Problem 56
Question
A person can see clearly up close but cannot focus on objects beyond 75.0 \(\mathrm{cm} .\) She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?
Step-by-Step Solution
Verified Answer
(a) Nearsighted, (b) Diverging lens, (c) Focal length: -0.75 m, Power: -1.33 D.
1Step 1: Identify Vision Problem
The person can only see objects clearly that are close and cannot focus on objects beyond 75.0 cm. This indicates that the person is nearsighted, or myopic, as they can see close objects clearly but not far ones.
2Step 2: Determine Lens Type
For correcting nearsightedness, diverging lenses are used. These lenses help in adjusting the focal point of the light so that distant objects can be focused correctly on the retina.
3Step 3: Calculate Focal Length
For nearsighted vision correction, the focal length of the lens should allow distant objects to come into focus at the near point of 75.0 cm, or 0.75 m. Using the lens formula for a lens correcting at infinity, which is \\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \where \(v\) is the image distance (set to be the near point, 75 cm or -0.75 m, because virtual images are formed on the same side of the lens as the object for diverging lenses) and \(u\) is at infinity (i.e., \( \frac{1}{u} = 0 \)). The formula simplifies to \\[ \frac{1}{f} = -\frac{1}{v} = -\frac{1}{0.75} \Rightarrow f = -0.75 \text{ m} \] \So, the focal length needed is -0.75 meters.
4Step 4: Calculate Power of Lens
The power \(P\) of a lens in diopters is given by the formula \\[ P = \frac{1}{f(\text{in meters})} \] \Thus, using the calculated focal length, \\[ P = \frac{1}{-0.75} = -1.33 \text{ diopters} \]. \The negative sign indicates a diverging lens.
Key Concepts
NearsightednessDiverging LensFocal LengthDiopters
Nearsightedness
Nearsightedness, also known as myopia, is a common vision issue where individuals can see objects close to them clearly but struggle to focus on distant objects. This occurs because the light entering the eye is focused in front of the retina instead of directly on it. This can make things like road signs or boards difficult to read unless they get closer. The eye shape, either being too long or the cornea being too curved, usually causes this condition. To correct nearsightedness, wearing lenses is one solution that can refocus the light precisely on the retina, providing clear vision at both near and far distances.
Diverging Lens
Diverging lenses, also known as concave lenses, are used to correct nearsightedness. These lenses have a distinct shape that is thinner at the center and thicker at the edges. When the light passes through a diverging lens, it spreads out or "diverges," creating a virtual image that the eye can focus on correctly. This adjustment helps direct light rays onto the retina rather than in front of it, which is necessary for clear distant vision.
- Diverging lenses are identified by their characteristic concave shape.
- They help correct the focal point of light so it aligns properly with the retina.
- The negative focal length or negative diopter indicates diverging lens power.
Focal Length
The focal length of a lens is an essential factor in correcting vision. In the case of diverging lenses for myopia, the focal length is negative. This negative measure reflects the lens’s ability to spread out light rays before they reach the eye.
In this scenario, where a person cannot see beyond 75.0 centimeters, adjustments in the focal length through corrective lenses bring distant objects into the range of clear vision. The lens formula, \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), is used here. Since the goal is to correct vision for objects at infinity, the formula simplifies, permitting us to find the negative focal length needed: \( f = -0.75 \) m. This calculation guides the choice of lens necessary to focus the light properly on the retina.
In this scenario, where a person cannot see beyond 75.0 centimeters, adjustments in the focal length through corrective lenses bring distant objects into the range of clear vision. The lens formula, \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), is used here. Since the goal is to correct vision for objects at infinity, the formula simplifies, permitting us to find the negative focal length needed: \( f = -0.75 \) m. This calculation guides the choice of lens necessary to focus the light properly on the retina.
Diopters
Diopters measure the refractive power of lenses and play a significant role in eye prescriptions. Specifically, in the context of nearsightedness correction, diopters express how much the lenses must diverge light to focus it on the retina.
- The calculation for diopter strength involves taking the inverse of the focal length in meters: \( P = \frac{1}{f} \).
- For our example, a focal length of \(-0.75\) meters results in a lens power of \(-1.33\) diopters.
- The negative sign in diopters indicates the use of a diverging lens.
Other exercises in this chapter
Problem 53
(a) Where is the near point of an eye for which a contact lens with a power of \(+2.75\) diopters is prescribed? (b) Where is the far point of an eye for which
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View solution Problem 58
A thin lens with a focal length of 6.00 \(\mathrm{cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object
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The focal length of a simple magnifier is 8.00 \(\mathrm{cm} .\) Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the m
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