The physics of eyesight correction
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The physics of eyesight correction
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[post_date] => 2025-01-09 21:24:26
[post_date_gmt] => 2025-01-10 02:24:26
[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 there are many factors that determine an individual’s ability to see, the most common and basic eyesight defects are those of myopia and hyperopia, or nearsightedness and farsightedness, respectively. These two terms describe the ability of the eye to focus an image. A nearsighted individual can clearly see things close to them, but their vision becomes increasingly blurred as distance to the object increases. A farsighted individual has the opposite issue: they are able to clearly see objects at a distance, but the object becomes blurrier the closer it is. Myopic vision generally results from the focal length of the eye being too short, while hyperopic vision is the result of the focal length being too long, as seen in figures 1 and 2.
As the eye can be conceived as a set of lenses through which light refracts and converges onto the retina, the complication of poor eyesight becomes a simple problem of optical physics to solve. A converging lens can be utilized to shorten the focal length in the case of hyperopia, and a diverging lens can be used to lengthen the focal length in myopic vision. A lens with a negative value of power will be a diverging lens, while a positive value is reserved for converging lenses. The degree of correction can additionally be adjusted by varying the magnitude of power of the lens.
Attribution: Gumenyuk I.S. CC-BY-SA 3.0/4.0
[post_title] => The physics of eyesight correction
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[question] => Severe myopia is defined as -6 diopters or worse. What would the minimum focal length of a lens have to be in order to correct for severe myopia?
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[answer] => 3
[description] => Reason for the Correct Answer:
A diopter is defined as the power of a lens when focal length is expressed in meters.
The power of lens is related to focal length via the equation: P = 1/f where P is the power of the lens, and f is the focal length.
Because treatment of myopia requires a diverging lens, power must be negative, and therefore the focal length would also be negative. This means we need a lens with focal length of at least -0.1667m.
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[answers] => Array
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[each_answer] => A. 0.1667 m
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[1] => Array
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[each_answer] => B. 0.1667 cm
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[2] => Array
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[each_answer] => C. -0.1667 m
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[3] => Array
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[each_answer] => D. -0.1667 cm
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => In humans, the total optical power of the relaxed eye is approximately 60 diopters. If a person is looking at an object 10m away, what would be the distance of the image from the lens to the retina of a relaxed eye, and would this be a real or virtual image?
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[answer] => 4
[description] => Reason for the Correct Answer:
While a virtual image can be produced by a converging lens such as the eye, the image is going to be focused on the opposite side of the lens (on the retina) and therefore is a real image.
We can substitute accordingly with the equations P= 1/f and 1/o + 1/i = 1/f to find the distance to the image from the lens. The distance to the object is always positive.
To solve for the distance of the image, we must solve for 1/i.
P=1/f ;
60 = 1/f
1/o + 1/i = 1/f
1/ 10m + 1/i = 60
0.1 + 1/i = 60
1/i = 59.9
i = 0.0167 m
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[answers] => Array
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[0] => Array
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[each_answer] => A. It would be a virtual image at -0.020m
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[1] => Array
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[each_answer] => B. It would be a virtual image at 0.020m
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[2] => Array
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[each_answer] => C. It would be a real image at -0.0167m
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[each_answer] => D. It would be a real image at 0.0167m
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => A pinhole occluder is an opaque disk with a very small aperture; patients can look through it and temporarily negate their myopia. Why would peering through a very small hole have this effect?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Dispersion of light is the variation of wave speed that produces the separation of white light into a full spectrum. It would not improve vision, and a pinhole would not induce this effect.
While the pinhole may cause diffraction, this would not help to negate the effects of myopia in any way.
The pinhole allows for light to come in almost entirely parallel to the lens, which would pass straight through and avoid refractive errors seen in myopia.
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[answers] => Array
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[0] => Array
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[each_answer] => A. The pinhole causes diffraction as light passes through it, which spread the light over the retina for a better image
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[1] => Array
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[each_answer] => B. The pinhole helps to disperse light as it passes through to properly focus it on the retina
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[2] => Array
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[each_answer] => C. The pinhole would help to focus more light as it enters the eye from the outer parts of the lens
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[each_answer] => D. The pinhole only allows light to pass through the center of the eye’s lens, and no refraction is needed
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[3] => Array
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[quiz_unique_key] => 2377279144
[question] => An astigmatism is a type of optical aberration of the lens of the eye, in which the curvature is irregularly distributed across the surface. What kind of optical deficit may result from an astigmatism?
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[answer] => 1
[description] => Reason for the Correct Answer:
Accommodation is the ability of the lens shape to change. When the flexibility of the lens declines due to age, the individual would experience “presbyopia.”
When the focal length of the eye is too long, this would be hyperopia.
When the focal length of the eye is too short, this would be myopia.
An irregularly shaped surface as seen in an astigmatism would cause blurry vision at any distance, as the eye would be incapable of focusing light into a sharp image.
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[answers] => Array
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[0] => Array
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[each_answer] => A. The eye would not properly accommodate from short to long distances
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[1] => Array
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[each_answer] => B. The focal length of the eye would be too long
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[2] => Array
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[each_answer] => C. The eye would be unable to focus an object into a sharp image
)
[3] => Array
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[each_answer] => D. The focal length of the eye would be too short
)
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[4] => Array
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[quiz_unique_key] => 2261298308
[question] => Frank had perfect vision, but he found if he wore two pairs of glasses together he could still see perfectly. This only worked if the first pair (for myopia) was 5cm away from his face, and the second (for hyperopia) was 3cm away from his face. If the focal length of the second pair is 4.5cm, where would the image of the first lens fall?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
The glasses for hyperopia would be convergent lenses, the glasses for myopia would be divergent lenses. There are two ways of approaching this problem, and we can essentially ignore “9.5cm” as an answer because we aren’t given any information on the focal length of the lens of the eye.
When combining lenses together, the image for the first can act as the object for the second. Realize that in the case of the second (convergent) lens, the image would be behind the lens, (so it can focus the image on the retina!)
If Frank could see perfectly after putting on 2 pairs of glasses, then the light is first refracted divergently, and subsequently refracted convergently, resulting in the image being back to perfectly parallel. If the light rays are parallel, then the object must be directly at the focal point of the converging lens.
The focal length should then equal the distance between the two glasses (5cm-3cm=2cm), plus the focal length of the diverging lens. Since the focal length is what we are trying to solve for, we can set up the equation as: (5cm-3cm)+fd = fc
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[answers] => Array
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[each_answer] => A. -2.5cm from the first lens
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[1] => Array
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[each_answer] => B. 3.5cm from the first lens
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[2] => Array
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[each_answer] => C. 4.5cm from the first lens
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[each_answer] => D. 9.5cm from the first lens
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