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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

A pencil that is 9.0 cm long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead 12.0 cm from the mirror surface and the end of the eraser 21.0 cm from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?

A layer of benzene (\(n\) = 1.502) that is 4.20 cm deep floats on water (\(n\) = 1.332) that is 5.70 cm deep. What is the apparent distance from the upper benzene surface to the bottom of the water when you view these layers at normal incidence?

(a) For a lens with focal length f, find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)

An object is placed 22.0 cm from a screen. (a) At what two points between object and screen may a converging lens with a 3.00-cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?

The cornea behaves as a thin lens of focal length approximately 1.8 cm, although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front surface is convex, with a radius of curvature of 5.0 mm. (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b) The closest distance at which a typical person can focus on an object (called the near point) is about 25 cm, although this varies considerably with age. Where would the cornea focus the image of an 8.0-mm-tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (\(Note:\) The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)

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