Optical Ed | Articles

Oblique Meridian

Brent McCardle

1. The approximate power on the vertical meridian for the Rx +3.00 -1.50 x 102 is ______. Give the power rounded to two decimal places, not to 1/8 diopter steps.

Vertical meridian is the power in the 90^th meridian. First we need to subtract our axis(102 in this problem) from the meridian we are looking for(90 in this problem). This will be $\alpha$ in our oblique meridian formula.

$\ F_{\theta} = [(\sin \alpha)^2 \times F_{cyl}] + F_{sph}$

$\ F_{\theta}$ is the power in the meridian you are trying to find
$\alpha$ is the difference in the axis and the meridian you are trying to find
$\ F_{cyl}$ is the cylinder of the prescription
$F_{sph}$ is the sphere power of the prescirption

$\ F_{\theta} = [(sin \12)^2 \times (-1.50)] + 3.00$

$\ F_{\theta} = [0.0432273\ \times\ -1.50] + 3.00$

$\ F_{\theta} = -0.06484095 + 3.00$

$\ F_{\theta} = +2.94\ D$

So you have a total power of +2.94 D on the 90^th meridian

2. If the Rx reads -10.25 -2.50 x 040, what is the approximate power on the horizontal meridian? Give the power rounded to two decimal places, not to 1/8 diopter steps.
Horizontal meridian is the power on the 180th meridian. In this case you can subtract 40 from 180 or 0 from 40 since it is either 0 or 180.
$\ sin(40)^2 = 0.413176$
$sin(140)^2 = 0.413176$

$\ F_{\theta} = [(\sin \40)^2 \times (-2.50)] + (-10.25)$

$\ F_{\theta} = [0.413176 \times (-1.50)] + (-10.25)$

$\ F_{\theta} = [-0.619764] + (-10.25)$

$\ F_{\theta} = -10.87\ D$