R, we must resolve the vertical and
horizontal decentration along the power
meridian of the cylinder. . R
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.
This is the point at which we
are finding prismatic effect.
We must find the perpendicular distance,
PR, of the point R from the cylinder axis.
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56. . Graphical construction
to find prismatic effect on an astigmatic lens
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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.
30º
30º
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.
First, construct an origin and
mark the nasal side of the lens
Next, determine the base direction of
the prismatic effect at R. Ask yourself,
“where does P lie with respect to R?”
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.
The simplest method is to use a graphical construction, details of which follow.
Next draw in the cylinder axis along
its prescribed axis direction, here, 30º.
If the cylinder had been negative in sign, P
would represent the position of the prism apex.
The base would have been DOWN & IN P N
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?
Now drop a perpendicular from R
to the cylinder axis meeting it at P. . R
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57. . Graphical construction
to find prismatic effect on an astigmatic lens
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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 vertical prismatic effect at R due to
the cylinder is given by:
PQ (cm) x Fcyl.
PQ is the effective decentration of the
plano-cylinder in the vertical meridian.
Since we require the vertical and
horizontal prismatic effects we can
now resolve PR into vertical and
horizontal components, PQ and QR.
QR is the effective decentration of the
plano-cylinder in the horizontal meridian.
The distance, PR, represents the
effective decentration of the plano-
cylinder along its power meridian. P N Q
The horizontal prismatic effect at R due
to the cylinder is given by:
QR (cm) x Fcyl. . R
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58. Ophthalmic Lenses & Dispensing
Graphical construction
to find prismatic effect on an astigmatic lens
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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.
In this example:
by measurement: PQ = 9.2mm
QR = 5.3mm P N Q
Hence the prism due to the cylinder is:
PV = x = 1.84 base UP
PH = x = 1.06  base OUT . R
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59. Ophthalmic Lenses & Dispensing
Graphical construction
to find prismatic effect on an astigmatic lens
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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 sphere OC . . R
So we have found:
Prism due to sphere = 4  base UP & 1.6  base OUT N . R P Q
Prism due
to cylinder
Prism due to cylinder = 1.84 base UP & 1.06  base OUT
Total prismatic effect = 5.84 base UP & 2.66  base OUT
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60. Lens thickness and weight
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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
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61. Lens thickness Edge thickness of minus lens t e r y s s = y2
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Edge thickness of minus lens t
This is a cross-sectional
view of a plano-concave lens e r y s
Sag at 60, n = 1.56
e is the edge thickness
We have: y = 30, r = 56
t is the centre thickness
First, find the radius of the surface from: r = 1000(n - 1) / Curve
Suppose we wanted to find the sag of a 10.00D curve at 60
diameter, the refractive index of the lens material being 1.56.
s = y2
r +  r2 - y2
The semi-diameter of the lens is 60 / 2 = 30mm =
302
56 + 
s is the sag of the concave surface
= 8.71mm.
= 1000( ) / 10
= 56mm
Centre of curvature of surface
From the geometry of the figure e = t + s
The quantity, s, is calculated from the sag formula, s = y2
r +  r2 - y2
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62. Lens thickness Edge thickness of minus lens so e = t + s2 - s1 t s1
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Edge thickness of minus lens t s1
This is a cross-sectional
view of a curved minus lens s2 e
e is the edge thickness
t is the centre thickness
s1 is the sag of the convex surface
s2 is the sag of the concave surface
From the geometry of the figure: t + s2 = e + s1
so e = t + s2 - s1
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63. Lens thickness Edge thickness of minus lens
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Edge thickness of minus lens
Consider a lens, made in glass, n = 1.523, with base and centre thickness of 1.0mm.
If the diameter is reduced to 40mm:
s1 = 1.5
s2 = 4.0
e = t +s2 - s1
= 3.5mm
-6.00 at 40 edge subs = 3.5 mm
-6.00 at 50 edge subs = 5.0mm
If the diameter is reduced to 50mm:
s1 = 2.4
s2 = 6.4
e = t +s2 - s1
= 5.0mm
-6.00 at 60 edge subs = 7.0mm
At diameter 60mm:
s1 = 3.5
s2 = 9.5
e = t +s2 - s1
= 7.0mm
7.0mm
5.0mm
3.5mm
The cross-sectional shape of a minus lens reminds us that
the smaller the lens diameter, the thinner will become the lens.
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64. . . Lens thickness Edge thickness of minus lens
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Edge thickness of minus lens
Consider a lens, made in glass, n = 1.523, with base and centre thickness of 1.0mm.
