/*! 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 107 A nearsighted person cannot read... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 26: Problem 107

A nearsighted person cannot read a sign that is more than \(5.2 \mathrm{~m}\) from his eyes. To deal with this problem, he wears contact lenses that do not correct his vision completely, but do allow him to read signs located up to distances of \(12.0 \mathrm{~m}\) from his eyes. What is the focal length of the contacts?

Short Answer

Expert verified
Focal length of the contacts is approximately 8.73 m.

Step by step solution

01

Understand the Problem

A nearsighted person has a far point of 5.2 meters without correction. With contact lenses, this person can see up to 12.0 meters. We are to find the focal length of these lenses.
02

Use the Lens Formula

The lens formula relates the object distance \(d_o\), image distance \(d_i\), and the focal length \(f\) of a lens: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] In this situation, the far point without lenses (5.2 meters) serves as the image distance \(d_i\), and the farthest point visible with lenses (12.0 meters) is the object distance \(d_o\).
03

Assign Known Values

Assign the known distances to the formula: \(d_o = 12.0\) m and \(d_i = -5.2\) m. The image distance \(d_i\) is negative because the image forms on the same side as the object for a nearsighted person.
04

Substitute Values into the Formula

Substitute the known values into the lens formula: \[\frac{1}{f} = \frac{1}{12.0} + \frac{1}{-5.2}\]
05

Calculate \(\frac{1}{f}\)

Calculate the right side of the equation: \[\frac{1}{f} = \frac{1}{12.0} - \frac{1}{5.2}\]Find common denominators and perform the subtraction.
06

Solve for \(f\)

After calculating the right side, invert the result to find \(f\): \(f = 1 / \text{(result from previous step)}\).
07

Conclusion

Compute the value of \(f\) and confirm the units in meters. The focal length of the contacts is the value you have calculated.

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.

Lens Formula
The lens formula is a fundamental concept in optics. It helps us relate the focal length of a lens to the distances of an object and its image. The formula is given by:
\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where:
  • \(f\) is the focal length of the lens.
  • \(d_o\) is the distance from the lens to the object (object distance).
  • \(d_i\) is the distance from the lens to the image (image distance).
The key thing about the lens formula is that it uses reciprocal values. You find the reciprocal of the focal length by adding the reciprocals of the object distance and the image distance. This equation is vital for designing optical systems, like glasses and cameras.
In our exercise, understanding the relationship between these distances and the focal length enables us to determine the power needed in lenses to correct vision.
Nearsightedness
Nearsightedness, or myopia, is a common vision condition where distant objects appear blurry while close objects can be seen clearly. This happens because the eye is elongated or the cornea is too curved, which causes light to focus in front of the retina rather than directly on it.
To address nearsightedness:
  • Corrective lenses such as glasses or contact lenses are used to diverge light rays and move the focus back onto the retina.
  • The far point (farthest distance an object can be seen clearly without correction) is closer than normal.
In the context of the exercise, the far point without lenses is 5.2 meters. This distance tells us how much vision correction is needed for the person in question.
Focal Length
Focal length is a critical term used when discussing lenses. It represents the distance from the lens to its focus, where parallel rays of light meet. A lens with shorter focal length bends light more than one with a longer focal length.
Key considerations for focal length include:
  • Lenses for nearsighted individuals will have a negative focal length.
  • Shorter focal length lenses provide stronger correction due to greater bending of light.
In our problem, the focal length helps determine how much correction the lens needs to provide to extend the far point from 5.2 meters to 12.0 meters. Once we calculate the focal length using the lens formula, we know exactly how much the contact lenses adjust the focus.
Vision Correction
Vision correction aims to adjust the eye's focusing ability to see clearly at various distances. For nearsightedness, lenses need to diverge light before it reaches the eye.
Methods of vision correction include:
  • Eyeglasses which can easily be adjusted for different lighting needs and distances.
  • Contact lenses which sit right on the eye for convenience and comfort.
Using the focal length found through calculations, we adjust the strength of these lenses. In the given exercise, contact lenses correct the person鈥檚 sight enough to see distant objects previously indistinct. The corrected vision now extends to 12.0 meters, indicating improved visual capacity.

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

The back wall of a home aquarium is a mirror that is a distance \(L\) away from the front wall. The walls of the tank are negligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. (a) Does the fish appear to be at a distance greater than, less than, or equal to \(\frac{1}{2} L\) from the front wall? (b) The mirror forms an image of the fish. How far from the front wall is this image located? Express your answer in terms of \(L\). (c) Assume that your answer to Question (b) is a distance \(D\). Does the image of the fish appear to be at a distance greater than, less than, or equal to \(D\) from the front wall? (d) Could the image of the fish appear to be in front of the mirror if the index of refraction of water were different than it actually is, and, if so, would the index of refraction have to be greater than or less than its actual value? Explain each of your answers. The distance between the back and front walls of the aquarium is \(40.0 \mathrm{~cm} .\) (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall. Verify that your answers are consistent with your answers to the Concept Questions.

Two identical diverging lenses are separated by \(16 \mathrm{~cm} .\) The focal length of each lens is \(-8.0 \mathrm{~cm} .\) An object is located \(4.0 \mathrm{~cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

Mars subtends an angle of \(8.0 \times 10^{-5} \mathrm{rad}\) at the unaided eye. An astronomical telescope has an eyepiece with a focal length of \(0.032 \mathrm{~m}\). When Mars is viewed using this telescope, it subtends an angle of \(2.8 \times 10^{-3} \mathrm{rad}\). Find the focal length of the telescope's objective lens.

A camper is trying to start a fire by focusing sunlight onto a piece of paper. The diameter of the sun is \(1.39 \times 10^{9} \mathrm{~m},\) and its mean distance from the earth is \(1.50 \times 10^{11} \mathrm{~m} .\) The camper is using a converging lens whose focal length is \(10.0 \mathrm{~cm}\) (a) What is the area of the sun's image on the paper? (b) If \(0.530 \mathrm{~W}\) of sunlight pass through the lens, what is the intensity of the sunlight at the paper?

At illustrates one approach to solving problems such as this one. A farsighted person has a near point that is \(67.0 \mathrm{~cm}\) from her eyes. She wears eyeglasses that are designed to enable her to read a newspaper held at a distance of \(25.0\) \(\mathrm{cm}\) from her eyes. Find the focal length of the eyeglasses, assuming that they are worn (a) \(2.2 \mathrm{~cm}\) from the eyes and (b) \(3.3 \mathrm{~cm}\) from the eyes.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Electrostatics

Read Explanation

Rotational Dynamics

Read Explanation

Modern Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Linear Momentum

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.