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. Question 37Photorefraction 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.)A.38 DB.42 DC.48 DD.52 D
Question
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. Question 37Photorefraction 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.)A.38 DB.42 DC.48 DD.52 D
Solution
To solve this problem, we need to use the thin lens formula, which is S = 1/o + 1/i.
In this case, o represents the object distance and i represents the image distance.
The object distance (o) is the distance from the camera to the child's eye, which is given as 50 cm.
The image distance (i) is the distance from the eye lens to the retina, which is given as 2.5 cm.
Substituting these values into the formula, we get:
S = 1/50 + 1/2.5
To simplify this, we need to find a common denominator. The common denominator of 50 and 2.5 is 50.
So, we rewrite the equation as:
S = 1/50 + 20/50
Adding these fractions together, we get:
S = 21/50
To convert this to diopters, we take the reciprocal of S (since 1 diopter is the reciprocal of 1 meter), which gives us:
Lens strength = 50/21 = 2.38 D
So, the closest answer is A. 38 D.
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