Answers to the oblique meridian formula problems.
EXERCISES:
5. What is the approximate power on the 180 meridian for the Rx +10.75
-2.00 x130?
-
180 - 130 = 50
-
sin 50 = 0.7660 . . .
-
sin 50
= 0.5868 . . .
-
x -2.00 = -1.1736 . . .
-
+ (+10.75) = +9.576 . . . = +9.58D of power on the 180 meridian.
6. The Rx is -5.50 +3.50 x 100. What is the approximate power on the 90th
meridian?
-
100 - 90 = 10
-
sin 10 = 0.1736 . . .
-
sin 10 = 0.0301 . . .
-
x +3.50 = +0.1055 . . .
-
+ (-5.50) = -5.3944. . . = -5.39D of power on the 90 meridian.
7. What is the power on the vertical meridian for the Rx -15.00 -3.00 x
055?
-
90 - 55 = 35
-
sin 35 = 0.5735 . . .
-
sin 35 = 0.3289 . . .
-
x -3.00 = -0.9869 . . .
-
+ (-15.00) = -15.986. . . = -15.99D of power on the 90 meridian.
8. If the Rx is -8.25 -1.25 x060, what is the approximate power on the
15th meridian?
-
60 - 15 = 45
-
sin 45 = 0.7071 . . .
-
sin 45 = 0.5 . . .
-
x -1.25 = -0.625.
-
+ (-8.25) = -8.875 . . . = -8.88D of power on the 15 meridian.
9. What is the approximate power on the horizontal meridian for the Rx
+1.00 -2.50 x 155?
-
180 - 155 = 25
-
sin 25 = 0.4226 . . .
-
sin 25 = 0.1786 . . .
-
x -2.50 = -0.4465 . . .
-
+ (+1.00) = +0.553 . . . = +0.55D of power on the 180 meridian.
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