/*! 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 89 A farsighted person can read pri... [FREE SOLUTION] | 91Ó°ÊÓ

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A farsighted person can read printing as close as \(25.0 \mathrm{cm}\) when she wears contacts that have a focal length of \(45.4 \mathrm{cm} .\) One day, she forgets her contacts and uses a magnifying glass, as in Figure \(26.39 b .\) Its maximum angular magnification is 7.50 for a young person with a normal near point of \(25.0 \mathrm{cm} .\) What is the maximum angular magnification that the magnifying glass can provide for her?

Short Answer

Expert verified
The maximum angular magnification is 2.

Step by step solution

01

Understanding the problem

The problem involves a farsighted person who uses a magnifying glass without her usual contact lenses. We need to find out the maximum angular magnification that the magnifying glass can provide for her.
02

Define known values

The focal length of the contact lenses is given as \( f = 45.4 \text{ cm} \), and the normal near point for a young person is \( 25.0 \text{ cm} \). The maximum angular magnification for a person with a close near point of \( 25.0 \text{ cm} \) is \( 7.50 \).
03

Identify effective focal length

When the farsighted person uses the magnifying glass, the far point (25 cm with contacts) becomes effectively \( 45.4 \text{ cm} \) without them due to her original vision without correction.
04

Calculate new angular magnification

Angular magnification \( M \) is given by \[ M = 1 + \frac{D}{f} \] where \( D \) is the near point when using the magnifying glass. Normally \( D = 25 \text{ cm} \), but for her it's effectively \( 45.4 \text{ cm} \) when using the magnifying glass. The maximum magnification can be re-calculated as \[ M = 1 + \frac{45.4}{45.4} \].
05

Simplify the expression and solve

Putting the values into the formula, we have \[ M = 1 + 1 = 2. \] This means that the maximum angular magnification that the magnifying glass can provide her is 2.

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Key Concepts

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

Angular Magnification
Angular magnification is an essential concept in optics, particularly when using devices like magnifying glasses. It describes how much larger an object appears when viewed through a lens or optical instrument compared to viewing it with the naked eye. This is usually expressed as a ratio. The formula used to calculate angular magnification is \[M = 1 + \frac{D}{f}\] where \(D\) is the distance of the near point and \(f\) is the focal length of the lens.
  • In simpler terms, angular magnification tells us how a lens increases the apparent size of an object based on its focal length.
A powerful magnifying glass or lens will have a higher angular magnification, allowing objects to appear significantly larger and more detailed. In the context of our exercise, it highlights the change in visual perception for someone with farsightedness when different assistive devices are used.
Farsightedness
Farsightedness, medically referred to as hyperopia, is a common vision condition where distant objects can be seen more clearly than close objects. This occurs when light entering the eye is focused behind the retina instead of on it. For people with farsightedness, reading or doing close work can be challenging without corrective lenses.
  • The most common solution for this is the use of convex lenses, which help converge the light rays before they hit the retina.
  • Contact lenses or glasses with the appropriate focal length can correct this issue for everyday activities.
In the problem given, the individual's usual contacts have a focal length of 45.4 cm to correct her farsightedness, allowing her to read at a comfortable distance. When the contacts are forgotten, a magnifying glass serves as a temporary aid by adjusting the angular magnification to make near objects more readable.
Focal Length
Focal length is a fundamental property of optical lenses and systems. It is the distance from the lens to the point where light rays converge to a focal point. This measure influences how a lens converges or diverges light, impacting how images appear.
  • A lens with a shorter focal length bends light more sharply and provides greater angular magnification.
  • Conversely, lenses with longer focal lengths have a gentler curvature and provide less magnification.
The focal length is crucial in determining the functionality of lenses in correcting vision issues like farsightedness. It also specifies the strength of a magnifying glass. In our exercise, converting the lens's focal length into a usable angular magnification for the individual demonstrates its practical application in optical devices.
Near Point
The near point represents the closest distance at which the eye can focus on an object clearly. For people with normal vision, this is typically about 25 cm. However, this distance can vary significantly based on individual vision ability, especially with conditions like farsightedness.
  • In young people with normal vision, the near point is around 25 cm, providing a baseline for angular magnification calculations.
  • For those with farsightedness, the near point can be much further, necessitating visual aids to help bring it closer.
The exercise highlights how the near point shifts based on lens correction. With the magnifying glass, the person’s effective near point becomes 45.4 cm, illustrating how corrective lenses alter this natural distance, thereby restoring the ability to perform tasks requiring detailed vision.

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Most popular questions from this chapter

A forensic pathologist is viewing heart muscle cells with a microscope that has two selectable objectives with refracting powers of 100 and 300 diopters. When he uses the 100 -diopter objective, the image of a cell subtends an angle of \(3 \times 10^{-3}\) rad with the eye. What angle is subtended when he uses the 300 -diopter objective?

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.

A woman can read the large print in a newspaper only when it is at a distance of \(65 \mathrm{cm}\) or more from her eyes. (a) Is she nearsighted (myopic) or farsighted (hyperopic), and what kind of lens is used in her glasses to correct her eyesight? (b) What should be the refractive power (in diopters) of her glasses (worn \(2.0 \mathrm{cm}\) from the eyes), so she can read the newspaper at a distance of \(25 \mathrm{cm}\) from her eyes?

The telescope at Yerkes Observatory in Wisconsin has an objective whose focal length is \(19.4 \mathrm{m} .\) Its eyepiece has a focal length of \(10.0 \mathrm{cm} .\) (a) What is the angular magnification of the telescope? (b) If the telescope is used to look at a lunar crater whose diameter is \(1500 \mathrm{m},\) what is the size of the first image, assuming that the surface of the moon is \(3.77 \times 10^{8} \mathrm{m}\) from the surface of the earth? (c) How close does the crater appear to be when seen through the telescope?

When light strikes the surface between two materials from above, the Brewster angle is \(65.0^{\circ} .\) What is the Brewster angle when the light encounters the same surface from below?

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