Chapter 34: Problem 11
A spherical, concave shaving mirror has a radius of curvature of 32.0 cm. (a) What is the magnification of a person's face when it is 12.0 cm to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.
Short Answer
Expert verified
The magnification is 4, the image is virtual, upright, and located 48 cm behind the mirror.
Step by step solution
01
Understanding the Problem
We need to find the magnification of a person's face using a concave mirror with a radius of curvature of 32.0 cm. The object (person's face) is 12.0 cm from the mirror. We also need to find the image position and determine if it is real or virtual.
02
Identify the Mirror Formula
The mirror formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. The magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). The focal length \( f \) of a concave mirror is half the radius of curvature \( R \), so \( f = \frac{R}{2} = \frac{32.0}{2} = 16.0 \) cm.
03
Apply the Mirror Formula
Using the mirror formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Plug in \( f = 16.0 \) cm and \( d_o = 12.0 \) cm. This gives \( \frac{1}{16} = \frac{1}{12} + \frac{1}{d_i} \). Rearrange to find \( d_i \): \( \frac{1}{d_i} = \frac{1}{16} - \frac{1}{12} \).
04
Solve for Image Distance \( d_i \)
Calculate \( \frac{1}{d_i} = \frac{3}{48} - \frac{4}{48} = -\frac{1}{48} \). Therefore, \( d_i = -48 \) cm. The negative sign indicates that the image is virtual and located 48 cm to the left of the mirror.
05
Calculate the Magnification
Using the magnification formula \( m = -\frac{d_i}{d_o} = -\frac{-48}{12} = 4 \). This means the image is 4 times the size of the object and is upright.
06
Principal-Ray Diagram Explanation
In a concave mirror, parallel rays converge at the focal point after reflecting. For an object inside the focal length, rays diverge after reflecting, and they appear to originate from a point behind the mirror, forming a virtual, magnified, and upright image.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radius of Curvature
When working with mirrors, particularly spherical ones like concave mirrors, the radius of curvature is a crucial measurement. It represents the radius of the sphere from which the mirror segment is derived. In simple terms, if you could imagine a giant sphere and carved a piece from its surface to create your mirror, the radius of that sphere is what we call the radius of curvature.
For instance, if a concave mirror has a radius of curvature of 32 cm, this indicates that the mirror is part of a sphere with a 32 cm radius. The radius of curvature is pivotal because it helps us determine the focal length (\( f \)). For a concave mirror, the focal length is half of the radius of curvature. Therefore, with a given radius of curvature of 32 cm, the focal length would be: \( f = \frac{32.0}{2} = 16.0 \text{ cm} \).
The significance of understanding and calculating the radius of curvature lies in its direct use in equations that solve for various properties of images formed by mirrors.
For instance, if a concave mirror has a radius of curvature of 32 cm, this indicates that the mirror is part of a sphere with a 32 cm radius. The radius of curvature is pivotal because it helps us determine the focal length (\( f \)). For a concave mirror, the focal length is half of the radius of curvature. Therefore, with a given radius of curvature of 32 cm, the focal length would be: \( f = \frac{32.0}{2} = 16.0 \text{ cm} \).
The significance of understanding and calculating the radius of curvature lies in its direct use in equations that solve for various properties of images formed by mirrors.
Mirror Formula
The mirror formula is a fundamental equation that relates the object distance (\( d_o \)), the image distance (\( d_i \)), and the focal length (\( f \)) of a mirror. It's expressed as:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
For instance, using this formula helps us determine where the image will be located relative to the mirror, as well as if it is real or virtual. Applying this in our exercise with a focal length of 16 cm and an object distance of 12 cm, we substitute these values into the formula to solve for the image distance \( d_i \):
\[ \frac{1}{16} = \frac{1}{12} + \frac{1}{d_i} \]
Rearranging gives:
\[ \frac{1}{d_i} = \frac{1}{16} - \frac{1}{12} \Rightarrow \frac{1}{d_i} = -\frac{1}{48} \]
Therefore, \( d_i = -48 \text{ cm} \). The negative sign indicates a virtual image, signifying that the image appears to be on the same side as the object.
The mirror formula extends its usefulness beyond mere calculations, as it provides insights into the nature and position of images formed by mirrors.
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
For instance, using this formula helps us determine where the image will be located relative to the mirror, as well as if it is real or virtual. Applying this in our exercise with a focal length of 16 cm and an object distance of 12 cm, we substitute these values into the formula to solve for the image distance \( d_i \):
\[ \frac{1}{16} = \frac{1}{12} + \frac{1}{d_i} \]
Rearranging gives:
\[ \frac{1}{d_i} = \frac{1}{16} - \frac{1}{12} \Rightarrow \frac{1}{d_i} = -\frac{1}{48} \]
Therefore, \( d_i = -48 \text{ cm} \). The negative sign indicates a virtual image, signifying that the image appears to be on the same side as the object.
The mirror formula extends its usefulness beyond mere calculations, as it provides insights into the nature and position of images formed by mirrors.
Magnification Formula
Magnification in mirrors tells us how much larger or smaller the image is compared to the object itself. This can be especially interesting with concave mirrors, as they can significantly enlarge images under certain conditions. The magnification formula is expressed as:
\( m = -\frac{d_i}{d_o} \)
In this exercise, with an image distance \( d_i = -48 \text{ cm} \) and an object distance \( d_o = 12 \text{ cm} \), the magnification \( m \) is calculated as:
\( m = -\frac{-48}{12} = 4 \)
This tells us that the image is four times larger than the object (the person's face, in this case) and is upright. The negative sign in the magnification formula is crucial as it indicates the orientation of the image — a positive magnification denotes an upright image while a negative magnification would imply an inverted one.
Understanding the magnification and its calculation helps us compare the actual size of objects with their images, offering practical insights into applications like makeup mirrors and telescopic devices.
\( m = -\frac{d_i}{d_o} \)
In this exercise, with an image distance \( d_i = -48 \text{ cm} \) and an object distance \( d_o = 12 \text{ cm} \), the magnification \( m \) is calculated as:
\( m = -\frac{-48}{12} = 4 \)
This tells us that the image is four times larger than the object (the person's face, in this case) and is upright. The negative sign in the magnification formula is crucial as it indicates the orientation of the image — a positive magnification denotes an upright image while a negative magnification would imply an inverted one.
Understanding the magnification and its calculation helps us compare the actual size of objects with their images, offering practical insights into applications like makeup mirrors and telescopic devices.