| Spherical equivalent. |
We are going to take a look at the prescription +1.50 -0.50 x090. What are the major power/meridian combinations?
Well, the +1.50 and the axis 090 go together. +1.50 -0.50 = +1.00, and 090 + 90 = 180, so the other leg of the optical cross is +1.00 at 180.
OK, in one major meridian we have a power of +1.50, and in the other major meridian we have a power of +1.00. In-between those two meridians the power on the lens is not really good, accurate power, but what is there ranges in-between +1.00 and +1.50.
Read page (53-54 / 66-68) in Optical Formulas Tutorial,
and then
look at the diagram on page (53 / 68). For this lens the
vertical (90th meridian) with the +1.00 power would have
that nice clear horizontal line focus at one meter, [since a +1.00D
power has a focal
length of 1/1D = 1m -- remember?] and the second meridian would have
that nice clear vertical line focus closer to the lens because the
power on the horizontal meridian is stronger.
Suppose we have a lens with the Rx +1.00 +2.00 x 180. Now the
+1.00 power, with a focal length of 1 meter, is on the horizontal
meridian of the lens, and it creates a vertical line image at one meter
from the lens.
If I have a +1.50 -0.50 x 180 lens outside in the nice bright sun I will not get a round image of the sun that I can use to start a fire. Instead, when I hold it one meter from the paper I will have a line image, and when I hold it 2/3 meter from the paper I will get another line image perpendicular to the first line image. What will I have in-between these two distances? I have a varying amount of fuzzy image, going from the first clear line, to elliptical in one direction, through round, to elliptical in the other direction, to the other clear line. A demonstration is here.
There is no good power between the two major meridians. That is why, when you look at a sphero-cylindrical lens in the focimeter, you do not get clear images for powers between the sphere lines and the cylinder lines. [Well, that is not the whole reason, but it is a part of one reason.]
The sun is a round object, but the images that we get from the toric lens are line images and, at one point, a fuzzy round image. The sphero-cylindrical lens distorts the image of the sun. At one particular length, however, the image of the sun IS round, even though it is fuzzy. This distance corresponds to what we call the circle of least confusion which is really the position of no distortion. This focal length can be converted to a power. The power that we get is called the spherical equivalent of the lens. We can find the spherical equivalent by taking the average power on the lens: the average of the two major meridian powers.
OK. So. The lens in our example has +1.50 in one meridian and +1.00
in the other. The average of +1.50 and +1.00 is [(+1.50) + (+1.00)] 
2 = +1.25. So the focal length of +1.25 is the spherical equivalent of
the lens, the focal length corresponding to the power of +1.25 is 0.8m,
so the circle of least confusion is located at 0.8m from the lens.
????
OK, here is another Rx: +2.00 +3.00 x 090. What are the major power/meridian combinations?
Well, we have +2.00 on the 090, and +5.00 on the 180.
All right, now where are the focal planes?
1/2 = 0.5 meters, and 1/5 = 0.2 meters. [Go back to the focal length lesson in the textbook if you do not remember what I just did.] So, one line focus will form at 0.5 meters, and the other will form at 0.2 meters. Notice the relationship that we found in that focal length lesson: the higher power has the shorter focal length.
Now, we need the position of the circle of least confusion. Since our powers are +2.00 and +5.00, the spherical equivalent, which is the average of these two powers, is [2+5]/2 = 7/2 = +3.50D. This corresponds to a focal length of 0.29 meters where there will be no distortion in the image. Every other focal length for this lens will result in a distorted image. This image will be out of focus and therefore not clear, but it will also not be distorted.
We actually use the spherical equivalent for a variety of reasons -- especially when we get into fitting contact lenses. And there is an easier way to find it. The sphere power is in effect all over the lens, and only the amount of cylinder changes from meridian to meridian. Therefore the average power will be the sphere plus the average of the cylinder, which is 1/2 of the cylinder. So, we have the spherical equivalent equal to the sphere plus 1/2 the cylinder.
Look at the example we were just working: The sphere plus 1/2 the cylinder is +2.00 + (1/2)(+3.00) = +2.00 + (+1.50) = +3.50, which is what we got the long way.
Lets do one more example of this whole concept, and then we will concentrate on the spherical equivalent. Suppose we are given a lens with the Rx -4.00 -1.00 x045.
Our major powers are -4.00 on the 045 meridian and -5.00 on the 135 meridian.
We have two virtual focal lengths of -0.25 meters and -0.20 meters. The spherical equivalent is (-4.00) + (1/2)(-1.00) = -4.00 - 0.50 = -4.50D. This gives a virtual distance of -0.22 meters for the plane where the image will appear to be with no distortion.
Notice, again, that the higher of the two powers goes with the shorter of the two focal lengths.
So, now we will do just straight spherical equivalents. Here are some for you do to:
Now do the exercises on page (54-55 / 68-69).