Note: If the overall diameter of the lens is 50mm but it has been decentred
4mm inwards, the edge thickness would change as follows: 50 . .
This is the optical
centre of the lens
This is the geometric
centre of the lens 29 21 25
The edge thickness on the
temporal side of the lens, eT,
is that of a 58 diameter lens.
The edge thickness on the
nasal side of the lens, eN,
is that of a 42 diameter lens.
eN = 3.7mm
eT = 6.2mm
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65. Lens thickness Thickness of plus lens This is a cross-sectional
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Thickness of plus lens
This is a cross-sectional
view of a plano-convex lens t s e
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
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66. Lens thickness Thickness of plus lens so t = e + s1 - s2 t
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Thickness of plus lens s1 t
This is a cross-sectional
view of a curved plus lens s2 e
s1 is the sag of the convex surface
s2 is the sag of the concave surface
e is the edge thickness
t is the centre thickness
From the geometry of the figure: t + s2 = e + s1
so t = e + s1 - s2
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67. Ophthalmic Lenses & Dispensing
Lens thickness
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Thickness of plus lens
If edged down to the smaller diameter of 40mm the edge thickness increases even more.
When this uncut is edged down to a diameter of 50mm the edge thickness increases.
At diameter 50mm:
F1 = t = 7.5mm
s1 = 4.0
s2 = 1.6
e = t - (s1 - s2)
= 5.1mm
+6.00 at 40 edge subs = 5.1mm
At diameter 65mm:
F1 = t = 7.5mm
s1 = 11.3
s2 =
e = t - (s1 - s2)
= 0.5mm
+6.00 at 65 edge subs = 0.5mm
At diameter 50mm:
F1 = t = 7.5mm
s1 = 6.4
s2 = 2.5
e = t - (s1 - s2)
= 3.6mm
+6.00 at 50 edge subs = 3.6mm
Consider a uncut, made in CR 39, n = 1.498, with base and edge thickness of 0.5mm.
7.5mm
7.5mm
7.5mm
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.
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.
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68. Lens thickness Thickness of plus lens
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Thickness of plus lens
The problems are even greater with plus astigmatic lenses
Consider the specification / x 180 which is to be edged to a 48 x 40 oval shape.
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
e thin
The thinnest points on the edge of the lens lie
at the extremities of the vertical meridian.
1.5 mm
We will custom-design the lens to
give 1.5mm at this point on the edge.
The centre thickness
will then be 3.5 mm. 40
+5.00
The horizontal diameter
of the lens is 48mm
The horizontal power
of the lens is +3.00
e thick
1.6 mm
The thickest points on the
edge of the lens lie at each end
of the horizontal meridian.
+3.00 48
The vertical diameter
of the lens is 40mm
The vertical power
of the lens is +5.00
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69. Lens thickness Thickness of plus lens
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Thickness of plus lens
The problems are even greater with plus astigmatic lenses
Now consider the specification / x 90 which is to be edged to the same oval shape.
e thick
3.3mm
The thickest points on the edge
of the lens now lie at each end
of the vertical 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
The centre thickness
will then be 4.5 mm.
+3.00 40
The horizontal diameter
of the lens is 48mm
The horizontal power
of the lens is +5.00
We will again custom-design the lens to give 1.5mm at this point on the edge.
1.5mm
e thin
The thinnest points on the edge of the lens now
lie at the extremities of the horizontal meridian.
+5.00 48
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.
The vertical diameter
of the lens is 40mm
The vertical power
of the lens is +3.00
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70. Ophthalmic Lenses & Dispensing
Lens thickness
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Variation in thickness of aspheric lens
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.
At diameter 65mm:
Spherical lens
F1 = t = 7.5mm
Aspheric lens
F1 = t = 6.4mm
Spherical Aspheric
+6.00 uncuts at 65
edge thickness =
At diameter 50mm:
Spherical lens
F1 = t = 7.5mm
Aspheric lens
F1 = t = 6.4mm
Spherical Aspheric
+6.00 uncuts at 50
edge thickness =
7.5
7.5
6.4
6.4
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71. Lens thickness Astigmatic prismatic lenses
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Astigmatic prismatic lenses
In the general case, a plus lens may have spherical, cylindrical and prismatic power.
We will now consider the thickness of circular plus lenses
which incorporate both 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.
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).
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.
