2020–2021 BCSC Basic and Clinical Science Course™

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2020–2021 BCSC Basic and Clinical Science Course™

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3 Clinical Optics

Introduction: Quick-Start Guide to Optics and How to Refract

Part 1: Introductory Optics

A Very Much Simplified Model Eye

We now have in hand the tools to illustrate the correction of simple axial refractive errors, which arise from a mismatch between the optical power of the anterior segment (cornea and crystalline lens) and the axial length of the eye.

Emmetropia

Suppose that all the refracting power of the eye is concentrated at a single plane, at the apex of the cornea, and assume the typical axial length of 24.0 mm. If this ultra-simplified model eye is to focus light from distant objects on the retina (Figure I-11A)—the power at the corneal apex must be given by the vergence equation: 0 + P = 1.33/0.024 m = 55.42 D. [The actual total nominal power of the human eye is in fact about 60 D, so our simplistic model is off by only about 10%.]

Myopia

Now suppose that our eye has the standard anterior segment with optical power 55.42 D but is 1.0 mm longer than normal—so the axial length is 25.0 mm. In such an eye, which is said to be myopic, light from a distant point source comes to a focus where the retina “should have been” (Figure I-11B); then the rays cross and continue for another millimeter before they form an (annoying) blur-circle on the retina. To correct the refractive error, we must shift the image location 1.0 mm toward the back of the eye. This can be done, say, with a concave (diverging, “minus” power) contact lens of power P placed adjacent to the corneal apex according to the vergence equation: 0 + (P + 55.42 D) = 1.33/.025 D = +53.20 D. Thus, the power of the required contact lens is P = 53.20 D – 55.42 D = –2.22 D. [In a more precise calculation, the correction would be about –3 D for each millimeter of excess axial length.]

Figure I-11 The position of light rays focusing. A, The normal eye. B, A myopic eye. C, A hyperopic eye.

(Illustration developed by Scott E. Brodie, MD.)

Hyperopia

Similarly, if the axial length of our ultra-simplified model eye were reduced by 1.0 mm, distant objects would form an image behind the retina (Figure I-11C). Such an eye is hyperopic. The required contact lens power would be given by the vergence equation: 0 + (P + 55.42 D] = 1.33/0.023 m = +57.83 D, or P = +57.83 D – 55.42 D = +2.41 D, a convex “plus” lens.

Thus, correction of axial refractive errors requires only the determination of the optimal spherical lens power needed to re-focus the light from a distant source on the retina.

Try It Yourself! I-1

Find a “trial lens set” with an assortment of + and – spherical lenses. (The signs of the lenses will be indicated on the handles or by the color of the lens frames). Work in groups of 3: 1 person should hold up a small light source (a penlight or the Finoff transilluminator on your instrument stand will work well); a second person should hold the lens or lenses at least 2 or 3 meters away; the third person should hold a small white card, which will serve as a screen to catch the images of the light formed by the lenses (Figure I-12). Start with, say, a +2.0 D sphere lens, and find the image (a small, round dot of light) about 66 cm farther away from the light source than the lens. (Why is the image not exactly 50 cm from the lens? Because your colleague holding the penlight is not infinitely far away!)

Move the light source closer to the lens, and find the new image: it will be farther away from the lens.

If you have the room, move the light source farther away from the lens than in step 1, and locate the new image: it will be closer to the lens than when you started.

Returning to the set-up in step 1, add a second + sphere lens (say, +1.0 D) and hold the lenses together to act as a single lens system. Locate the image: it will be closer to the lenses than when you started.

Remove the additional +1.0 D lens, and replace it with a –1.0 D lens. Now locate the image: it will be farther away from the lenses than when you started.

Compare the location of the image formed by your +2.0 D and +1.0 D lens combination with the location of the image formed by the actual +3.0 D lens from the trial set. They should agree closely.

Try comparing the +2.0 and –1.0 combination with an actual +1.0 lens.

Verify this “lens arithmetic” with other combinations of + and/or – lenses.

Figure I-12 Three residents forming a “human optical bench.” One holds a penlight, the second compares a +3.00 D sphere with a +4.00 D sphere held together with a –1.00 D sphere, and the third holds a small white card. Notice that the images on the white card are the same.

(Courtesy of Scott E. Brodie, MD, PhD.)

A net minus lens will not form an image; the white card will show only a dimmer circle of light inside the shadow of the frame than outside it. Why? (Because the minus lens diverges light, reducing the intensity of the illumination/cm² of the light that passes through it.)

Excerpted from BCSC 2020-2021 series : Section 3 - Clinical Optics. For more information and to purchase the entire series, please visit https://www.aao.org/bcsc.

3 Clinical Optics

Introduction: Quick-Start Guide to Optics and How to Refract

Part 1: Introductory Optics

A Very Much Simplified Model Eye

Emmetropia

Myopia

Figure I-11 The position of light rays focusing. A, The normal eye. B, A myopic eye. C, A hyperopic eye.

Hyperopia

Try It Yourself! I-1

Figure I-12 Three residents forming a “human optical bench.” One holds a penlight, the second compares a +3.00 D sphere with a +4.00 D sphere held together with a –1.00 D sphere, and the third holds a small white card. Notice that the images on the white card are the same.

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