/*! 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 31 A short straight object of lengt... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 35: Problem 31

A short straight object of length \(L\) lies along the central axis of a spherical mirror of focal length \(f\), a distance \(o\) from the mirror. (a) Show that its image in the mirror has a length \(L^{\prime}\) where $$L^{\prime}=L\left(\frac{f}{o-f}\right)^{2}$$ (Hint: Locate the two ends of the object.) (b) Show that the longitudinal magnification \(m^{\prime}\left(=L^{\prime} / L\right)\) is equal to \(m^{2}\), where \(m\) is the lateral magnification.

Short Answer

Expert verified
The image length is \(L^{\prime} = L\left( \frac{f}{o-f} \right)^2\) and longitudinal magnification \(m^{\prime = m^2\)\.

Step by step solution

01

Understanding the Problem

The problem involves a short straight object placed along the central axis of a spherical mirror. The goal is to determine the image length and the longitudinal magnification.
02

Recall the Mirror Equation

The mirror equation relates the object distance (\(o\)), image distance (\(i\)), and focal length (\(f\)): \[ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \]
03

Express Image Distance

Rearrange the mirror equation to express the image distance (\(i\)): \[ i = \frac{of}{o-f} \]
04

Locate Image Positions of Two Ends

Consider the ends of the object located at \(o\) and \(o + L\): \[ i_1 = \frac{of}{o-f} \] and \[ i_2 = \frac{(o + L)f}{o + L - f} \]
05

Approximate for Small Lengths

If \(L\) is small compared to \(o\), approximate the difference between the image distances of the two ends using a binomial expansion: \[ i_2 \approx \frac{of}{o-f} + \frac{Lf^2}{(o-f)^2} \]
06

Calculate the Image Length

The length of the image (\(L^{\prime}\)) is the difference \[(i_2 - i_1)\]: \[ L^{\prime} = \frac{Lf^2}{(o-f)^2} \]
07

Express Longitudinal Magnification

Longitudinal magnification (\(m^{\prime\)) is defined as: \[ m^{\prime} = \frac{L^{\prime}}{L} = \left(\frac{f}{o-f}\right)^2 \]
08

Relate to Lateral Magnification

Recall the lateral magnification (\(m\)): \[ m = -\frac{i}{o} = \frac{f}{o-f} \]. Thus, longitudinal magnification is: \[ m^{\prime} = m^2 \]

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.

spherical mirror
A spherical mirror can be either concave or convex. These mirrors have a surface that forms a part of a sphere. When light rays reflect off these mirrors, they follow specific paths determined by the shape of the mirror. The key parts of a spherical mirror include:

  • Center of Curvature (C): The center of the sphere from which the mirror is a section.
  • Principal Axis: A straight line passing through the center of curvature and the center of the mirror.
  • Focal Point (F): The point where parallel rays of light either converge (concave) or appear to diverge (convex) after reflecting off the mirror.
  • Focal Length (f): The distance between the focal point and the mirror's surface.
Understanding these terms helps in analyzing the behavior of light and the image formation in spherical mirrors.
mirror equation
The mirror equation is crucial for determining the relationship between object distance ( o ), image distance ( i ), and focal length ( f ). The equation is expressed as:

\[ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \]
This means that for any object placed at a distance o from the mirror, there is a corresponding image formed at distance i .
The equation allows us to predict where the image will form if we know the object distance and the focal length. Rearranging the equation to solve for the image distance gives:

\[i = \frac{of}{o-f}\]
So, if we know the focal length and the object distance, we can easily calculate where the image will appear.
lateral magnification
Lateral magnification ( m ) describes how the size of the image compares to the size of the object in a mirror. It tells us how much larger or smaller the image is compared to the actual object.
  • Formula: \[m = \frac{h_i}{h_o} = \frac{-i}{o}\] where h_i is the image height, h_o is the object height, i is the image distance, and o is the object distance.
If the magnification is positive, the image is upright compared to the object. If the magnification is negative, the image is inverted.

In our problem, we found that the longitudinal magnification ( m' ) is related to the lateral magnification by:
  • Formula for longitudinal magnification: \[m' = m^2\]
This shows that as the image gets larger or smaller laterally, its length changes quadratically. This helps in understanding the detailed nature of image formation and precise calculations required for spherical mirrors.

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

You may have observed people with cameras behaving strangely: (a) At a conference in North Carolina, one of the physics graduate students (who should have known better!) tried to take a picture of the overheads projected on a white screen. Since the room was darkened, he used a flash. Explain why this is a bad idea and what his pictures are likely to show. (b) Someone mentioned to the student that he probably should not be using his flash, so he turned it off. He then proceeded to try and take pictures of the participants in the darkened room! Explain why this is a bad idea and what his pictures are likely to show. (c) A woman on an airplane at night with a camera was impressed with the view of the city lights in the dark as the plane flew over Washington, D.C. She stood back in the aisle with her camera and tried to take a picture through the window using her flash. Explain why this is a bad idea and what her pictures are likely to show.

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?

Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of \(32.0^{\circ}\) with the normal to the surface, while in the glass it makes an angle of \(21.0^{\circ}\) with the normal. What is the index of refraction of the glass?

The index of refraction of benzene is \(1.8\). What is the critical angle for a light ray traveling in benzene toward a plane layer of air above the benzene?

Light traveling in water of refractive index \(1.33\) is incident on a plate of glass with index of refraction 1.53. At what angle of incidence is the reflected light fully polarized?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Mechanics and Materials

Read Explanation

Electricity and Magnetism

Read Explanation

Work Energy and Power

Read Explanation

Fields in Physics

Read Explanation

Famous Physicists

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.