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Chapter 34: Problem 8

An object is 24.0 \(\mathrm{cm}\) from the center of a silvered spherical glass Christmas tree ornament 6.00 \(\mathrm{cm}\) in diameter. What are the position and magnification of its image?

Short Answer

Expert verified
The image is at 1.60 cm from the mirror and is inverted with a magnification of -0.0667.

Step by step solution

01

Understand the Terms

The problem involves finding the position and magnification of an image through a spherical mirror. The diameter of the ornament is given, allowing us to calculate the radius. The object distance is provided.
02

Calculate the Radius of the Ornament

The diameter of the spherical ornament is 6.00 cm. The radius \( R \) is half of the diameter: \( R = \frac{6.00}{2} = 3.00 \) cm.
03

Determine the Focal Length

For a spherical mirror, the focal length \( f \) is half of the radius of curvature, \( f = \frac{R}{2} \). Since this is a silvered ornament, it acts as a concave mirror: \( f = \frac{3.00}{2} = 1.50 \) cm.
04

Use the Mirror Formula

The mirror formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o = 24.0 \) cm (object distance) and \( f = 1.50 \) cm. Thus, \( \frac{1}{1.50} = \frac{1}{24.0} + \frac{1}{d_i} \).
05

Solve for Image Position \( d_i \)

Rearrange the mirror formula to find \( d_i \): \( \frac{1}{d_i} = \frac{1}{1.50} - \frac{1}{24.0} \). Calculate to find \( d_i \).
06

Perform the Calculations for \( d_i \)

\( \frac{1}{1.50} = 0.6667 \) and \( \frac{1}{24.0} = 0.0417 \). Thus, \( \frac{1}{d_i} = 0.6667 - 0.0417 = 0.625 \). Solve for \( d_i \): \( d_i = \frac{1}{0.625} = 1.60 \) cm.
07

Calculate Magnification

Magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). Substitute \( d_i = 1.60 \) cm and \( d_o = 24.0 \) cm: \( m = -\frac{1.60}{24.0} \).
08

Perform the Calculation for Magnification

Solve the expression for magnification: \( m = -0.0667 \). The negative sign indicates the image is inverted.

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Key Concepts

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

Focal Length
Spherical mirrors come in two types: concave and convex. The focal length is a critical property of these mirrors, which is the distance from the mirror's surface to its focal point.
  • The focal point is where light rays parallel to the mirror's principal axis converge (for concave mirrors) or appear to diverge from (for convex mirrors).
  • To calculate the focal length (\( f \)), you use the radius of curvature (\( R \)) of the mirror.
For spherical mirrors, the focal length is half of the radius of curvature:\[f = \frac{R}{2}\]In the given problem, the radius (\( R \)) is determined by dividing the diameter of the ornament by 2, which results in 3.00 cm. Thus, the focal length is calculated as 1.50 cm, showing how the focal length links directly to the size of the mirror.
Mirror Formula
The mirror formula is an essential equation that relates an object's distance from the mirror, the image's distance from the mirror, and the mirror's focal length. It is expressed as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]Where:
  • \( f \) is the focal length.
  • \( d_o \) is the object distance (distance from the object to the mirror).
  • \( d_i \) is the image distance (distance from the image to the mirror).
Using the mirror formula in the example, we substitute \( f = 1.50 \) cm and \( d_o = 24.0 \) cm. This setup starts the problem-solving process for finding \( d_i \), giving you insight into the mirror's focal length helping decide where the image will form. This formula is a vital tool for determining both the image position and its nature (real or virtual).
Image Magnification
Magnification in mirrors describes how much larger or smaller the image is compared to the actual object. It also indicates the orientation of the image.
  • The magnification (\( m \)) is calculated using the formula:\[m = -\frac{d_i}{d_o}\]
  • The negative sign indicates that if the image is inverted, it will have a negative magnification.
  • A magnification greater than 1 means the image is larger, while less than 1 suggests it is smaller compared to the object.
In this example, the object is 24.0 cm away, and the image is 1.60 cm on the same side due to the concave nature of the mirror. Thus, the magnification becomes \( m = -0.0667 \), showing that the image is significantly smaller than the object and also inverted. This understanding of magnification provides deeper insight into the nature of images produced by spherical mirrors.

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Most popular questions from this chapter

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 \(\mathrm{cm}\) to the right of the lens. A diverging lens is now placed 15.0 \(\mathrm{cm}\) to the right of the converging lens, and it is found that the screen must be moved 19.2 \(\mathrm{cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

A converging lens with a focal length of 7.00 \(\mathrm{cm}\) forms an image of a \(4.00-\mathrm{mm}\)-tall real object that is to the left of the lens. The image is 1.30 \(\mathrm{cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

A person can see clearly up close but cannot focus on objects beyond 75.0 \(\mathrm{cm} .\) She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?

Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 \(\mathrm{mm}\) is 160 \(\mathrm{mm}\) from its second focal point. The eyepiece has a focal length of 26.0 \(\mathrm{mm}\) . (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its nicroscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 \(\mathrm{mm}\) apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

Rear-View Mirror. A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 \(\mathrm{cm} .\) (a) Another car is behind your car, 9.00 m from the mirror, and this car is viewed in the mirror by your passenger. If this car is 1.5 \(\mathrm{m}\) tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?
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