Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A concave mirror has a radius of curvature of 34.0 cm. (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33), what is its focal length?

You are in your car driving on a highway at 25 m\(/\)s when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 cm) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 m\(/\)s when the truck is 2.0 m from the mirror, what is the speed of the truck relative to the highway?

The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) 12.0 cm; (c) 2.00 cm.

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.

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns (\(\mu\)m) is typical near the center of the eye. We shall model the eye as a sphere 2.50 cm in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0 \(\mu\)m in diameter. (a) What is the smallest object you can resolve at a near point of 25 cm? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes (1\(^\circ\) = 60 min), and compare it with the typical experimental value of about 1.0 min. (\(Note\): There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free