Passage For a person with perfect vision, light from an object is properly refracted by the eye lens to converge on a single point at the retina, forming a clear image of the object. Vision defects result from eye shape abnormalities or errors in the refractive power of the eye lens. Myopia (nearsightedness) occurs when light from a distant object is incorrectly focused in front of the retina. Hyperopia (farsightedness) occurs when light rays from a nearby object are focused beyond the retina.Many optical techniques are available to measure the refractive error of an individual to determine the necessary correction. Photorefraction is a photographic technique often used with young children because it does not require the individual to be still for a lengthy duration. When the patient is looking at the camera, a flash photograph is taken of the eye to determine the amount of light that is reflected off the retina and captured by the camera lens.In healthy eyes, all the light from the flash that enters the eye is reflected off the retina and returns back to the camera's light source. Because the camera lens does not receive this light, the pupil is completely dark in the resulting image. A myopic eye cannot properly focus the light at the retina. Due to the geometry of the eye and its lens, some of the light is reflected to the top portion of the camera lens. The camera captures an image of a pupil with a crescent of light at the top. In a hyperopic eye, the crescent appears at the bottom of the pupil. Ray diagrams for photorefraction are shown in Figure 1. Figure 1 Paths of light in photorefraction for different eyes: (A) Healthy, (B) Myopic, and (C) Hyperopic. HC. Howland, "Optics of photorefraction: orthogonal and isotropic methods." ©1983 Optical Society of America. -Photorefraction with a camera at a distance of 50 cm away from a child produces a completely dark image of the pupil. If the child's retina is 2.5 cm from the eye lens, what is the lens strength of the eye? (Note: Use the thin lens formula, S = 1/o + 1/i.)

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11ecba2d_12d0_2c1b_a3bf_db728e272bff_MD0008_00 Lens strength is a measure of a lens's refractive power; the greater the lens strength, the more light will bend. The inverse of a lens's focal length is equal to its lens strength (S = 1/f); the closer the image is focused to the lens, the greater the lens strength. The given thin lens equation relates the strength of a lens (S) to the distance to the object (o) and to the created image (i): 11ecba2d_12d0_2c1c_a3bf_eb9742db1908_MD0008_00 According to the passage, a completely dark photographed pupil is produced from an eye lens that does not require correction because the image is properly focused at the retina. Therefore, the image distance is equal to the lens-retina distance (i = 2.5 cm). Because the child is looking at the camera during photorefraction, the object distance is equal to the distance to the camera (o = 50 cm). In calculations involving optics, image distance i and object distance o are both positive for converging (convex) lenses, like the eye lens. Lens strength is measured in units of diopters (D), which are equal to inverse meters (m⁻¹). Therefore, these lengths are converted from cm to meters: o = 0.5 m and i = 0.025 m. Using these values in the above equation, 11ecba2d_12d0_532d_a3bf_49ca332d45a2_MD0008_00 11ecba2d_12d0_532e_a3bf_f7b4bf58e8d9_MD0008_00 (Choice A) The image distance would be negative for a diverging lens. Incorrectly using i = −2.5 cm would result in a lens strength of −38 D. (Choice D) If the inverses of i and o are not used in the thin lens equation, the incorrectly calculated lens strength would be 52 cm. Educational objective: Lens strength (refractive power) is the inverse of the lens's focal length and is measured in units of diopters (D). A diopter is equal to an inverse meter (m⁻¹). Lens strength can be determined from object distance and image distance using the thin lens equation: S = 1/o + 1/i.

Passage For a Person with Perfect Vision, Light from an Object