Analysis of image production by the human eye
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Analysis of image production by the human eye
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[ID] => 559838
[post_author] => 12815
[post_date] => 2025-01-09 11:24:34
[post_date_gmt] => 2025-01-09 16:24:34
[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
While optical instruments can analyze the properties of light, such as interference and refraction with the interferometer and refractometer, respectively, a more familiar role of optical instruments is for image enhancement. There are two types of lens, convex and concave, but converging lens are much more prevalent since they can function as an image producer and magnifier.
The cornea and lens form a system that acts approximately like a single thin lens. Nonetheless, refraction primarily occurs at the cornea and then secondarily at the lens, so most of the power of the eye can be attributed to the cornea. For clear vision, a real image must be projected precisely onto the retina. Because the lens-to-retina distance does not change, the image distance remains the same at 2.0 cm for objects at all distances. The eye manages around this restriction by varying the power (and focal length) of the lens for objects at various distances, i.e. accommodation. It allows a person with normal vision to see objects clearly at distances ranging from 25 cm to essentially infinity.
In the laboratory, the most common instrument for image enhancement is the compound microscope, which is the archetype of a multiple lens system. The first lens is called the objective lens and has typical magnification values from 5× to 100×. In standard microscopes, they are mounted such that when you switch between objectives the sample remains in focus. The second, called the eyepiece or ocular, has several lenses which slide inside a cylindrical barrel, but the lenses in the eyepiece have been simplified to a single lens in Figure 2. The focusing ability is provided by the movement of both the objective lens and the eyepiece. A final consideration is that the final image be produced in a location far enough to be easily viewed, since the eye cannot focus on images that are too close.
[post_title] => Analysis of image production by the human eye
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[question] => For any multiple lens system, the image produced by the first lens becomes the object for the second lens, i.e. objective lens and eyepiece. Based on Figure 2, which of the following accurately describes the images produced by the objective lens and the final image by the eyepiece?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
The diagram indicates that the first image produced by the objective is produced within the focal length of the eyepiece and that it is a real and inverted image.
When an object sits outside the focal length of a converging lens, the image produced is a real, inverted, and enlarged one.
That image becomes the object for the eyepiece. The object for the eyepiece sits within its focal length.
When an object is located within the focal length of a converging lens, the image produced is a virtual, upright, and enlarged one.
Because the image is already inverted, you would produce an image not inverted or “upright”. The final image is a virtual, inverted, and enlarged image.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Objective: real, inverted, enlarged image; Eyepiece: real, inverted, enlarged image
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[1] => Array
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[each_answer] => B. Objective: real, inverted, enlarged image; Eyepiece: virtual, inverted, enlarged image
)
[2] => Array
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[each_answer] => C. Objective: real, inverted, reduced image; Eyepiece: virtual, inverted, enlarged image
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[3] => Array
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[each_answer] => D. Objective: real, inverted, reduced image; Eyepiece: virtual, upright, reduced image
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => What is the image size of a 6-foot tall man (1.8 m) who is standing at a distance approximately 10 feet away (3.0 m) on the human retina?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
The object and the image form similar triangles in relation to the lens such that hi/ho = -di/do = m.
In the question stem, the object height and distance are given, which are 1.8 meters and 3.0 meters, respectively. The image distance can be found in the passage in Figure 1, which is 0.02 meters.
Plug the values into the equation to get hi/1.8 = 0.02/3.0.
To solve for hi, isolate it and solve: (1.8*0.02)/3.0 = hi, which is 0.036/3 or 0.012 meters.
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[answers] => Array
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[0] => Array
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[each_answer] => A. 0.038 m
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[1] => Array
(
[each_answer] => B. 0.38 m
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[2] => Array
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[each_answer] => C. 0.012 m
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[3] => Array
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[each_answer] => D. 0.12 m
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => Accommodation is the process of adjusting the focal length to view objects at varying distances. What is the range in power of the eye in diopters when viewing from the closest (d = 25 cm) to the farthest possible distance from the eye?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
Power is the inverse of the focal length: P = 1/f, which are in units of diopters or inverse meters.
Substitute in power into the thin lens formula: 1/f = 1/i + 1/o, so P = 1/i + 1/o.
Calculate the power when viewing from the closest distance: P = 1/.25 + 1/0.02 = 4 + 50 = 54
Calculate the power when viewing from the farthest distance: P = 1/∞ + 1/0.02 = 0 + 50 = 50
The difference in power is 54-50 or 4 diopters.
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[answers] => Array
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[0] => Array
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[each_answer] => A. 4 diopters
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[1] => Array
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[each_answer] => B. 0.02 diopters
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[2] => Array
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[each_answer] => C. 25 diopters
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[3] => Array
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[each_answer] => D. 3 diopters
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[3] => Array
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[quiz_unique_key] => 2377279144
[question] => The critical differences with the telescope are that it has an objective lens with a longer focal length and has an eyepiece with a focal point coinciding with the focal point of the objective lens. Given such changes, what final image would you predict the telescope to produce when viewing a faraway object?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
When parallel rays of light come from infinity through a converging lens, they converge at the focal point.
The image produced from converging parallel rays is a real, inverted, and similarly sized one. This image serves as the object of the eyepiece.
The focal point of the objective is within the focal length of the eyepiece. When an object is placed within the focal length of a converging lens, the image produced is a virtual, upright, and enlarged one.
Once again, since the image is already inverted, producing an “upright” image means not inverting. The final image for the telescope is a virtual, inverted, and enlarged image just like the microscope.
)
[answers] => Array
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[0] => Array
(
[each_answer] => A. A real, inverted, enlarged image
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[1] => Array
(
[each_answer] => B. A real, inverted, reduced image
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[2] => Array
(
[each_answer] => C. A virtual, upright, reduced image
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[3] => Array
(
[each_answer] => D. A virtual, inverted, enlarged image
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)
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[4] => Array
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[quiz_unique_key] => 2261298308
[question] => Which of the following statements accurately describes the phenomenon of chromatic abberation?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
The focal length is dependent on the refractive index, and different wavelengths of light experience different indices of refraction.
Dispersion is the change of index of refraction with wavelength. The refractive index tends to decrease with increasing wavelength and subsequently increase with frequency.
Blue experiences a greater index of refraction and travels more slowly in the material, while red experiences the lowest index of refraction and travels more quickly.
Since blue experiences the greater refractive index, blue will converge closer to the lens than red will.
)
[answers] => Array
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[0] => Array
(
[each_answer] => A. Dispersion is the change of the index of refraction with frequency.
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[1] => Array
(
[each_answer] => B. Blue travels the fastest through the lens because it experiences the lowest index of refraction.
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[2] => Array
(
[each_answer] => C. Shorter wavelengths experience a greater index of refraction, so blue converges closer to the lens than red.
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[3] => Array
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[each_answer] => D. The focal length depends on refraction, so red experiences the greatest refraction and has the shortest focal length.
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