/*! 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 5 A concave spherical mirror has a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 5

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 -mm-tall object whose distance from the mirror is (a) \(15.0 \mathrm{cm},(\mathrm{b}) 10.0 \mathrm{cm},(\mathrm{c}) 2.50 \mathrm{cm},\) and (d) 10.0 \(\mathrm{m}\)

Short Answer

Expert verified
(a) Image at 7.5 cm, -4.0 mm; (b) Image at 10 cm, -8.0 mm; (c) Image at -2.5 cm, 8.0 mm; (d) Image at 5 cm, -0.04 mm.

Step by step solution

01

Determine Focal Length

The focal length \( f \) of a concave mirror is half of the radius of curvature. Given the radius of curvature \( R = 10.0 \text{ cm} \), the focal length is \( f = \frac{R}{2} = 5.0 \text{ cm} \).
02

Use Mirror Equation for Location of Image

The mirror equation is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) where \( d_o \) is the object distance and \( d_i \) is the image distance.
03

Apply Mirror Equation for (a)

For part (a), \( d_o = 15.0 \text{ cm} \). Plug into the mirror equation: \( \frac{1}{5.0} = \frac{1}{15.0} + \frac{1}{d_i} \). Solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{5.0} - \frac{1}{15.0} = \frac{2}{15} \), so \( d_i = 7.5 \text{ cm} \).
04

Calculate Magnification for (a)

The magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). For \( d_i = 7.5 \text{ cm} \) and \( d_o = 15.0 \text{ cm} \), \( m = -\frac{7.5}{15.0} = -0.5 \). Image height is \( m \times 8.0 \text{ mm} = -4.0 \text{ mm} \).
05

Apply Mirror Equation for (b)

For part (b), \( d_o = 10.0 \text{ cm} \). Plug into the mirror equation: \( \frac{1}{5.0} = \frac{1}{10.0} + \frac{1}{d_i} \). Solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{5.0} - \frac{1}{10.0} = \frac{1}{10} \), so \( d_i = 10.0 \text{ cm} \).
06

Calculate Magnification for (b)

The magnification is \( m = -\frac{d_i}{d_o} = -\frac{10.0}{10.0} = -1 \). Image height is \( -1 \times 8.0 \text{ mm} = -8.0 \text{ mm} \).
07

Apply Mirror Equation for (c)

For part (c), \( d_o = 2.50 \text{ cm} \). Plug into the mirror equation: \( \frac{1}{5.0} = \frac{1}{2.50} + \frac{1}{d_i} \). Solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{5.0} - \frac{1}{2.50} = -\frac{2}{5} \), so \( d_i = -2.5 \text{ cm} \).
08

Calculate Magnification for (c)

The magnification is \( m = -\frac{d_i}{d_o} = -\frac{-2.5}{2.5} = 1 \). Image height is \( 1 \times 8.0 \text{ mm} = 8.0 \text{ mm} \).
09

Apply Mirror Equation for (d)

For part (d), \( d_o = 1000 \text{ cm} \). Plug into the mirror equation: \( \frac{1}{5.0} = \frac{1}{1000} + \frac{1}{d_i} \). Solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{5.0} - \frac{1}{1000} \approx \frac{1}{5.0} \), so \( d_i \approx 5.0 \text{ cm} \).
10

Calculate Magnification for (d)

The magnification is \( m = -\frac{d_i}{d_o} \approx -\frac{5.0}{1000} = -0.005 \). Image height is \( -0.005 \times 8.0 \text{ mm} = -0.04 \text{ mm} \).

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 Determination
The focal length of a concave mirror is an essential concept to grasp when working with optical systems. It is the distance from the mirror to the focal point, where parallel rays of light converge after reflection. To determine the focal length ( f ) of a concave mirror, use the formula:
  • \( f = \frac{R}{2} \)
  • where \( R \) is the radius of curvature.
In this exercise, we are given a radius of curvature of \( R = 10.0 \) cm. By applying the formula, we find that the focal length is 5.0 cm. Understanding and finding the focal length is the first critical step when dealing with mirrors as it influences how you calculate other properties like image formation and magnification.
Mirror Equation
The mirror equation is a powerful tool used to relate the object distance ( d_o ), image distance ( d_i ), and focal length ( f ) of a mirror. This equation is formulated as:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
  • where \( f \) is the focal length, \( d_o \) is the distance from the object to the mirror, and \( d_i \) is the distance from the image to the mirror.
By rearranging this equation, you can solve for any one of these variables if the other two are known. In practice, the mirror equation enables you to predict where the image will be formed based on the object's position relative to the mirror. It's crucial for understanding how images are constructed in concave mirror systems.
Image Magnification
Image magnification is the ratio of the height of the image to the height of the object. This concept is essential in understanding the scale and orientation of the image produced by a mirror. The magnification ( m ) can be calculated using the formula:
  • \( m = -\frac{d_i}{d_o} \)
  • where \( d_i \) is the image distance and \( d_o \) is the object distance.
The negative sign indicates that the image is inverted relative to the object. If magnification is less than one, the image is smaller than the object. If more than one, the image is larger. This calculation allows you to not only determine the size but also the orientation of the image produced by a concave mirror.
Radius of Curvature
The radius of curvature ( R ) is the radius of the sphere from which the mirror is a segment. It's a crucial characteristic that influences the mirror's overall properties, especially in defining its focal length. The radius of curvature is directly related to the focal length by the formula:
  • \( R = 2f \)
  • where \( f \) is the focal length.
In practical terms, the radius of curvature helps to identify how severely the mirror will curve. A smaller radius means a more sharply curved mirror, leading to different image formation characteristics compared to a mirror with a larger radius. Understanding the radius of curvature is fundamental for analyzing and predicting the behavior of a concave mirror in optical systems.

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

If you place a concave glass lens into a tank of a liquid that has an index of refraction that is greater than that of the lens, what will happen? A. The lens will no longer be able to create any images. B. The focal length of the lens will become longer. C. The focal length of the lens will become shorter. D. The lens will become a converoing lens

\(\cdot\) The front, convex, surface of a lens made for eyeglasses has a radius of curvature of \(11.8 \mathrm{cm},\) and the back, concave, surface has a radius of curvature of 6.80 \(\mathrm{cm} .\) The index of refraction of the plastic lens material is 1.67 . Calculate the local length of the lens.

\(\cdot\) An insect 3.75 \(\mathrm{mm}\) tall is placed 22.5 \(\mathrm{cm}\) to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude \(13.0 \mathrm{cm},\) and the index of refraction of the lens material is 1.70 . (a) Calcu- late the location and size of the image this lens forms of the insect. Is it real or virtual? erect or inverted? (b) Repeat part (a) if the lens is reversed.

A spherical fishbowl. A small tropical fish is at the center of a water-filled spherical fishbowl \(28.0 \mathrm{~cm}\) in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

(. A converging meniscus lens (see Fig. 24.31\()\) with a refrac- tive index of 1.52 has spherical surfaces whose radii are 7.00 \(\mathrm{cm}\) and 4.00 \(\mathrm{cm} .\) What is the position of the image if an object is placed 24.0 \(\mathrm{cm}\) to the left of the lens? What is the magnification?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Magnetism

Read Explanation

Quantum Physics

Read Explanation

Particle Model of Matter

Read Explanation

Electrostatics

Read Explanation

Conservation of Energy and 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.