Variation in edge thickness of circular lenses
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72. Ophthalmic Lenses & Dispensing
Lens thickness
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Astigmatic prismatic lenses
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.
spherical element
prismatic element
cylindrical element
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73. Lens thickness Astigmatic prismatic lenses
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Astigmatic prismatic lenses
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 spherical element
is +2.00
The prismatic element
is 4 base OUT
The cylindrical element
is x 90
The centre thickness
of this component is
1.9mm
The centre thickness
of this component is
3.6mm
The centre thickness
of this component is
2.4mm
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74. Lens thickness Astigmatic prismatic lenses
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Astigmatic prismatic lenses
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 spherical element
+2.00, tC = 1.9
The prismatic element
4 base OUT, tC = 2.4
The cylindrical element
+4.00 x 90, tC = 3.6
Note that the edge thickness
of this lens is zero at the nasal edge.
The total centre thickness of
the lens is seen to be 7.9mm.
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75. Lens thickness Astigmatic prismatic lenses
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Astigmatic prismatic lenses
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 spherical element
is +2.00
The cylindrical element
is x 90
The prismatic element
is 4 base DOWN
The centre thickness
of this component is
1.9mm
The centre thickness
of this component is
3.6mm
The centre thickness
of this component is
2.4mm
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76. Lens thickness Astigmatic prismatic lenses
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Astigmatic prismatic lenses
R / x 90 with 4 base DOWN
60mm and ethin = 0.
The thickness of the specification is made up as follows:
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.
The spherical element
+2.00, tC = 1.9
The prismatic element
4 base DOWN, tC = 2.4
The cylindrical element
+4.00 x 90, tC = 3.6
If this lens is to have zero edge
thickness it is seen that the centre
thickness must be reduced.
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77. To obtain zero edge thickness the centre thickness of the lens
Lens thickness
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Astigmatic prismatic lenses
R / x 90 with 4 base DOWN
60mm and ethin = 0.
To obtain zero edge thickness, the centre thickness can be reduced by 2.0mm.
The spherical element
+2.00, tC = 1.9
Now the edge thickness
of the lens is zero.
The thickness here can
be reduced by 2.0mm
The cylindrical element
+4.00 x 90, tC = 3.6
To obtain zero edge thickness
the centre thickness of the lens
need be only 5.9mm.
The prismatic element
4 base DOWN, tC = 0.4
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78. Ophthalmic Lenses & Dispensing
Lens thickness
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Astigmatic prismatic lenses
R / x 90 with 4 base DOWN
60mm and ethin = 0.
Note the position on the edge of the lens where the edge thickness is zero.
In fact, there are two points on the edge where the edge thickness is a minimum.
emin
emin
160 20
160 20
180 meridian
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79. Lens thickness Astigmatic prismatic lenses 
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Astigmatic prismatic lenses
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.
The cylindrical element
Note first, that the thinnest point
on the edge of the cylinder lies at
the extremities of the minus cyl axis.
prism apex
The prismatic element
emin
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. 
minus cyl axis
Also note that the thinnest
point on the edge of a plano
prism lies at the prism apex.
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80. Lens thickness Astigmatic prismatic lenses z 
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Astigmatic prismatic lenses
The position on the edge of an astigmatic, prismatic lens where
the thickness is a minimum can be determined as follows.
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.
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).
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.)
First, calculate the quantity: z =
200P dF
Next, draw a horizontal line, AB, 10 units long to represent the minus cylinder axis direction.
Then, construct a perpendicular, BC, from B C
where P is the prism power
d is the lens diameter
and F is the power of the cylinder D z F  
A B E
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81. Lens thickness Astigmatic prismatic lenses  = 90
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Astigmatic prismatic lenses
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.
Next, draw a horizontal line, AB, 10 units long to represent the minus cylinder axis direction.
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 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).
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.
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.)
Then, construct a perpendicular, BC, from B
where P is the prism power, 4.
d is the lens diameter, 60
and F is the power of the cylinder, 4
First, calculate the quantity: z =
200P dF
z =
200x4
60x4 C = D F
z = 3.33
 = 90 
A B E
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82. Lens weight Material nd Density Density
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Density
One of the physical properties which is published by lens
material manufacturers is the density of the material.
Material nd Density
Glasses
White
White
White
White
White
Plastics CR
mid-index
Polycarbonate
high index
high index
high index
The density is expressed in
grams per cubic cm (g/cm3)
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83. Lens weight
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Density
10mm
10mm
CR 39 material has a density of 1.32.
This means that 1cm3 of CR 39 weighs 1.32g.
Water has a density of exactly 1.0
so 1cm3 of water weighs 1gram.
