/*! 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 51 The right-side decides to do an ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 51

The right-side decides to do an experiment to determine the focal length of this mirror. He holds a plane mirror next to the rearview mirror and views an object that is \(163 \mathrm{cm}\) away from each mirror. The object appears \(3.20 \mathrm{cm}\) wide in the plane mirror, but only \(1.80 \mathrm{cm}\) wide in the rearview mirror. What is the focal length of the rearview mirror? mirror of Mike's car says that objects in the mirror are closer than they appear. Mike

Short Answer

Expert verified
Answer: The focal length of the rearview mirror is approximately 314.88 cm.

Step by step solution

01

Write down the mirror and magnification formulas.

We have two formulas relating the object distance (p), the image distance (q), and the focal length (f) in a mirror setup: 1. The mirror formula: \(\frac{1}{p} + \frac{1}{q} = \frac{1}{f}\) 2. The magnification formula: \(M = -\frac{q}{p}\)
02

Find the magnification in both mirrors.

For the plane mirror, the magnification is equal to the ratio of the image width to the object width. Therefore, \(M_{plane} = \frac{3.20 \mathrm{cm}}{163 \mathrm{cm}}\) For the rearview mirror, the magnification is equal to the ratio of the image width to the object width. Therefore, \(M_{rearview} = \frac{1.80 \mathrm{cm}}{163 \mathrm{cm}}\)
03

Calculate M (the ratio between the sizes of images).

To find the ratio between the magnifications of both mirrors, divide \(M_{rearview}\) by \(M_{plane}\): \(M = \frac{M_{rearview}}{M_{plane}} = \frac{1.80 \mathrm{cm} / 163 \mathrm{cm}}{3.20 \mathrm{cm} / 163 \mathrm{cm}}\)
04

Simplify the expression for M.

Cancelling out the common factors, we get: \(M = \frac{1.80}{3.20}\)
05

Identify which mirror is convex and which is concave.

Since the rearview mirror shows the objects smaller, it must be a convex mirror (positive M), while the plane mirror has zero magnification.
06

Relate M to the mirror formula for the rearview mirror.

Now we can rewrite the mirror formula in terms of M for the rearview mirror (keep in mind that for convex mirrors, q is negative): \(\frac{1}{p} - \frac{1}{M\cdot p} = \frac{1}{f}\)
07

Find f for the rearview mirror.

Substitute the values we know (p = 163 cm, M = 1.80/3.20) into the formula: \(\frac{1}{163} - \frac{1}{(\frac{1.80}{3.20}) \cdot 163} = \frac{1}{f}\) Now solve the equation for f. First, rearrange to make f the subject: \(f = \frac{1}{\frac{1}{163} - \frac{1}{(\frac{1.80}{3.20}) \cdot 163}}\)
08

Calculate the numerical value of f.

Now, calculate the value of f: \(f = \frac{1}{\frac{1}{163} - \frac{1}{(\frac{1.80}{3.20}) \cdot 163}} = 314.878 \mathrm{cm}\) So, the focal length of the rearview mirror is approximately 314.88 cm.

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!

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

Show that the deviation angle \(\delta\) for a ray striking a thin converging lens at a distance \(d\) from the principal axis is given by \(\delta=d l f .\) Therefore, a ray is bent through an angle \(\delta\) that is proportional to \(d\) and does not depend on the angle of the incident ray (as long as it is paraxial). [Hint: Look at the figure and use the small-angle approximation \(\sin \theta=\tan \theta \approx \theta(\text { in radians) } .]\)
What is the index of refraction of the core of an optical fiber if the cladding has \(n=1.20\) and the critical angle at the core-cladding boundary is \(45.0^{\circ} ?\)
Apply Huygens's principle to a 5 -cm-long planar wavefront approaching a reflecting wall at normal incidence. The wavelength is \(1 \mathrm{cm}\) and the wall has a wide opening (width \(=4 \mathrm{cm}\) ). The center of the incoming wavefront approaches the center of the opening. Repeat the procedure until you have wavefronts on both sides of the wall. Without worrying about the details of edge effects, what are the general shapes of the wavefronts on each side of the reflecting wall? Repeat Problem 2 for an opening of width $0.5 \mathrm{cm} .$

A glass prism bends a ray of blue light more than a ray of red light since its index of refraction is slightly higher for blue than for red. Does a diverging glass lens have the same focal point for blue light and for red light? If not, for which color is the focal point closer to the lens?
An object is placed \(10.0 \mathrm{cm}\) in front of a lens. An upright, virtual image is formed \(30.0 \mathrm{cm}\) away from the lens. What is the focal length of the lens? Is the lens converging or diverging?
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Fields in Physics

Read Explanation

Famous Physicists

Read Explanation

Solid State Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Thermodynamics

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.