Page 108 - The Indian Optician Digital Edition May-Junel 2022

Adding the components separately gives: the resultant cylinder, CR, is given by:
MRE C C 45 C = √(C + C 45 2 ) = √(-0.2094 + -0.6222 ) =
2
2
2
0
0
R
Lens 1 +0.50 +0.50 +0.866 +0.656.
Lens 2 -2.00 +2.00 -3.464 The resultant axis is found from:
Sum -1.50 +2.50 -2.598 tan θ = (C – C ) /C = +0.8654/-0.6222, so
0
R
R
45
To convert back to sphero-cylindrical form, θ = -54.28°
R
the resultant cylinder, CR, is given by: and since this axis is negative, we must add
C = √(C + C 2 ) = √(+2.5 + -2.598 ) = +3.605. 180 to it to obtain 125.7°, so θ = 125.7°.
2
2
2
R
R 0 45
The resultant axis is found from: The resultant sphere, S , is given by:
R
tan θ = (C – C ) /C = +1.105/-2.598, so total MRE - C /2 = +1.25 - +0.656/2 = +0.92.
R
45
R
R
0
θ = -23.04. and the sum of the lenses is +0.92 / +0.66 x
R
125.7.
and since this axis is negative, we must add
180 to it to obtain 156.96°, so θ = 156.96°.
R
The resultant sphere, S , is given by:
R
total MRE - C /2 = -1.50 - 3.61/2 = -3.30.
R
and the sum of the lenses is -3.30 / +3.61 x 157
as found before.
EXAMPLE
Find the sphero-cylinder equivalent to the
following pair of sphero-cylinders:
-2.75/+1.00 x10 and +4.25/-1.50 x 20
Lens 1: (-2.75/+1.00 x 10)
MRE = S + C/2 = -2.75 + 1/2 = -2.25

C = C cos 2θ = 1 cos 20 = +0.9397
0
C = C sin 2θ = 1 sin 20 = +0.3420
45
Lens 2: (+4.25/-1.50 x 20)

MRE = S + C/2 = +4.25 + -1.5/2 = +3.50
C = C cos 2θ = 1 cos 20 = -1.1491
0
C = C sin 2θ = 1 sin 20 = -0.9642
45
Adding the components separately gives:

MRE C C 45
0
Lens 1 -2.25 +0.9397 +0.3420

Lens 2 +3.50 -1.1491 -0.9642
Sum +1.25 -0.2094 -0.6222

To convert back to sphero-cylindrical form,

104 THE INDIAN OPTICIAN MAY-JUNE 2022 LENS TALK