/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 132 A millipede sits \(1.0 \mathrm{~... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 132

A millipede sits \(1.0 \mathrm{~m}\) in front of the nearest part of the surface of a shiny sphere of diameter \(0.70 \mathrm{~m}\). (a) How far from the surface does the millipede's image appear? (b) If the millipede's height is \(2.0\) \(\mathrm{mm}\), what is the image height? (c) Is the image inverted?

Short Answer

Expert verified
(a) Image appears 0.138 m in front of the surface inside the sphere. (b) Image height is 0.424 mm. (c) The image is inverted.

Step by step solution

01

Identify the problem type

This problem involves reflection in a concave spherical mirror. We need to find image distance, image height, and determine if the image is inverted.
02

Calculate the radius of curvature

The diameter of the sphere is given as 0.70 m, so the radius of curvature \( R \) is half of the diameter: \( R = \frac{0.70}{2} = 0.35 \) m.
03

Use the mirror formula

The mirror formula is \( \frac{1}{f} = \frac{1}{R} \), which gives the focal length \( f \). With \( R = 0.35 \) m, the focal length is \( f = \frac{0.35}{2} = 0.175 \) m.
04

Apply the mirror equation

Using the mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance (1.0 m) and \( d_i \) is the image distance, we solve for \( d_i \). \[ \frac{1}{0.175} = \frac{1}{1.0} + \frac{1}{d_i} \] \[ \frac{1}{d_i} = \frac{1}{0.175} - \frac{1}{1.0} = 5.71 - 1 = 4.71 \] \( d_i = \frac{1}{4.71} \approx 0.212 \) m. Since the sign of \( d_i \) is positive, the image appears 0.212 m away from the mirror.
05

Calculate the image distance from the surface

The image distance from the mirror is \( d_i \), which is from the center. To find the distance from the surface: distance from surface = \( d_i - R \). So, \( 0.212 - 0.35 = -0.138 \) m, meaning the image is 0.138 m in front of the surface inside the sphere.
06

Calculate the magnification

The magnification \( m \) is given by \( m = \frac{h_i}{h_o} = \frac{d_i}{d_o} \), where \( h_i \) is the image height and \( h_o \) is the object height (2.0 mm or 0.002 m). \[ m = \frac{0.212}{1.0} = 0.212 \] Thus, \( h_i = m \times h_o = 0.212 \times 0.002 = 0.000424 \) m (0.424 mm).
07

Determine if the image is inverted

In concave mirrors, a real image for objects beyond the focal point is inverted. Since the image formed is real and in front of the mirror surface, it is inverted.

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 91影视!

Key Concepts

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

Understanding Concave Mirrors
Concave mirrors are spherical mirrors with a reflecting surface curving inward, resembling a portion of the interior of a sphere. These mirrors gather incoming light and converge it at a particular point, known as the focal point. Concave mirrors are widely used in applications that require focusing light, such as in telescopes and headlights.

Key features of concave mirrors include:
  • The principal axis: a straight line passing through the center of curvature and the midpoint of the mirror.
  • The focal point: where reflected light converges.
  • The radius of curvature (R): the radius of the sphere from which the mirror segment is derived.
Understanding these basic components helps in determining how an image is formed and where it appears.
The Mirror Formula Explained
The mirror formula is a crucial equation used to relate the object distance, image distance, and the focal length of a spherical mirror. It is expressed as: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]Where:
  • \(f\) is the focal length of the mirror.
  • \(d_o\) is the distance from the object to the mirror.
  • \(d_i\) is the distance from the image to the mirror.
This equation helps us find unknown values when two parameters are known. To solve for the image distance, rearrange the formula and plug in the known values.

Using the mirror formula in exercises helps solve for image characteristics like position and size, making it invaluable for studying optics.
Calculating Image Distance
To find out how far an image appears from the surface of the concave mirror, we use the mirror formula in combination with other known measurements. For instance, a common problem may provide you with the object distance (\(d_o\)) and the focal length (\(f\)) and ask for the image distance (\(d_i\)).

