please tell me why I got the answer to this incorrect

[kristiekyle]

You are fitting a patient with 44eye size 24DBL (I know its wide, but its the question in the book). The distance pd is only 62mm. If by some unfortunate error the lenses are NOT, how much prism is created in each eye?
RX: O.U. -2.25 -1.50 X 45

a. 0.9 BI
b. 0.5 BI
c. 0.9 BO
d. 0.5 BO

I answered (b), however the book with sample questions says its (a)

please tell me why

[Sibirer]

You are fitting a patient with 44eye size 24DBL (I know its wide, but its the question in the book). The distance pd is only 62mm. If by some unfortunate error the lenses are NOT, how much prism is created in each eye?
RX: O.U. -2.25 -1.50 X 45

a. 0.9 BI
b. 0.5 BI
c. 0.9 BO
d. 0.5 BO

I answered (b), however the book with sample questions says its (a)

please tell me why

I am assuming the question reads the lenses were NOT DECENTERED. The PD would be wide by 6mm or 3 mm ou. 0.3 x 3= 0.9 ;prism diopters. A minus lens centered too wide would produce base in prism. ( The thickness would be towards the bridge.)

[tx11]

remember to add the amount of cylinder power that is on the 180 line to the sphere power when calculating the amount of prism. Sometimes it helps to draw a picture (power cross with meridians).

[Barry Santini]

The answer is a; 0.9D BI.

B

[Bok]

-2.25/-1.50x45 - concerned with horozontal prismatic error. your axis is not on the horozontal meridian so calculate the power on the 180 which would be -3.00 RE decenter that by 3 so your calc now runs as.

0.3 x 3 = 0.9 BI

So the correct answer is A.

To calculate your power on the 180 run the following

DT = (sin alpha)^2 * Dc + DS, where

DT = off-axis power
a (alpha) = the angle between the main axis
Dc = cylinder
DS = sphere