Page 126 - The Indian Optician Digital Edition March-April 2024

FIGURE 3 - +4.00 DC
X 30 LENSES
THE PRISMATIC
EFFECT IN THE
AREAS SHADED
YELLOW MUST LIE
BASE DOWN AND
OUT ALONG 120.

Hence, the total prismatic effect = 1Δ base 180 and +1.00 D along 90. We must find the
down and 2Δ base out. decentrations along these principal meridians.

Alternatively, treating the lens as a pair of c = 6 sin 30 = 3mm up
V
crossed cylinders:
and c = 6 cos 30 = 5.2mm in.
H
+2.00 DC x 180/+4.00 DC x 90.
Hence, P = c F = 0.3 x 1 = 0.3Δ base up
V
V
V
We have, F = +2.00 and F = +4.00. P = c F = 0.52 x 2 = 1.04Δ base out.
V
H
H
H
Then, P = c F = 0.5 x 2 = 1Δ base down V The single resultant prismatic effect is
V
V
V
and P = c F = 0.5 x 4 = 2Δ base out as before. found from
H
H
H
We may compound these prisms into their P = √(0.32 + 1.042) = 1.08Δ base up @ 163°54'.
R
single resultant effect if required, Part 3 will continue with prismatic effects
P = √(12 + 22) or 2.24 base down @ 26° 34', in the general case of sphero-cylindrical
R
the base setting being found from prescriptions which have oblique cylinders.
tan (P /P ) as usual. REFERENCES
-1
H
V
It should be noted that the equivalent 1. Jalie M. (2024), The prismatic effect of
single decentration (7.07mm) lies along 45° decentration, Part 1, The Indian Optician, Jan/
whereas the resulting prismatic effect lies Feb 2024.
along 26°34'. As with plano-cylinders we can 2. Jalie M. (2021), Principles of Ophthalmic Lenses,
only find the prismatic effect along the power 6 Ed. ABDO, Godmersham, Kent.
th
meridians of the lens. If the decentration is
given in any other meridian, it must first be (to be continued)
resolved along the power meridians as will be
shown in detail in Part Three of this series.
EXAMPLE:
Find the prismatic effects introduced by
decentring the lens R -2.00 DS/+3.00 DC x 180,
6mm up along 30.
The powers of the lens are -2.00 D along

LENS TALK THE INDIAN OPTICIAN | MAR-APR 2024 | 122