Ophthalmic Lenses

In this section, we build upon the basic principles of first-order optics to show how both simple lenses and complex optical systems are modeled. We also demonstrate how imaging problems are solved.

Vergence

We begin by considering the concept of vergence. Light rays emanating from a single object point spread apart and are referred to as divergent. Light rays traveling toward an image point, after passing through an optical lens, come together and are referred to as convergent. If rays are diverging, the vergence is negative; if rays are converging, the vergence is positive. Consider a lens placed close to an object point (Fig 1-26A). The lens collects a large fraction of the light radiating from the object point. When the lens is moved away from the object point, it collects a smaller portion of the light radiated by the object point. The rays that reach the lens are less divergent than they were when the lens was closer to the object (Fig 1-26B). Close to the object point, the light is more divergent; farther from the object point, the light is less divergent. Similarly, close to an image point, light is more convergent; farther from the image point, light is less convergent.

Figure 1-26 A, Close to an object point, light is strongly divergent, so a lens placed close to the object point collects a large fraction of the light radiated from the point. B, Farther from an object point, light is much less divergent, so a lens collects a much smaller portion of the light radiated by the object point. (Illustration developed b y Kevin M. Miller, MD, and rendered

b y C. H. Wooley.)

Vergence is inversely proportional to distance from the object or image point. Vergence is the reciprocal of the distance. The distances used most often in ophthalmology are 4 m, 2 m, 1 m, 0.5 m, 0.33 m, 0.25 m, and 0.2 m. For convenience, the reciprocal meter (m–1) is given another name, the diopter (D). The reciprocals of these distances are, respectively, 0.25 m–1 (or 0.25 D), 0.5 m–1 (or 0.5 D), 1 m–1 (or 1 D), 2 m–1 (or 2 D), 3 m–1 (or 3 D), 4 m–1 (or 4 D), and 5 m–1 (or 5 D).

As light travels away from an object point or toward an image point, its vergence constantly changes. To calculate the vergence of light at any point, one must know the location of the object or image point. Conversely, if one knows the vergence at a selected point, the position of the object or image point can be determined.

Reduced Vergence

Reduced vergence is vergence multiplied by the refractive index of the medium. This term is confusing because reduced vergence is numerically larger than vergence. For example, 1 m in front of an image point, light traveling in glass (n = 1.5) has a vergence of +1.0 D but a reduced vergence of +1.5 D. Confusing or not, however, the term reduced vergence is too well entrenched to be changed.

The LME can be interpreted in terms of reduced vergence. Light from an object point diverges, but the degree of divergence decreases as the light moves farther from the object point. Eventually, the light encounters the refracting surface, and just as it reaches the surface, it has a reduced vergence of n/o. The refracting surface suddenly changes the light’s vergence by an amount equal to its power. As the light leaves the refracting surface, it has a reduced vergence of (n/o) + P, but because the light is converging to an image point, this must equal n′/i.

Calculations using the LME are inconvenient because they involve reciprocal distances. Vergence is a way to simplify the calculations. By means of reduced vergence, the LME,

can be written in a very simple form:

U + P = V

where U is reduced object vergence and V is reduced image vergence.

Consider an object in air 50 cm in front of a +5 D refracting surface with n = 1.5. Where is the image? Light diverging from the object has a negative vergence. When the light reaches the lens, it has a reduced vergence of –2 D. The lens adds +5 D, for a final reduced vergence at the lens of +3 D. The plus sign indicates that the light converges as it leaves the lens. Dividing the reduced vergence by the index of the glass gives a vergence of +2 D, so the image is 50 cm behind the refracting surface.

The most common mistake in working with vergence calculations is ignoring the negative sign for divergent light. One way to avoid this mistake is to deal with the signs first, rather than with the numbers. For example, to solve the previous problem, many people would begin by converting distance to diopters—that is, the object is 50 cm from the lens, so the vergence is 2 D. After this conversion has been performed, it is easy to forget about the minus sign. It is better to deal with the sign first. In this problem, begin by noting that light diverges from the object and has a negative value; then write down the negative sign and convert distance to vergence (–2 D). Always write the sign in front of the vergence, even when the sign is positive, as in the preceding example (+5 D and +3 D). If you encounter difficulties with a vergence calculation, check the signs first. The problem is most likely a dropped minus sign. (See Clinical Example 1-6.)

Clinical Example 1-6

Imagine you are having a difficult time outlining the borders of a subretinal neovascular membrane on a fluorescein angiogram. You pull out a 20 D indirect ophthalmoscopy lens and use it as a simple magnifier. If you hold the lens 2.5 cm in front of the angiogram, where is the image?