The process involves these steps:
  • Calculate the focal length using the given radius of curvature: \(f = \frac{R}{2}\).
  • Rearrange the mirror formula to solve for \(d_i\): \(\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\).
  • Solve the equation for \(d_i\) to find the distance from the mirror.
  • Subtract \(R\) from the result to find the distance from the surface of the mirror.
This method helps determine where the image "forms" relative to the mirror, which is crucial for understanding reflection and image properties.
Image Height Calculation
The height of an image formed by a concave mirror can be determined by calculating the magnification. Magnification relates the image height to the object height and the image distance to the object distance. The formula for magnification \(m\) is: \[m = \frac{h_i}{h_o} = \frac{d_i}{d_o}\]Where:
  • \(h_i\) is the height of the image.
  • \(h_o\) is the height of the object.
  • \(d_i\) and \(d_o\) are the image and object distances, respectively.
To find the image height \(h_i\), rearrange the formula to: \(h_i = m \times h_o\). By knowing the object's height and the distances involved, this calculation becomes straightforward.

This concept is vital for predicting how large or small the reflected image will appear compared to the actual object, and whether the image is upright or inverted.

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 grasshopper hops to a point on the central axis of a spherical mirror. The absolute magnitude of the mirror's focal length is \(40.0 \mathrm{~cm}\), and the lateral magnification of the image produced by the mirror is \(+0.200 .\) (a) Is the mirror convex or concave? (b) How far from the mirror is the grasshopper?

Two-lens systems. In Fig. \(34-45\), stick figure \(O\) (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to \(O\), which is at object distance \(p_{1}\). Lens 2 is mounted within the farther boxed region, at distance \(d .\) Each problem in Table \(34-9\) refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by \(\mathrm{C}\) for converging and D for diverging; the number after C or \(\mathrm{D}\) is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance \(i_{2}\) for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification \(M\) for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object \(O\) or noninverted (NI), and (e) on the same side of lens 2 as object \(O\) or on the opposite side. \(\begin{array}{lllll}\mathbf{8 5} & +4.0 & \text { C, } 6.0 & 8.0 & \text { D, } 6.0\end{array}\)

Figure \(34-56\) shows a beam expander made with two coaxial converging lenses of focal lengths \(f_{1}\) and \(f_{2}\) and separation \(d=f_{1}+f_{2}\). The device can expand a laser beam while keeping the light rays in the beam parallel to the central axis through the lenses. Suppose a uniform laser beam of width \(W_{i}=2.5 \mathrm{~mm}\) and intensity \(I_{i}=9.0 \mathrm{~kW} / \mathrm{m}^{2}\) enters a beam expander for which \(f_{1}=12.5\) \(\mathrm{cm}\) and \(f_{2}=30.0 \mathrm{~cm}\). What are (a) \(W_{f}\) and (b) \(I_{f}\) of the beam leaving the expander? (c) What value of \(d\) is needed for the beam expander if lens 1 is replaced with a diverging lens of focal length \(f_{1}=-26.0 \mathrm{~cm} ?\)

A concave shaving mirror has a radius of curvature of \(35.0 \mathrm{~cm}\). It is positioned so that the (upright) image of a man's face is \(2.50\) times the size of the face. How far is the mirror from the face?

An eraser of height \(1.0 \mathrm{~cm}\) is placed \(10.0 \mathrm{~cm}\) in front of a two-lens system. Lens 1 (nearer the eraser) has focal length \(f_{1}=\) \(-15 \mathrm{~cm}\), lens 2 has \(f_{2}=12 \mathrm{~cm}\), and the lens separation is \(d=12 \mathrm{~cm} .\) For the image produced by lens 2, what are (a) the image distance \(i_{2}\) (including sign), (b) the image height, (c) the image type (real or virtual), and (d) the image orientation (inverted relative to the eraser or not inverted)?
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Geometrical and Physical Optics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Famous Physicists

Read Explanation

Nuclear Physics

Read Explanation
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.