/*! 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 12 Consider a concave mirror that h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 12

Consider a concave mirror that has a focal length \(f\). In terms of \(f\), determine the object distances that will produce a magnification of (a) \(-1,(b)-2,\) and \((c)-3\).

Short Answer

Expert verified
(a) 2f, (b) 1.5f, (c) 1.33f.

Step by step solution

01

Understanding Magnification

The magnification (m) for a mirror is given by the equation \( m = -\frac{q}{p} \), where \( q \) is the image distance and \( p \) is the object distance. If \( m = -1 \), \(-2\), or \(-3\), it indicates an inverted image that's the same size, twice the size, or three times the size of the object respectively.
02

Mirror Formula

The mirror formula relates the object distance \( p \), the image distance \( q \), and the focal length \( f \) by \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \). We'll use this and the magnification formula to find \( p \) for each magnification value.
03

Case (a): Magnification -1

If \( m = -1 \), then \( -\frac{q}{p} = -1 \), implying \( q = p \). Substituting \( q = p \) in the mirror formula: \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{p} = \frac{2}{p} \]This gives us \( p = 2f \) for \( m = -1 \).
04

Case (b): Magnification -2

For \( m = -2 \), \(-\frac{q}{p} = -2\), implying \( q = 2p \). Substituting \( q = 2p \) in the mirror formula:\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{2p} = \frac{3}{2p} \]Rearranging gives \( p = \frac{3}{2}f \) for \( m = -2 \).
05

Case (c): Magnification -3

For \( m = -3 \), \(-\frac{q}{p} = -3\), implying \( q = 3p \). Substituting \( q = 3p \) in the mirror formula:\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{3p} = \frac{4}{3p} \]Rearranging gives \( p = \frac{4}{3}f \) for \( m = -3 \).

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.

Focal Length
Learning about concave mirrors begins with understanding focal length. The focal length, denoted as \( f \), is the distance between the mirror's surface and its focal point鈥攖he point where parallel rays of light converge after reflecting off the mirror. For concave mirrors, the focal length is a positive value. The concept of focal length is crucial when calculating the placement of images formed by the mirror. Remember, the focal length helps relate the object and image distances using the mirror formula.
  • Concave mirrors have a focal point in front of the mirror.
  • The focal length is half the mirror's radius of curvature.
  • Understanding focal length is essential for solving mirror problems.
Magnification
Magnification describes how much larger or smaller the image is compared to the object. In the context of mirrors, the magnification is given by the formula \( m = -\frac{q}{p} \), where \( q \) is the image distance and \( p \) is the object distance. The negative sign indicates that the image is inverted relative to the object. For magnifications like
\( m = -1, -2, -3 \), the image is inverted and of different sizes:
  • \( m = -1 \): Image is the same size as the object.
  • \( m = -2 \): Image is twice as large as the object.
  • \( m = -3 \): Image is three times the size of the object.
Understanding magnification helps predict physical properties of images in mirror reflection.
Mirror Formula
The mirror formula is a critical tool in optics. This equation relates the object distance \( p \), image distance \( q \), and focal length \( f \) through the equation: \[\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\]This formula can be used to find unknown distances, whether it's the distance of the object from the mirror or the distance at which the image is formed. By substituting known values, we can manipulate the equation to solve for unknown variables.
The mirror formula underpins all calculations related to concave mirrors:
  • It's used in conjunction with magnification to solve many real-world problems.
  • This relation is fundamental for applications in designing optical devices.
  • It simplifies finding out which position should an object be placed to form desired image sizes.
Object Distance
Object distance, denoted as \( p \), is the distance from the object to the mirror surface. In problems involving concave mirrors, determining the correct object distance is crucial to forming images of desired characteristics. By using the mirror formula and magnification, we can derive object distances needed for specific magnifications. The relation between the object distance and the focal point helps predict where an image will form, and its size, relative to the object.
Consider these object distances for specific magnifications:
  • \( p = 2f \) for \( m = -1 \): Produces an inverted image of the same size as the object.
  • \( p = \frac{3}{2}f \) for \( m = -2 \): Forms a twice magnified inverted image.
  • \( p = \frac{4}{3}f \) for \( m = -3 \): Results in a three times larger inverted image.
Understanding object distance in relation to focal length is vital for controlling the nature of the image formed by a concave mirror.

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 person is swimming \(1.0 \mathrm{~m}\) beneath the surface of the water in a swimming pool. A child standing on the diving board drops a ball into the pool directly above the swimmer. The swimmer sees the ball dropped from a height of \(3.0 \mathrm{~m}\) above the water. From what height was the ball actually dropped?

The focal length of a mirror can be determined entirely from the shape of the mirror. In contrast, to determine the focal length of a lens, we must know both the shape of the lens and its index of refraction - as well as the index of refraction of the surrounding medium. For instance, when a thin lens is immersed in a liquid, we must modify the thin-lens equation to take into account the refractive properties of the surrounding liquid: $$\frac{1}{f}=\left(\frac{n}{n_{\mathrm{liq}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right),$$ where \(n_{\text {liq }}\) is the index of refraction of the liquid and \(n\) is the index of refraction of the glass. If you place a concave mirror with a focal length of \(1 \mathrm{~m}\) into water, what will happen? A. The mirror will no longer be able to focus light. B. The focal length of the mirror will decrease. C. The focal length of the mirror will increase. D. Nothing will happen.

The left end of a long glass rod \(8.00 \mathrm{~cm}\) in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of \(4.00 \mathrm{~cm} .\) An object in the form of an arrow \(1.50 \mathrm{~mm}\) tall, at right angles to the axis of the rod, is located on the axis \(24.0 \mathrm{~cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image upright or inverted?

A converging lens has a focal length of \(14.0 \mathrm{~cm} .\) For each of two objects located to the left of the lens, one at a distance of \(18.0 \mathrm{~cm}\) and the other at a distance of \(7.00 \mathrm{~cm},\) determine (a) the image position, (b) the magnification, (c) whether the image is real or virtual, and (d) whether the image is upright or inverted. Draw a principal-ray diagram in each case.

A concave spherical mirror has a radius of curvature of \(10.0 \mathrm{~cm}\). Calculate the location and size of the image formed of an \(8.00-\mathrm{mm}-\) tall object whose distance from the mirror is (a) \(15.0 \mathrm{~cm},\) (b) \(10.0 \mathrm{~cm}\), (c) \(2.50 \mathrm{~cm},\) and (d) \(10.0 \mathrm{~m}\)
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Thermodynamics

Read Explanation

Mechanics and Materials

Read Explanation

Particle Model of Matter

Read Explanation

Kinematics Physics

Read Explanation

Energy Physics

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.