10mm
Density expresses the weight in grams of one cubic centimetre of the material.
1cm3 of crown glass weighs 2.54g so
this glass is 2.54x heavier than water
and about twice as heavy as CR 39!
1cm3 of CR 39 weighs 1.32 grams so
CR 39 is 1.32x heavier than water.
Note that the weight is found by multiplying the volume by the density of the material.
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84. Lens weight
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Density
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.
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.
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85. Lens weight Weight of plus lens
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Weight of plus lens
A plano-convex lens can be considered to be made up from two separate components.
The weight of a circular spectacle lens is found by multiplying
the volume of the lens by the density of the lens material. y
Consider, first, the cylindrical plate e
The volume of this plate is .y2.e
a spherical cap…
…combined with a cylindrical plate.
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86. Lens weight Weight of plus lens y s Now consider the spherical cap y r
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Weight of plus lens y y s r
Now consider the spherical cap
The volume of the cap is .s2(3r - s) / 3
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87. Lens weight Weight of plus lens The volume of the cap
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Weight of plus lens
The volume of the cap
is .s2(3r - s) / 3
The volume of a plano-convex lens is the sum of the
volumes of the cylindrical plate and the spherical cap
i.e., volume = . (s2 (3r - s) / 3 + y2.e).
The volume of the
plate is .y2.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 .(s12 (3r1 - s1) / 3 - s22 (3r2 - s2) / 3 + y2.e)
where s1 and r1 relate to the convex surface and s2 and r2 relate to the concave surface.
The volume of a plano-convex lens is the sum of the
volumes of the cylindrical plate and the spherical cap.
spherical cap cylindrical plate
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.
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88. Lens weight Weight of minus lens e
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Weight of minus lens e
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.
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.
The volume of the plate is .y2.e and the volume of the cap is .s2(3r - s) / 3
hence, the volume of the lens is .(y2.e - s2(3r - s) / 3)
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89. Lens weight Weight of minus lens
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Weight of minus lens
… so the volume of a minus meniscus lens is .(s12 (3r1 - s1) / 3 - s22 (3r2 - s2) / 3 + y2.e)
where s1 and r1 relate to the convex surface and s2 and r2 relate to the concave surface.
In the case of a curved form minus lens, the volume of the convex
spherical cap which represents the front surface must be added...
This is exactly the same result as that obtained for plus meniscus lenses.
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90. Lens weight Weight of minus lens .s12 (3r1 - s1) / 3 + .y2.e
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Weight of minus lens
.s12 (3r1 - s1) / 3
+ .y2.e
- .s22 (3r2 - s2) / 3
The total volume of a curved lens is made up from the volume of the convex spherical cap...
…the volume of a flat cylindrical plate which represents the edge thickness...
…minus the volume of the concave spherical cap.
This statement is true for both plus and minus lenses.
The volume of meniscus lenses is .(s12 (3r1 - s1) / 3 - s22 (3r2 - s2) / 3 + y2.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.
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91. Lens weight Weight of minus lens
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Weight of minus lens
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.
r1 = cm
s1 = 0.349cm
The convex spherical cap has a radius of mm
and a sag at ø60 of 3.49mm.
r2 = 5.23cm
s2 = 0.946cm
The concave spherical cap has a radius of 52.3mm
and a sag at ø60 of 9.46mm.
The edge thickness of the lens is 6.97mm
e = cm
y = 3.0cm
The volume of the lens is .(s12 (3r1 - s1) / 3 - s22 (3r2 - s2) / 3 + y2.e)
= .( (3x ) / (3x ) / x0.697)
= 10.85cm3
Weight = volume x density = x 2.54 = 27.56g.
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92. Lens design and performance
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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
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93. Lens design and performance
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Off-axis performance of spectacle lenses
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/+0.25.
The refracted pencil is afflicted with aberrational astigmatism.
When the eye rotates 40º the effect of the lens is +4.00/+1.25.
+4.00 / +1.25
40º
+4.00 /
20º
+4.00
The form of this lens is quite shallow,
the back surface power is only -1.50D.
+4.00
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94. Lens design and performance
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Field diagrams
In practice, the off-axis powers are measured from the vertex sphere,
an imaginary spherical surface concentric with the eye’s centre of rotation.
+4.00 / +1.25
40º
+4.00 /
20º
+4.00
vertex sphere
+4.00
For a 40º rotation of the eye,
the sagittal oblique vertex sphere power is +4.00
and the tangential oblique vertex sphere power is +5.25
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95. . . .. . Lens design and performance Field diagrams S T +4.00
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Field diagrams
ocular
rotation
The off-axis performance is usually shown in the
form of a field diagram where oblique vertex sphere
powers are plotted against ocular rotation.
