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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.
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.
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.

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Most popular questions from this chapter

To determine whether a frog can judge distance by means of the amount its lens must move to focus on an object, researchers covered one eye with an opaque material. An insect was placed in front of the frog, and the distance that the frog snapped its tongue out to catch the insect was measured with high-speed video. The experiment was repeated with a contact lens over the eye to determine whether the frog could correctly judge the distance under these conditions. If such an experiment is performed twice, once with a lens of power -9-D and once with a lens of power -15-D, in which case does the frog have to focus at a shorter distance, and why? (a) With the -9-D lens; because the lenses are diverging, the lens with the longer focal length creates an image that is closer to the frog. (b) With the -15-D lens; because the lenses are diverging, the lens with the shorter focal length creates an image that is closer to the frog. (c) With the -9-D lens; because the lenses are converging, the lens with the longer focal length creates a larger real image. (d) With the -15-D lens; because the lenses are converging, the lens with the shorter focal length creates a larger real image.

A thin lens with a focal length of 6.00 cm is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 cm from the eye, and that the lens is very close to the eye.

A candle 4.85 cm tall is 39.2 cm to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 cm in front of her eye? (b) If the object is 1.00 mm high, what is the height of its image formed by the magnifier?

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