/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 The relaxed eyes of a patient ha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 38

The relaxed eyes of a patient have a refractive power of 48.5 diopters. (a) Is this patient nearsighted or farsighted? Explain. (b) If this patient is nearsighted, find the far point. If this person is farsighted, find the near point. (For the purposes of this problem, treat the eye as a single-lens system, with the retina \(2.40 \mathrm{cm}\) from the lens.)

Short Answer

Expert verified
The patient is farsighted with a near point greater than 25 cm.

Step by step solution

01

Determine if the patient is nearsighted or farsighted

The total refractive power of a relaxed normal eye is about 60 diopters. If the patient's refractive power is 48.5 diopters, this suggests that the eye has less power than normal, meaning it cannot focus on nearby objects effectively. Hence, the patient is farsighted.
02

Calculate the near point for a farsighted person

For a normal eye, the near point is typically 25 cm. To find the near point for this patient, first calculate the focal length using the formula: \[ f = \frac{1}{P} \] where \( P \) is the refractive power. Substitute \( P = 48.5 \) diopters to find \( f \).
03

Use the lens formula to determine the near point

The lens formula is \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where \( f \) is the focal length, \( v \) is the image distance (distance from lens to retina, 2.4 cm), and \( u \) is the object distance (near point distance we need to find). Solve for \( u \) to find the near point, using the focal length calculated in the previous step.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Refractive Power
Refractive power is a measure of how much a lens can bend light to bring it to a focus. It's expressed in diopters
  • A higher refractive power indicates a stronger lens that can significantly bend light.
  • For the human eye, a refractive power of about 60 diopters is typically normal when the eye is relaxed.
For a patient with a refractive power of 48.5 diopters, like in our example, the eye does not have enough power to focus on close objects. This scenario suggests farsightedness, where distant objects are clear, but nearby objects appear blurry. In optics, understanding refractive power is crucial for diagnosing visual conditions and prescribing corrective lenses.
Farsightedness
Farsightedness, also known as hyperopia, is a common vision condition where distant objects appear clearer than close ones. This occurs when the refractive power of the eye is lower than normal.
  • The eye lens doesn't bend light sufficiently to bring close objects into focus on the retina.
  • Instead, the image is formed behind the retina.
This condition can be diagnosed using tests that measure the eye's refractive power. Farsightedness can be corrected with converging lenses, which increase the eye's total refractive power, allowing images to focus correctly on the retina for clear vision.
Lens Formula
In optics, the lens formula is fundamental for understanding how lenses form images. The formula is given by: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Where
  • \( f \) is the focal length of the lens,
  • \( v \) is the image distance (for the eye, typically the retina's distance from the lens),
  • \( u \) is the object distance (distance of the object from the lens).
This formula helps determine either the position of the image or the object when the other quantities are known. For individuals with visual impairments like farsightedness, the lens formula allows optometrists to calculate the required strength of corrective lenses to achieve normal vision.
Focal Length
Focal length is the distance from the lens to the point where parallel rays of light converge to a single focus. In simplest terms, it determines how "strong" a lens is. Shorter focal lengths imply stronger lenses with greater refractive ability. In mathematical terms, focal length \( f \) is calculated using:\[ f = \frac{1}{P} \]where \( P \) is the refractive power in diopters.
  • For a normal eye with 60 diopters, the focal length is approximately 1.67 cm.
  • For a patient with 48.5 diopters, the focal length is longer at around 2.06 cm.
This difference indicates why the eye cannot focus on nearby objects effectively, as seen in farsighted individuals. Assessing focal length aids in specifying the type and power of lenses that will correct a patient鈥檚 vision accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A converging lens with a focal length of \(4.0 \mathrm{cm}\) is to the left of a second identical lens. When a feather is placed \(12 \mathrm{cm}\) to the left of the first lens, the final image is the same size and orientation as the feather itself. What is the separation between the lenses?

The medium-power objective lens in a laboratory microscope has a focal length \(f_{\text {objective }}=4.00 \mathrm{mm}\). (a) If this lens produces a lateral magnification of \(-40.0,\) what is its "working distance"; that is, what is the distance from the object to the objective lens? (b) What is the focal length of an eyepiece lens that will provide an overall magnification of \(125 ?\)

A simple camera telephoto lens consists of two lenses. The objective lens has a focal length \(f_{1}=+39.0 \mathrm{cm}\). Precisely \(36.0 \mathrm{cm}\) behind this lens is a concave lens with a focal length \(f_{2}=\) \(-10.0 \mathrm{cm} .\) The object to be photographed is \(4.00 \mathrm{m}\) in front of the objective lens. (a) How far behind the concave lens should the film be placed? (b) What is the linear magnification of this lens combination?

\- A telescope has lenses with focal lengths \(f_{1}=+30.0 \mathrm{cm}\) and \(f_{2}=+5.0 \mathrm{cm}\) (a) What distance between the two lenses will allow the telescope to focus on an infinitely distant object and produce an infinitely distant image? (b) What distance between the lenses will allow the telescope to focus on an object that is \(5.0 \mathrm{m}\) away and to produce an infinitely distant image?

A person's prescription for his new bifocal eyeglasses calls for a refractive power of -0.0625 diopter in the distance-vision part and a power of +1.05 diopters in the close-vision part. Assuming the glasses rest \(2.00 \mathrm{cm}\) from his eyes and that the corrected near-point distance is \(25.0 \mathrm{cm},\) determine the near and far points of this person's uncorrected vision.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Circular Motion and Gravitation

Read Explanation

Particle Model of Matter

Read Explanation

Space Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.