Anatomy & Refractive States of the Eye
Identifying ametropia with the optical cross

 

The formula presented on the previous page is only accurate when it is applied to a thin lens in air. For our purposes, we will consider all lenses to be optically thin and to be surrounded by air.

Take a moment to complete the three practice exercises found on page 14. Simply combine the front and back surfaces algebraically to determine the total power of the lens.

Cylinder and the Optical Cross
Page 15 shows three sets of power crosses. The lens in the first example is ground on a +6.00 D spherical base curve. The back curve, however, shows two powers, -7.50 D along the horizontal or 180 degree meridian, and a -8.00 D along the 90th or vertical meridian. Since the back surface contains two powers, it means the cylinder is on the back surface of the lens. So by combining the front and back powers of this lens along the 90 degree meridian, we get +6.00 plus -8.00 to equal -2.00. Along the 180 degree meridian we get a +6.00 plus -7.50 to equal -1.50. Therefore the total power of this lens as shown in the example is -1.50 in the 180 and -2.00 in the 90. The written prescription looks like: -1.50 - 0.50 x 180. Since this lens contains only minus power, -1.50 D along the 180 degree meridian and -2.00 D along the 90 degree meridian, this Rx is designed to correct compound myopic astigmatism. This ametropia is illustrated on page 15 as both points of light coming to focus in front of the retina.

The middle example on page 15 shows the spherical base curve of + 8.00 D, and two back curves of -5.25 D and -4.00 D. Like the previous example, we combine these numbers algebraically, combining each principal meridian separately. Looking at the 90 degree meridian, + 8.00 plus - 5.25 equals +2.75. In the 180 degree meridian +8.00 plus -4.00 equals + 4.00. The written Rx looks like this: +4.00 -1.25 x 180. Since the lens contains only plus power, +4.00 D in the 180 degree meridian and + 2.75 D in the 90 degree meridian, it is designed to correct the ametropia compound hyperopic astigmatism where both points of light come to focus behind the retina.

The lens in the third example on page 15 is ground on a + 4.00 D base curve, the back curves are -7.75 D and -8.50 D resulting in a total power of -3.75 D in the 90th meridian and -4.50 along the 180th meridian. The prescription, as indicated, would be written as
-3.75 - 0.75 x 90 on a +4.00 base curve.

Take a moment and complete the three practice exercises found on page 16. These are similar to the previous exercises with spherical lenses. Only now we’re combining the front & back surfaces of compound lenses. The answers to the practice exercises are found on the bottom of page 19.