Introduction to the Evaluation of Astigmatism

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Astigmatism is really a hard notion to explain to patients and students. Browsing more than the net is unlikely to lead to a satisfactory straightforward definition. In reasonably basic terms, what it implies is that the optical program focuses lines oriented in one particular path at a different point to these in all other directions. In optics, these directions are named meridians. If the optical system is circular the meridians are analogous to the lines of longitude of the earth exactly where the optical centre is the North Pole and the lines extend about the globe to the other pole.

In astigmatism, some of these lines are far more tightly curved than other people. The lines with the highest and lowest curvature are called the principal meridians and in normal astigmatism, these are often at 90 degrees to 1 one more. An example of a shape which would have this impact when produced as a lens is a barrel. Picture that the optical centre is at a single point on the widest point of the barrel. Lines drawn out at each and every clock hour would each and every have a various curve – the most curved would be the meridian which goes about the barrel and the least curved would be the meridian heading towards the top and bottom of the barrel. In its intense kind the barrel becomes a cylinder and the up and down meridian has no curve at all – this shape, as a lens, is termed a cylindrical lens.

You can immediately see from this analogy that the meridians can only take values from to 180 degrees : 190 degrees is the same as ten degrees.

In ophthalmology, when we consider astigmatism, we do so in one of two approaches. An optical element can be described either as the sum of 2 cylinders such as 10D x 90 degrees + 20D x degrees or as a sphere (regular lens) plus cylinder – in this case 10D + 10D x degrees (known as the sphero-cylindrical type). These forms are simple to interconvert and each totally describe the element provided that it displays regular astigmatism. Transposing the spherocylindrical type is a trivial task- in this case it is 20D – 10D x 90 degrees. Just don’t forget that the quantity of cylinder always stays the exact same.

The spherical equivalent of a sphero-cylindrical lens is equal to the sphere plus half the cylinder. This is the best approximation to the lens by a spherical lens only As a standard verify of your arithmetic bear in mind that when you transpose the cylinder, the spherical equivalent always stays the exact same. Also, if you add thin lenses in make contact with with a single yet another, the spherical equivalents often add algebraically.

Addition and subtraction of thin lenses in speak to with a single another is effortless if the principal meridians of the astigmatism are aligned. In that case, the spheres and cylinders just add collectively. For example, take two lenses

+ 1 x 90 degrees
3 +two x degrees

transposing the second one particular we get

5 -two x 90 degrees

adding together now is

5 -1x 90 degrees

checking the spherical equivalents we have .five + 4 from the combination and the outcome is also 4.5D

It gets much more difficult when you have to add lenses collectively that are not aligned You may well guess that you can use vector addition from high college physics but try this :

+ 1 x 90 degrees and
+ 1 x degrees

Vector addition would yield 1.4 by 45 degrees but the appropriate answer is 1D sphere.

To realize what is going on here, have a look at the prime half of this diagram I have illustrated the addition of a 2D x degrees (blue) and 2D x 45 degrees (red). Along the x axis is the meridians and the y axis shows the energy for each and every meridian. The green line is the net power. Naturally if these sine waves have been degrees and 90 degrees they would be totally out of phase and add to a straight line (the sphere) of 2D. The matrix mathematics to solve directly is hinted at below.

Any lens can be deemed to be a matrix like this and if several lenses are in contact, the matrix elements just add up. You then need to pull the 3 parameters out the matrix and resolve for S, C and theta. This is basic 1st year university mathematics.