Light from the angiogram enters the 20 D lens with a reduced vergence of

It exits the lens with a reduced vergence of

–40 D + 20 D = –20 D

The light is divergent as it exits the lens; thus, the virtual image you see is on the same side of the lens as the angiogram. It is located (1/20 D) = 0.05 m = 5 cm in front of the lens. Because the image is twice as far from the lens as is the object, the transverse magnification is 2×.

Thin-Lens Approximation

The LME concerns a single refracting surface, but, of course, lenses have 2 surfaces. According to the LME, when light from an object strikes the front surface of a lens, its (reduced) vergence changes by an amount equal to the power of the front surface, Pf. The vergence continues to change as the light moves from the front to the back surface; this is known as the vergence change on transfer, Pt. The back lens surface changes the vergence by an amount equal to the back-surface power, Pb. Thus,

The powers of the front and back lens surfaces are easily calculated, but the vergence change on transfer is difficult to calculate. However, because the vergence change on transfer is small in a thin lens, it is ignored to arrive at the thin-lens approximation. The total lens power is the sum of the frontand back-surface powers. Thus,

This is the thin-lens equation (TLE). Although the TLE and LME appear to be the same, there is an important difference. In the LME, P is the power of a single surface; in the TLE, P is the combined power of the front and back surfaces.

For example, if a +5 D thin lens has water (n = 1.33) in front and air in back and an object is 33

cm in front of the lens, where is the image? Light from the object strikes the lens with a reduced vergence of (–1.33/0.33 m) = –4 D. The lens changes the vergence by +5 D, so light leaves the lens with a vergence of +1 D, forming an image 1 m behind the lens.

The transverse magnification is the ratio of reduced object vergence to reduced image vergence. In the preceding example, the magnification is –4×, indicating that the image is inverted and 4 times as large as the object.

Lens Combinations

Most optical systems consist of several lenses. For instance, consider an optical system consisting of 2 thin lenses in air. The first lens is +5 D, the second lens is +8 D, and they are separated by 45 cm. If an object is placed 1 m in front of the first lens, where is the final image and what is the transverse magnification?

In paraxial optics, the way to analyze a combination of lenses is to look at each lens individually. The TLE shows that the first lens produces an image 25 cm behind itself with a magnification of – 0.25. Light converges to the image and then diverges again. The image formed by the first lens becomes the object for the second lens. The image is 20 cm in front of the second lens; thus, light strikes the second lens with a vergence of –5 D and forms an image 33 cm behind the second lens. The transverse magnification for the second lens alone is (–5 D/3 D) = –1.66. The total magnification is the product of the individual magnifications –1.66 × –0.25 = 0.42.

It is absolutely essential to calculate the position of the image formed by the first lens. Only after locating the first image is it possible to calculate the vergence of light as it reaches the second lens.

Any number of lenses can be analyzed in this way. Locate the image formed by the first lens and use it as the object for the second lens. Repeat the process for each subsequent lens. The overall transverse magnification is the product of the transverse magnifications produced by each individual lens.

Virtual Images and Objects

Many people find the subject of virtual images and virtual objects to be the most difficult aspect of geometric optics. Virtual images and objects can be understood with the use of a few simple rules. The trick is to not “overthink” the subject.

Consider an object 10 cm in front of a +5 D thin lens in air (Fig 1-27A). Light strikes the lens with a vergence of –10 D and leaves with a vergence of –5 D. In this case, unlike in all the previous examples, light emerges with a negative vergence, which means that light is still diverging after crossing the lens. No real image is produced. The reader can easily verify this by repeating the basic imaging demonstration with a +5 D spherical convex trial lens. Notice that an image does not appear, no matter where the paper is held.

Figure 1-27 Light exits the +5 D lens with a vergence of –5 D (A), producing a virtual image 20 cm in front of the lens (B). The virtual image becomes the object for the +6 D lens, which in turn produces a real image 50 cm to the right of the lens

(C). (Illustration developed b y Kevin M. Miller, MD, and rendered b y C. H. Wooley.)

Now, suppose a +6 D thin lens is placed 5 cm behind the first lens. Will an image form? If so, what are its characteristics? Light has a vergence of –5 D, but as the light crosses the 5 cm to the second lens, its vergence changes (the vergence change on transfer). In order to determine the vergence at the second lens, it is necessary to find the location of the image formed by the first lens. However, if the first lens does not form an image, how can the vergence at the second lens be calculated?

The solution is to use a mathematical trick. Light leaving the first lens has a vergence of –5 D. The