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Chapter 34: Problem 9

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 cm. Reflection from the surface of the shell forms an image of the 1.5-cm-tall coin that is 6.00 cm behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

Short Answer

Expert verified
The coin is 3.6 cm from the mirror; the image is 2.5 cm tall, upright, and virtual.

Step by step solution

01

Understand the given values and formula needed

We are given the radius of curvature \( R = 18.0 \) cm and the image distance \( i = -6.0 \) cm (since the image is located behind the glass shell, the distance is negative). We need to find the object distance \( o \) and use the mirror equation: \[ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \] where \( f \) is the focal length, \( f = \frac{R}{2} \) for a spherical mirror, which gives \( f = 9.0 \) cm.
02

Substitute known values to find object distance

Substitute \( f = 9.0 \) cm and \( i = -6.0 \) cm into the mirror equation: \[ \frac{1}{9} = \frac{1}{o} + \frac{1}{-6} \]. Solve for \( \frac{1}{o} \): \[ \frac{1}{o} = \frac{1}{9} + \frac{1}{6} \]. Convert to a common denominator to get \( \frac{1}{o} = \frac{2}{18} + \frac{3}{18} = \frac{5}{18} \). Therefore, \( o = \frac{18}{5} = 3.6 \) cm.
03

Calculate the magnification and image size

The magnification \( m \) is given by the ratio \( m = -\frac{i}{o} \). Substitute \( i = -6.0 \) cm and \( o = 3.6 \) cm: \[ m = -\frac{-6}{3.6} = \frac{6}{3.6} = 1.67 \]. The image height is the object height multiplied by the magnification: \( 1.5 \times 1.67 = 2.5 \) cm.
04

Determine the orientation and nature of the image

Since the magnification is positive \( (1.67) \), the image is upright. The image distance is negative, indicating the image is virtual.

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Key Concepts

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

Convex Surfaces
Convex surfaces are outward-curving surfaces found on objects like spherical mirrors or lenses. In the context of spherical mirrors, a convex mirror reflects light outward. Convex mirrors are unique because they always form virtual images, meaning you cannot project these images onto a screen simply because the light rays spread out and appear to originate from a single point behind the mirror.
These types of mirrors give a wide field of view, which is why they are commonly used as rearview mirrors in vehicles. Key characteristics of images formed by convex mirrors include:
  • Images are always smaller than the actual object.
  • Images are virtual, meaning they cannot be projected.
  • Images are upright, appearing the same way up as the object.
All these definitions and properties apply due to the mirror's shape and the way it interacts with light.
Mirror Equation
The mirror equation is a fundamental formula used to determine the relationship between the object distance, image distance, and the focal length of a spherical mirror. The formula is:\[\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\]where:
  • \( f \) is the focal length,
  • \( o \) is the object distance (distance from the object to the mirror),
  • \( i \) is the image distance (distance from the image to the mirror).
For a convex mirror, the focal length \( f \) is half of the radius of curvature, and it's always positive, but the image distance \( i \) is negative because the image forms on the opposite side of the mirror. This is crucial in defining where images form relative to the object and the mirror itself.
Virtual Images
In optics, a virtual image is one that cannot be captured directly on a screen because the light rays do not actually meet but appear to diverge from a common point. Convex mirrors are perfect for creating virtual images due to their shape.

These images are always formed on the side opposite to where the object exists. They appear to originate from a point behind the mirror where the reflected rays seem to diverge. For our specific problem with the coin and the spherical glass shell, the image is virtual because:
  • The image distance is negative.
  • The light rays are diverging.
  • It appears upright and smaller.
This concept is critical when considering mirrors in practical applications like vehicle mirrors, providing a direct visualization that helps in safe driving.
Magnification
Magnification is a measure of how much larger or smaller the image is compared to the object itself. It is determined by the formula:\[m = -\frac{i}{o}\]Where:
  • \( m \) is the magnification,
  • \( i \) is the image distance, and
  • \( o \) is the object distance.
In our convex mirror example, the magnification is calculated to be positive, which indicates that the image is upright relative to the object. If the magnification is greater than 1, as in this exercise where it's 1.67, the image appears larger compared to the object, though it remains virtual and upright. Understanding magnification is essential in applications where image size needs to be more accurately perceived, such as in telescopes and cameras.
Optics
Optics is the branch of physics that studies the behavior and properties of light, including its interactions with matter. It encompasses everything from the design of lenses and mirrors to understanding complex light phenomena. Within the context of convex mirrors and our exercise example, optics helps explain why images form the way they do. It involves:
  • Understanding light reflection and refraction processes.
  • Applying formulas like the mirror equation to practical tasks.
  • Exploring how lenses and mirrors alter visual perception.
Studying optics allows us to create technologies that enhance human vision, develop precise scientific instruments, and design the cameras and microscopes used in various scientific and everyday applications. It's a foundational aspect of understanding how we perceive the world visually.

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Most popular questions from this chapter

An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

A camera lens has a focal length of 180.0 mm and an aperture diameter of 16.36 mm. (a) What is the \(f\)-number of the lens? (b) If the correct exposure of a certain scene is \(1\over 30\)s at \(f/\)11, what is the correct exposure at \(f/\)2.8?

Given that frogs are nearsighted in air, which statement is most likely to be true about their vision in water? (a) They are even more nearsighted; because water has a higher index of refraction than air, a frog's ability to focus light increases in water. (b) They are less nearsighted, because the cornea is less effective at refracting light in water than in air. (c) Their vision is no different, because only structures that are internal to the eye can affect the eye's ability to focus. (d) The images projected on the retina are no longer inverted, because the eye in water functions as a diverging lens rather than a converging lens.

A speck of dirt is embedded 3.50 cm below the surface of a sheet of ice \((n = 1.3092)\). What is its apparent depth when viewed at normal incidence?

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_1\) = +12.0 cm and \(R_2\) = +28.0 cm. The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first.
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