. . S T
For 20º rotation the
effect is +4.00/+0.25.
400
300
200
100 00
FIELD DIAGRAM
+4.00 / +1.25
40º ..
For 40º rotation the
effect is +4.00/+1.25.
For 0º rotation the
effect is
+4.00 /
20º
+4.00
vertex sphere .
oblique vertex
sphere powers
+4.00
S = plot of sagittal powers
T = plot of tangential powers
In practice, field diagrams
are plotted by the computer.
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96. Lens design and performance
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Best form lenses
T & S
400
300
200
100 00
FIELD DIAGRAM
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.
-6.00
The field diagram for a point focal
lens would look like this.
+4.00
Point focal lens
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.
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97. Lens design and performance
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Best form lenses
MOP = BVP S T
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.
400
300
200
100 00
FIELD DIAGRAM
-4.00
The field diagram for a Percival
form lens would look like this.
+4.00
Percival lens
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.
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98. Lens design and performance
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Best form lenses S T
400
300
200
100 00
FIELD DIAGRAM
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.
-5.00
The field diagram for a Minimum
T-Error form would look like this.
+4.00
Minimum T-Error lens
Minimum T-Error forms have the advantage that they
perform well over a wide range of vertex distances.
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99. Lens design and performance
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Best form lenses
Effects of changes in vertex distance
with Minimum T-Error forms S T
T & S S T
400
300
200
100 00
400
300
200
100 00
Point focal at long
vertex distance
400
300
200
100 00
Percival at short
vertex distance
Minimum T-error
at design vertex distance
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100. Lens design and performance
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Aspheric spectacle lenses
1.0mm
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. 70
-5.25
6.6mm
+4.00
Point focal lens
made in CR 39.
Weight = 20.3g
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101. Ophthalmic Lenses & Dispensing
Lens design and performance
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Aspheric spectacle lenses
T & S
1.0mm
400
300
200
100 00
FIELD DIAGRAM 70
-5.25
6.6mm
+4.00
Point focal lens
Weight = 20.3g
The field diagram for this point-focal design would look like this.
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102. Lens design and performance
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Aspheric spectacle lenses S
1.0mm
400
300
200
100 00
FIELD DIAGRAM 70 T
-1.50
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.
6.0mm
+4.00
-1.50 base curve
Weight = 18.1g
The field diagram for this flatter form +4.00Dlens would look like this.
The off-axis performance of this flatter curved form is unacceptable!
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103. Lens design and performance
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Aspheric spectacle lenses
T & S
1.0mm
400
300
200
100 00
FIELD DIAGRAM
for aspheric form with convex
hyperboloidal surface, p = -1.8. 70
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.
-1.50
5.4mm
+4.00
-1.50 base curve
Weight = 16.0g
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!
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104. Lens design and performance
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Tscherning’s Ellipses - point focal lenses
Lens power (F)
Back
curve
(F2)
-25
-30 -5
-10
-15
-20
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, F2 is given by:
F2 = {14F -375±( F-140F2)} / 28
-10
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.
-12.25
Ostwalt DV
Wollaston DV
+2.25
Ostwalt
point-focal
form
Wollaston
-10.00D lenses
-24.50
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.
-35
-40
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105. Lens design and performance
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Tscherning’s Ellipses - point focal lenses
Lens power (F)
Back
curve
(F2)
-25
-30 -5
-10
-15
-20
For example, lenses made in a material
of refractive index 1.5 and mounted 26.67
mm in front of the eye’s centre of rotation,
will be free from astigmatism for near
vision at 25cm when the inside curve , F2
is given by:
F2 = {14F -335±( F-140F2)} / 28
-10
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.
Ostwalt NV
Wollaston NV
-9.10
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
-10.00D 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.
-35
-40
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106. Lens design and performance
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Tscherning’s Ellipses - point focal lenses
Lens power (F)
Back
curve
(F2)
-25
-30 -5
-10
-15
-20
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:
Fmax =
F min =
2L´2 (n - 1)
2-3n
(n -1) - n3
n + 2
(n -1) +
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 Fmax Fmin
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.
-31.36
-22.22
+7.23
+7.66
1.50
1.70
-35
-40
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107. Lens design and performance
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Tscherning’s Ellipses - minimum tangential error forms
Lens power (F)
Back
curve
(F2)
-25
-30 -5
-10
-15
-20
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:
F2 = {49F ± ( F-1295F2)} / 88.
