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Chapter 34: Problem 127

A concave mirror has a radius of curvature of \(24 \mathrm{~cm}\). How far is an object from the mirror if the image formed is (a) virtual and \(3.0\) times the size of the object, (b) real and \(3.0\) times the size of the object, and (c) real and \(1 / 3\) the size of the object?

Short Answer

Expert verified
(a) 8 cm, (b) 16 cm, (c) 4 cm.

Step by step solution

01

Understand the Mirror Formula

The mirror formula relates the object distance \( u \), the image distance \( v \), and the focal length \( f \) of a mirror as follows: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \).
02

Calculate the Focal Length

For a concave mirror, the focal length \( f \) is half of the radius of curvature. Given a radius of curvature of \( 24 \mathrm{~cm} \), the focal length is \( f = \frac{24}{2} = 12 \mathrm{~cm} \).
03

Consider Magnification for (a) Virtual Image

The magnification \( m \) is given by \( m = \frac{v}{u} \). For a virtual image that is 3 times the size of the object, \( m = -3 \). Use this to find the object distance \( u \).
04

Calculate Object Distance for (a)

With \( m = -3 \), we have \( v = -3u \). Substitute into the mirror formula: \( \frac{1}{12} = \frac{1}{u} + \frac{1}{-3u} \), solve for \( u \) resulting in \( u = 8 \mathrm{~cm} \).
05

Consider Magnification for (b) Real Image

For a real image that is 3 times the size of the object, \( m = 3 \). Hence, \( v = 3u \). Use this in the mirror formula.
06

Calculate Object Distance for (b)

Substitute \( v = 3u \) into the mirror formula: \( \frac{1}{12} = \frac{1}{u} + \frac{1}{3u} \), solving for \( u \) results in \( u = 16 \mathrm{~cm} \).
07

Consider Magnification for (c) Real Image

For a real image that is \( \frac{1}{3} \) the size of the object, \( m = \frac{1}{3} \). Hence, \( v = \frac{u}{3} \). Use this in the mirror formula.
08

Calculate Object Distance for (c)

Substitute \( v = \frac{u}{3} \) into the mirror formula: \( \frac{1}{12} = \frac{1}{u} + \frac{3}{u} \), solving for \( u \) results in \( u = 4 \mathrm{~cm} \).

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Key Concepts

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

Mirror Formula
The mirror formula is a key concept in understanding how an object and its image relate to a mirror's properties. This formula is expressed mathematically as \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), where:
  • \( f \) is the focal length of the mirror.
  • \( u \) is the object distance, i.e., how far the object is from the mirror.
  • \( v \) is the image distance, or how far the image appears from the mirror.
Using this equation allows us to determine one quantity if two are known. For example, knowing the focal length and object distance, we can calculate the image distance. The signs of \( u \) and \( v \) are important, based on the convention that distances are positive if they are in front of the mirror and negative if behind the mirror.
Focal Length
The focal length of a mirror, symbolized by \( f \), is the distance from the mirror to the focal point, where light converges or appears to diverge. For a concave mirror, this point is in front of the mirror. The focal length is directly influenced by the radius of curvature \( R \), which is the radius of the sphere of which the mirror forms a part. It is exactly half the radius:
  • \( f = \frac{R}{2} \)
For instance, if the radius of curvature is \( 24 \, \text{cm} \), then \( f = \frac{24}{2} = 12 \, \text{cm} \). Knowing this helps us apply the mirror formula effectively to solve problems related to mirror imaging.
Magnification
Magnification gives us an idea of how much larger or smaller the image is compared to the original object. It is represented by \( m \), and can be calculated with the formula \( m = \frac{v}{u} \), where both \( v \) and \( u \) relate to their respective distances.
  • If \( m \) is positive, the image is upright and virtual.
  • If \( m \) is negative, the image is inverted and real.
In the problem, if a virtual image is three times as large as the object, then \( m = -3 \), indicating image size and orientation. For real images of different sizes, use \( m = 3 \) for larger and \( m = \frac{1}{3} \) for smaller real images.
Radius of Curvature
The radius of curvature \( R \) of a mirror is vital to understanding its reflecting properties. It represents the radius of the entire spherical surface of which the mirror is a segment.
  • A small radius indicates a tightly curved surface, leading to a shorter focal length.
  • A larger radius means a less curved mirror, leading to a longer focal length.
As such, \( R \) directly influences the focal length by the relationship \( f = \frac{R}{2} \). Knowing \( R \) allows calculations regarding how light will reflect and focus, essential for determining distances in mirror equations.

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Most popular questions from this chapter

A cheese enchilada is \(4.00 \mathrm{~cm}\) in front of a converging lens. The magnification of the enchilada is \(-2.00 .\) What is the focal length of the lens?

Spherical refracting surfaces. An object \(O\) stands on the central axis of a spherical refracting surface. For this situation, each problem in Table \(34-5\) refers to the index of refraction \(n_{1}\) where the object is located, (a) the index of refraction \(n_{2}\) on the other side of the refracting surface, (b) the object distance \(p,(\mathrm{c})\) the radius of curvature \(r\) of the surface, and (d) the image distance \(i\). (All distances are in centimeters.) Fill in the missing information, including whether the image is (e) real (R) or virtual (V) and (f) on the same side of the surface as object \(O\) or on the opposite side. \(\begin{array}{lllll}7 & 1.5 & 1.0 & +10 & -6.0\end{array}\)

One end of a long glass rod \((n=1.5)\) is a convex surface of radius \(6.0 \mathrm{~cm}\). An object is located in air along the axis of the rod, at a distance of \(10 \mathrm{~cm}\) from the convex end. (a) How far apart are the object and the image formed by the glass rod? (b) Within what range of distances from the end of the rod must the object be located in order to produce a virtual image?

An eraser of height \(1.0 \mathrm{~cm}\) is placed \(10.0 \mathrm{~cm}\) in front of a two-lens system. Lens 1 (nearer the eraser) has focal length \(f_{1}=\) \(-15 \mathrm{~cm}\), lens 2 has \(f_{2}=12 \mathrm{~cm}\), and the lens separation is \(d=12 \mathrm{~cm} .\) For the image produced by lens 2, what are (a) the image distance \(i_{2}\) (including sign), (b) the image height, (c) the image type (real or virtual), and (d) the image orientation (inverted relative to the eraser or not inverted)?

A glass sphere has radius \(R=5.0 \mathrm{~cm}\) and index of refraction 1.6. A paperweight is constructed by slicing through the sphere along a plane that is \(2.0 \mathrm{~cm}\) from the center of the sphere, leaving height \(h=3.0 \mathrm{~cm}\). The paperweight is placed on a table and viewed from directly above by an observer who is distance \(d=8.0 \mathrm{~cm}\) from the tabletop (Fig. 34-39). When viewed through the paperweight, how far away does the tabletop appear to be to the observer?
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