/*! 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 122 An object is in front of a conve... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 26: Problem 122

An object is in front of a converging lens \((f=0.30 \mathrm{m})\). The magnification of the lens is \(m=4.0 .\) (a) Relative to the lens, in what direction should the object be moved so that the magnification changes to \(m=-4.0 ?\) (b) Through what distance should the object be moved?

Short Answer

Expert verified
Move the object 0.30 m away from the lens to achieve the new magnification.

Step by step solution

01

Understanding the Problem

We are given a converging lens with a focal length \(f = 0.30 \text{ m}\) and an initial magnification \(m = 4.0\). We need to determine how to change the position of the object to achieve a new magnification of \(m = -4.0\). We will use the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), and the magnification formula \(m = -\frac{d_i}{d_o}\).
02

Calculate Initial Distance

Using the formula \(m = -\frac{d_i}{d_o}\), we have \(4 = -\frac{d_i}{d_o}\) which implies \(d_i = -4d_o\). Substituting into the lens formula: \(\frac{1}{0.30} = \frac{1}{d_o} - \frac{1}{4d_o}\). Solving this gives \(d_o = \frac{0.30}{3} = 0.10 \text{ m}\).
03

Calculate Final Distance for New Magnification

For a magnification \(m = -4.0\), we use \(-4 = -\frac{d_i}{d_o}\), leading to \(d_i = 4d_o\). Substituting into the lens formula \(\frac{1}{0.30} = \frac{1}{d_o} + \frac{1}{4d_o}\). Solving this equation gives \(d_o = 0.40 \text{ m}\).
04

Determine Direction and Distance to Move

Initially, \(d_o = 0.10 \text{ m}\) and finally \(d_o = 0.40 \text{ m}\). The object should be moved away from the lens. The object needs to be moved through a distance \(0.40 - 0.10 = 0.30 \text{ m}\).

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 vital tool in optics that relates the focal length of a lens to the distances of the object and its image. It is expressed as: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] where:
  • \(f\) is the focal length of the lens.
  • \(d_o\) is the distance from the object to the lens.
  • \(d_i\) is the distance from the image to the lens.
This formula helps determine the position and nature of the image formed by a lens. A converging lens focuses incoming light to a point, creating a real or virtual image. The sign conventions for the distances are crucial:
  • Real images have positive \(d_i\), and virtual images have negative \(d_i\).
  • If the object is on the same side as the incoming light, \(d_o\) is positive.
Using this formula, you can solve problems like the one in the exercise to find out how the position changes when you alter magnification.
Magnification
Magnification describes how much larger or smaller the image is compared to the object. It is calculated by the formula: \[m = -\frac{d_i}{d_o}\] where:
  • \(m\) is the magnification factor.
  • \(d_i\) is the image distance.
  • \(d_o\) is the object distance.
The negative sign in the formula indicates that a real image is inverted. Magnification greater than one means the image is larger than the object, while magnification less than one means it is smaller.
  • If \(m\) is positive, the image is erect (upright).
  • If \(m\) is negative, the image is inverted.
In the exercise, initially, the magnification is 4 (positive), indicating an erect image, whereas changing it to -4 makes the image inverted, as specified by the formula.
Focal Length
Focal length is a crucial parameter of any lens that determines where the light converges to a point. For converging lenses, it is the distance from the lens to the focal point, where parallel rays of light meet. The focal length, \(f\), affects how large the image appears and how far it is from the lens.
  • Shorter focal lengths bring images closer to the lens and make them appear larger.
  • Longer focal lengths push images farther away and make them smaller.
In optics exercises, the focal length is often a given parameter used alongside the lens and magnification formulas.
For the problem in the exercise, the focal length is given as 0.30 m. This value is crucial in calculating the initial and final positions of the object to achieve the desired magnification changes. By understanding focal length, you can predict how changing the object's position affects the image formed by a lens.

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) For a diverging lens \((f=-20.0 \mathrm{cm}),\) construct a ray diagram to scale and find the image distance for an object that is \(20.0 \mathrm{cm}\) from the lens. (b) Determine the magnification of the lens from the diagram.

Bill is farsighted and has a near point located \(125 \mathrm{cm}\) from his eyes. Anne is also farsighted, but her near point is \(75.0 \mathrm{cm}\) from her eyes. Both have glasses that correct their vision to a normal near point \((25.0 \mathrm{cm}\) from the eyes), and both wear the glasses \(2.0 \mathrm{cm}\) from the eyes. Relative to the eyes, what is the closest object that can be seen clearly (a) by Anne when she wears Bill's glasses and (b) by Bill when he wears Anne's glasses?

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

Violet light and red light travel through air and strike a block of plastic at the same angle of incidence. The angle of refraction is \(30.400^{\circ}\) for the violet light and \(31.200^{\circ}\) for the red light. The index of refraction for violet light in plastic is greater than that for red light by 0.0400. Delaying any rounding off of calculations until the very end, find the index of refraction for violet light in plastic.

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?
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Quantum Physics

Read Explanation

Scientific Method Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Classical Mechanics

Read Explanation

Fluids

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.