This equation also plots as an ellipse with limits and
-10
Ostwalt DV
Wollaston DV
-11.88
+1.18
Ostwalt
Min T-error
Wollaston
-10.00D lenses
-24.83
-35
-40
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108. Ophthalmic Lenses & Dispensing
Iso -V-prism theory
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Spherical lenses
Plano-cylindrical lenses
Sphero-cylindrical lenses
Iso-V-differential prism zones
Graphical construction for iso-V-prism zones
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109. Iso-V-prism zone theory
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Spherical lenses
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.
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.
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.
R 20 mm wide
L 10 mm wide
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110. Iso-V-prism zone theory
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Plano-cylindrical lenses A A´
If the vertical prismatic effect is denoted by PV then the oblique prism, PR, which
gives rise to PV, along the power meridian of the cylinder, is found from:
PR = PV / sin (90 + )
Where  is the axis direction of the cylinder.
Hence the 1 iso-V-prism zone is a band 20mm wide along the 60º meridian.
The result could be depicted 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.
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.
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:
PR = PV / sin (90 + ) = 1 / sin 150 = 2 base 150.
This could have been determined by a graphical construction as follows.
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.
Their dimensions can be determined as follows.
1 base UP is produced by 2 base 150 J J´ K K´ 2 1
1 iso-V-prism zone for x 60
is a band, 20mm wide, lying along 60.
Power meridian
Note that along JJ´ the prism base is DOWN
whereas along KK´ the prism base is UP.
20mm
Along the axis meridian, AA´,
the prismatic effect is zero.
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111. . Iso-V-prism zone theory Sphero-cylindrical lenses
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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 120.
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).
These two points are plotted, one on each principal meridian of the lens,
using the same method which was just described for plano-cylinders.
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 30.
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.
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.
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.
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.
In all other cases the band will not be parallel with the
cylinder axis but will lie along some other meridian.
+1.50 O
-2.00 . J 1
base UP
2 base UP at 30 10
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112. . . Iso-V-prism zone theory Sphero-cylindrical lenses
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Sphero-cylindrical lenses
Choosing a suitable scale, point, J´, can be plotted on the 120
meridian of the lens, 7.7mm down from the optical centre, O.
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.
Note that the eye must rotate DOWN along
120 in order to encounter prism 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).
Consider now the 120 meridian of the lens. The prism along 120 which
will give rise to 1 base UP, is 1 / sin 120 =  base UP at 120.
Note that the vertical prismatic effect at any point on this line is base UP.
We can now construct the 1 base UP iso-V-prism line through the points JJ´.
+1.50
1 base UP
1.15 base
UP at 120 1
base UP
-2.00 . J O
7.7 . J´
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113. . . Iso-V-prism zone theory Sphero-cylindrical lenses
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Sphero-cylindrical lenses
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 ´.
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´.
It will be realised that when the eye views through any point
on this line it will encounter only horizontal prismatic effect.
Note that the vertical prismatic effect at
any point on the line KK´ is base DOWN.
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´.
We can now draw a line, HH´, through O parallel to JJ´. K J J´ K´
The result could be depicted as follows.
12.2mm
67.6º H H´
+1.50
1 base UP K K´
1 base
DOWN
-2.00
If the eye remains within the band
it will not encounter more than
1 of vertical prismatic effect. . J
67.6º O . J´
1 iso-V-prism zone for / x 30.
By measurement, the band is 12.2mm
wide and lies along the meridian, 67.6º.
12.2mm
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114. Iso-V-prism zone theory
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Iso-V-differential-prism zones
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!
The 1 iso-V-prism zone for a -2.00D lens is simply a horizontal band 10mm wide.
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!
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.
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.
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.
10mm
The eye must remain within this zone so as not to
encounter more than 1 of vertical differential prism.
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115. Iso-V-prism zone theory
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Iso-V-differential-prism zones
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.
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.
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.
0.00
-2.00
-2.00 x 180
Differential prescription
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116. Iso-V-prism zone theory
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Iso-V-differential-prism zones
Consider the prescription:
R /+2.00 x L / x 105
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 24º.
In the general case, the direction of the iso-V-differential prism
band may not coincide with either cylinder axis direction.
32mm
24º
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117. Ophthalmic lenses and dispensing
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Ophthalmic lenses and dispensing
This presentation was created by Professor Mo Jalie
© Reed Educational and Professional Publishing Ltd © Original articles, Optician
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