2. Assume you have perfect eyesight, meaning that you can focus on objects rangi
ID: 3161491 • Letter: 2
Question
2. Assume you have perfect eyesight, meaning that you can focus on objects ranging from 25cm in front of your eye (typical reading distance) up to infinite distances. If the diameter of your eyeball (ie. the distance between the lens and the retina) is 2.439cm, then (a) what is the focal length fand focal power P of the eye in the near-accommodated state? (b) what is the focal length fand focal power P of the eye in the far-accommodated state? (c) what is the range of accommodation of your eye in diopters?Explanation / Answer
In order to understand common vision defects, it is helpful to understand how the “normal” eye works. The muscles that control the focal length of the eye-lens are fully relaxed when the eye is focused on a very distant object. The muscles contract and shorten the focal length of the eye-lens to bring nearer objects into focus. The two limits of this range are called the far-point, which is the distance of an object that is in focus when the muscles are relaxed, and the near-point, which is the distance at which an object is in focus when the muscles have shortened the focal length of the eye-lens as much as possible.
When an object is located at the near-point (the closest point at which an object can be brought into clear focus on the retina), the focal length of the cornea and the eye-lens must be changed so that the image is formed on the retina, which is still 20 mm away. If we take 25 cm as a typical value for the near-point, the algebraic form of the ray-tracing rules shows that the focal length of the cornea and the eye-lens must now be about 18.52 mm. The strength of the cornea and the eye-lens must now be about 1/0.01852 = 54 diopters. Thus the muscles of the eye must provide an accommodation range of 4 diopters. (This required range is actually independent of the size of the eye or the focal length of the eye lens when it is focused on a very distant object. It depends only on the value for the near-point.)
The distance between the eye-lens and the retina is about 20 mm. When an object is very far away from the eye, the image is located essentially at the focal point. Therefore the focal length of the cornea and the eye-lens should be about 20 mm when the muscles of the eye are relaxed. Since the strength of a lens is the reciprocal of its focal length in meters, the strength of the cornea and the eye-lens in this situation is about 1/0.020 = 50 diopters.
*value given in the question is quite absurd.
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