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Chapter 34: Problem 8

An object is 18.0 cm from the center of a spherical silvered-glass Christmas tree ornament 6.00 cm in diameter. What are the position and magnification of its image?

Short Answer

Expert verified
The image position is -1.38 cm, and the magnification is 0.077.

Step by step solution

01

Recognize the Problem Type

The problem involves determining the image position and magnification for an object and a spherical mirror (silvered-glass ornament). Knowing the properties of a spherical mirror, we identify that the given ornament behaves as a convex mirror.
02

Determine the Mirror's Focal Length and Radius of Curvature

Since the ornament is a sphere, the radius of curvature \( R \) can be found using its diameter. Given the diameter is 6.00 cm, the radius is \( R = \frac{6.00}{2} = 3.00 \) cm. The focal length \( f \) of a spherical mirror is \( \frac{R}{2} \). For a convex mirror, \( f = -\frac{R}{2} = -1.50 \) cm.
03

Apply the Mirror Equation

Use the mirror equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f = -1.50 \) cm, \( d_o = 18.0 \) cm, and \( d_i \) is the image distance we need to find. Solve for \( \frac{1}{d_i} \): \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{-1.50} - \frac{1}{18.0} \]
04

Calculate the Image Distance

Compute \( \frac{1}{d_i} \):\[ \frac{1}{d_i} = -\frac{2}{3} - \frac{1}{18} = -\frac{12}{18} - \frac{1}{18} = -\frac{13}{18} \]Thus, \( d_i = -\frac{18}{13} \approx -1.38 \) cm. The negative sign indicates the image is virtual and located behind the mirror.
05

Calculate the Magnification

The magnification \( m \) is given by \( m = \frac{-d_i}{d_o} \). Substitute \( d_i = -1.38 \) cm and \( d_o = 18.0 \) cm to get:\[ m = \frac{-(-1.38)}{18.0} = \frac{1.38}{18.0} \approx 0.077 \]The positive magnification indicates the image is upright.

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Key Concepts

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

Convex Mirror
A convex mirror is a type of spherical mirror where the reflecting surface is curved outward. This shape causes reflected light to spread out, which is why convex mirrors are known as diverging mirrors. Convex mirrors create images that have some unique properties which are very useful in everyday life.
  • They always form virtual images. This means the reflected light appears to originate from a point behind the mirror.
  • The images are diminished, meaning they are smaller than the actual object.
  • These images are typically upright.
Convex mirrors are commonly used in places where a wide field of view is needed, like in vehicle side mirrors and security mirrors. Their ability to give a wider perspective is due to the way they reflect light.
Mirror Equation
The mirror equation is a fundamental formula used to relate the object distance (\(d_o\)), the image distance (\(d_i\)), and the focal length (\(f\)) of a spherical mirror. The equation is:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This equation is crucial as it helps determine where the image will form relative to the mirror. For convex mirrors, the focal length is always negative because the focus is behind the mirror.
  • The equation provides a systematic way to find unknown values, like the image distance when the object distance and focal length are known.
  • Using this equation, students can predict whether an image is real or virtual based on the signs of the image distance.
Mastering the usage of the mirror equation is essential for anyone studying optics, as it builds the foundation for understanding how mirrors and lenses work.
Virtual Image
A virtual image, unlike a real image, cannot be projected onto a screen. This is because the light rays that form a virtual image do not actually meet; they only appear to diverge from a common point behind the mirror. In the case of convex mirrors, all the images are virtual.
  • Virtual images are always the same orientation as the object 鈥� they are upright.
  • The size of a virtual image in a convex mirror is always smaller than the object.
  • These images appear to be located behind the mirror surface.
Understanding virtual images is important for applications in various optical devices. Since these images are non-projectable, they are often used in applications where direct sight or reflection is the goal.
Magnification Calculation
Magnification is a measure of how much larger or smaller the image is compared to the object itself. The formula to determine magnification (\(m\)) in mirrors is:\[ m = \frac{-d_i}{d_o} \]Where \(-d_i\) is the image distance, and \(d_o\) is the object distance.
  • A positive magnification indicates the image is upright compared to the object.
  • A magnification of less than 1 means the image is smaller than the object, which is always true for convex mirrors.
  • Conversely, a negative magnification would indicate an inverted image, although not relevant for convex mirrors since their images are upright.
Through magnification calculations, students can quantify how convex mirrors affect the size of images, which is instrumental in designing optical systems like cameras or security devices.

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Most popular questions from this chapter

A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 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 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?

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?

A converging lens forms an image of an 8.00-mm-tall real object. The image is 12.0 cm to the left of the lens, 3.40 cm tall, and erect. What is the focal length of the lens? Where is the object located?

A camera with a 90-mm-focal-length lens is focused on an object 1.30 m from the lens. To refocus on an object 6.50 m from the lens, by how much must the distance between the lens and the sensor be changed? To refocus on the more distant object, is the lens moved toward or away from the sensor?

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?
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