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BIO 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 focallength contact lens is needed, and what is its power in diopters?

Short Answer

Expert verified
(a) Nearsighted, (b) Diverging lens, (c) Focal length = -0.75 m, Power = -1.33 D.

Step by step solution

01

Identify the Vision Problem

When a person cannot focus on objects beyond a certain point, they have difficulty seeing distant objects clearly, which means they are nearsighted. Nearsightedness is also called myopia.
02

Determine the Type of Lens Needed

Nearsighted individuals require diverging lenses to correct their vision. Diverging lenses help in adjusting the focal point so that distant objects can be brought into clear focus.
03

Calculate the Focus Length

The person can see clearly up to 75.0 cm. Therefore, the far point is 75.0 cm or 0.75 meters. Since an ideal lens should make objects at infinity appear at this far point, we use the lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( u \) is infinity (\( \frac{1}{\infty} = 0 \)), and \( v = -0.75 \), since the image formed is virtual and on the same side as the object. Thus, \( \frac{1}{f} = 0 - \left( -\frac{1}{0.75} \right) = \frac{1}{0.75} = \frac{4}{3} \). Therefore, \( f = 0.75 \) meters.
04

Compute Lens Power in Diopters

The power of a lens in diopters is given by \( P = \frac{1}{f} \), where \( f \) is the focal length in meters. From the previous step, \( f = -0.75 \), so \( P = -\frac{1}{0.75} = -\frac{4}{3} \approx -1.33 \) diopters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Myopia
Myopia, commonly known as nearsightedness, is a prevalent visual impairment where individuals can see near objects clearly but struggle to focus on distant ones. This condition occurs when the eyeball is slightly elongated, causing light entering the eye to focus in front of the retina rather than directly on it. As a result, distant images appear blurry.
Myopia is typically diagnosed through a standard eye exam, and its severity can vary. Factors contributing to the development of myopia include genetics and lifestyle, such as prolonged periods focusing on close objects like screens and books. Corrective lenses or surgery can effectively manage myopia, enhancing distance vision.
Role of Diverging Lenses in Myopia Correction
Diverging lenses, also known as concave lenses, are essential in correcting myopia. These lenses are characterized by their thinner center compared to their edges, allowing them to spread light rays outwards. This adjustment helps the eye focus images on the retina, thus clarifying distant objects.
When light passes through a diverging lens, it diverges, or spreads out, effectively extending the focus to the retina. This optical adjustment is crucial for myopic individuals, enabling them to see distant objects with improved clarity. Choosing the right diverging lens involves assessing the individual's specific refractive error to customize the lens curvature and thickness accordingly.
Calculating Lens Power in Diopters
Lens power, measured in diopters, quantifies a lens's ability to converge or diverge light. It is a fundamental concept in ophthalmology. For myopia correction, lens power helps determine the strength of the necessary diverging lenses, with negative values indicating their divergent nature.
The formula for lens power is simple: \( P = \frac{1}{f} \), where \( f \) is the focal length in meters. In cases of myopia, this value is negative, reflecting the diverging nature of the lens needed. For instance, a lens with a power of -1.33 diopters indicates a focal length of -0.75 meters, suitable to correct the specific myopic condition described.
Focal Length Calculation for Lenses
The focal length is a key parameter in designing lenses, indicative of how strongly a lens can focus or disperse light. When correcting myopia, the focal length is the distance from the lens where light converges to a focal point. It is calculated using the formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( u \) is the object distance (infinity for distant objects), and \( v \) is the image distance or far point of a myopic eye.
For a myopic eye viewing distant objects through corrective lenses, the far point is negative, representing a virtual image formed on the same side as the object. Hence, for a person with a 75 cm far point, the calculation yields a focal length \( f = -0.75 \) meters. This negative value signifies a diverging lens, essential for adjusting the focus to the neural retina.

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Most popular questions from this chapter

A certain microscope is provided with objectives that have focal lengths of \(16 \mathrm{~mm}, 4 \mathrm{~mm},\) and \(1.9 \mathrm{~mm}\) and with eyepieces that have angular magnifications of \(5 \times\) and \(10 \times .\) Each objective forms an image \(120 \mathrm{~mm}\) beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable.

The characteristics that follow are characteristics of two of Galileo's surviving double-convex lenses. The numbers given are magnitudes only; you must supply the correct signs. \(L_{1}:\) front radius \(=950 \mathrm{~mm},\) rear radius \(=2700 \mathrm{~mm},\) refractive index \(=\) \(1.528 ; L_{2}\) :front radius \(=535 \mathrm{~mm},\) rear radius \(=50,500 \mathrm{~mm},\) refractive index \(=1.550 .\) (a) What is the largest angular magnification that Galileo could have obtained with these two lenses? (b) How long would this telescope be between the two lenses?

A refracting telescope has an objective lens of focal length 16.0 in. and eyepieces of focal lengths \(15 \mathrm{~mm}, 22 \mathrm{~mm}, 35 \mathrm{~mm},\) and \(85 \mathrm{~mm}\) What are the largest and smallest angular magnifications you can achieve with this instrument?

A compound microscope has an objective lens of focal length \(10.0 \mathrm{~mm}\) with an eyepiece of focal length \(15.0 \mathrm{~mm},\) and it produces its final image at infinity. The object to be viewed is placed \(2.0 \mathrm{~mm}\) beyond the focal point of the objective lens. (a) How far from the objective lens is the first image formed? (b) What is the overall magnification of this microscope?

The crystalline lens of the human eye is double convex and has a typical range of optical power from 115 diopters to 150 diopters. (a) What is the range of focal lengths the eye can achieve? (b) At minimum power, where does it focus the image of a very distant object? (c) At maximum power, where does it focus the image of an object at the near point of \(25 \mathrm{~cm} ?\)

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