Page 110 - The Indian Optician Digital Edition September-October 2022
P. 110
Thus, the 2Δ iso-V-prism zone is a band
11.1mm wide lying along the 13.9° meridian.
This result could have been found by means
of a graphical construction whereby point J
would be marked off 20mm up along 30 from
O and point J′, 5.77mm up along 120 from O and
the line connecting these points is the 2Δ iso-V-
prism line.
Points K and K′ have been marked off
equidistant from the origin to represent the 2Δ
iso-V-prism line along which the vertical prism is
2Δ base down. The dashed line which is parallel
to, and equidistant from, the lines JJ′ and KK′ FIGURE 8 - 2Δ ISO-V-PRISM ZONE FOR
and passes through the origin, O, represents the SPHERO-CYLINDER, +3.00 /+2.00 DC X 60
0Δ iso-V-prism line. Along this line the vertical
prism is zero.
zone for the lens +3.00 / +2.00 x 60.
It is an easy matter to confirm that at the
point J, the vertical prism is 2Δ base up. The (x, y) The method is illustrated in Figure 8. Along
coordinates for point J are given by the 60 meridian where the power is +3.00D, the
value of P which gives rise to 2Δ base down is
x = OJ cos 30 = 20 cos 30 = 17.32mm 60
given by
and y = OJ sin 30 = 10mm. P = 2 / sin 60,
60
Using the sign convention for prismatic effects, from which, P = 2.31Δ base 60.
y = -10mm. 60
Now using c = P / F the eye can roam
The vertical prism at J is given by 60 60 60
2.31/ 3 = 0.77cm or 7.7mm upwards along 60
P = y S + C cos θ (x sin θ + y cos θ), before it meets 2Δ of base down prism. The
V iso-V-prism line must pass through point J lying
= -1 x -2 - 2 cos 30(1.732 sin 30 - 1 cos 30) 7.7mm up along 60 from O.
= +2Δ, i.e., 2Δ base up. Along the 150 meridian, where the power is
EXAMPLE: +5.00, the value of P150 which gives rise to 2Δ
base down is given by
Find the dimension of the 2Δ iso-V-prism
P = 2 / sin 150,
150
FIGURE 7 - 2Δ from which, P = 4Δ base 150.
ISO-V-PRISM 150
ZONE FOR Since c = P / F the eye can roam
SPHERO- 150 150 150
CYLINDER, -2.00 4 / 5 = 0.8 cm or 8mm up along 150 before it
/ -2.00 DC X 30 meets 2Δ base down. The iso-V-prism line must
pass through point J′, lying 8mm up along 150
from O.
The perpendicular distance of point J from
the cylinder axis (JH in Figure 8) can be found as
follows.
tan β = OJ′ / OJ = 8 / 7.7 = 1.039
from which angle β = 46.1°.
106 | THE INDIAN OPTICIAN | SEPT-OCT 2022 LENS TALK 108 | THE INDIAN OPTICIAN | SEPT-OCT 2022
