/*! 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 80 A farsighted man uses eyeglasses... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A farsighted man uses eyeglasses with a refractive power of 3.80 diopters. Wearing the glasses 0.025 m from his eyes, he is able to read books held no closer than 0.280 m from his eyes. He would like a prescription for contact lenses to serve the same purpose. What is the correct contact lens prescription, in diopters?

Short Answer

Expert verified
3.57 diopters

Step by step solution

01

Understand the Problem

The farsighted man uses glasses with a refractive power of 3.80 diopters to read books held no closer than 0.280 meters from his eyes. His glasses are worn 0.025 meters from his eyes. We need to find the refractive power for contact lenses that allow him to read books at the same distance.
02

Calculate the Lens-to-Focal Point Distance for Glasses

Since the glasses are placed 0.025 meters from the eyes and the closest reading distance is 0.280 meters, the lens-to-focal point distance is the difference: \(0.280 \, \text{m} - 0.025 \, \text{m} = 0.255 \, \text{m}\).
03

Determine the Actual Focal Length for Glasses

Use the lens formula \(f = \frac{1}{P}\), where \(P = 3.80\) diopters for the glasses. Thus, \(f_{glasses} = \frac{1}{3.80} \approx 0.263 \, \text{m}\).
04

Compare with Reading Distance

The focal length \(f_{glasses} = 0.263\) m should focus the image at the reading distance minus the distance of glasses from the eye: \(0.255\) m. We don't need further adjustments since they are close enough, indicating slight accommodation is involved.
05

Calculate Needed Power for Contact Lenses

Use the reading distance directly for contact lenses (since they are on the eye): \(0.280\) m. Thus, we use \(P = \frac{1}{0.280} \approx 3.57\) diopters.

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 91Ó°ÊÓ!

Key Concepts

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

Farsightedness
Farsightedness, also known as hyperopia, is a common vision condition where distant objects can be seen more clearly than nearby ones.
This occurs when the light entering the eye focuses behind the retina instead of directly on it. The primary cause of farsightedness is either the eyeball being too short or the cornea having too little curvature. Individuals with farsightedness often struggle with tasks like reading, as they have difficulty seeing objects up close.
Symptoms may include eye strain, headaches, or squinting when focusing on close work. To address this issue, corrective lenses are used. These lenses help to bend the light rays together more sharply, allowing them to focus on the retina instead of behind it. In our example, the man used eyeglasses with corrective power to assist with his vision while reading at a close distance.
Refractive Power
Refractive power is a measure of how much a lens can bend light to focus on a specific point. It is expressed in units called diopters (D).
A lens with a higher diopter value has a greater refractive power and can bend light more effectively. The formula for calculating refractive power is given as: \[ P = \frac{1}{f} \]where \( P \) is the power in diopters and \( f \) is the focal length in meters. A positive refractive power indicates converging lenses, usually used for correcting farsightedness.
In the exercise, the man uses glasses with a refractive power of 3.80 diopters to correct his farsightedness, by bringing the focal point of light onto his retina.
Contact Lenses Prescription
When prescribing contact lenses, the refractive power needs to be accurately calculated because they sit directly on the eye.
Unlike eyeglasses, which are positioned some distance away from the eyes, contact lenses eliminate this gap, and thus, adjustments are necessary.In the problem, the glasses have a refractive power of 3.80 diopters.{The man wearing these glasses can read at a minimum distance of 0.280 meters. To maintain the same vision ability with contact lenses, we directly use this reading distance in our calculations.} The formula for calculating the refractive power needed for the contact lenses is:\[ P = \frac{1}{0.280} \approx 3.57 \, \text{diopters} \]This power compensates for the loss of distance between the lens and the eye present with glasses. Thus, the correct prescription for the contact lenses in this scenario is 3.57 diopters.

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 near point of a naked eye is 32 cm. When an object is placed at the near point and viewed by the naked eye, it has an angular size of 0.060 rad. A magnifying glass has a focal length of 16 cm, and is held next to the eye. The enlarged image that is seen is located 64 cm from the magnifying glass. Determine the angular size of the image.

The drawing shows a rectangular block of glass \((n=1.52)\) surrounded by liquid carbon disulfide \((n=1.63) .\) A ray of light is incident on the glass at point A with a \(30.0^{\circ}\) angle of inci- dence. At what angle of refraction does the ray leave the glass at point \(\mathrm{B} ?\)

A ray of light is traveling in glass and strikes a glass–liquid interface. The angle of incidence is \(58.0^{\circ}\) , and the index of refraction of glass is \(n=1.50 .\) (a) What must be the index of refraction of the liquid so that the direction of the light entering the liquid is not changed? (b) What is the largest index of refraction that the liquid can have, so that none of the light is transmitted into the liquid and all of it is reflected back into the glass?

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 passes through the lens, what is the intensity of the sunlight at the paper?

A converging lens \((f=25.0 \mathrm{cm})\) is used to project an image of an object onto a screen. The object and the screen are 125 \(\mathrm{cm}\) apart, and between them the lens can be placed at either of two locations. Find the two object distances.

See all solutions

Recommended explanations on Physics Textbooks

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.