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 36What is the camera lens' refractive index if incident light 50° from the normal is refracted to 35° from the normal?A.sin 50°/sin 35°B.sin 35°/sin 50°C.1/sin 50°D.1/sin 35°
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 36What is the camera lens' refractive index if incident light 50° from the normal is refracted to 35° from the normal?A.sin 50°/sin 35°B.sin 35°/sin 50°C.1/sin 50°D.1/sin 35°
Solution
The refractive index (n) of a medium can be calculated using Snell's law, which states that the ratio of the sine of the angle of incidence (i) to the sine of the angle of refraction (r) is equal to the refractive index of the second medium (n2) relative to the first medium (n1). In mathematical terms, this is expressed as:
n1sin(i) = n2sin(r)
Assuming the incident light is coming from air (which has a refractive index of approximately 1), the equation simplifies to:
n = sin(i) / sin(r)
In this case, the angle of incidence (i) is 50° and the angle of refraction (r) is 35°. Therefore, the refractive index of the camera lens can be calculated as:
n = sin(50°) / sin(35°)
So, the correct answer is A. sin 50°/sin 35°.
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