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Chapter 32: Problem 27

Convex mirrors are often used in side view mirrors on cars. Many such mirrors display the warning "Objects in mirror are closer than they appear." Assume a convex mirror has a radius of curvature of \(14.0 \mathrm{~m}\) and that there is a car that is \(11.0 \mathrm{~m}\) behind the mirror. For a flat mirror, the image distance would be \(11.0 \mathrm{~m}\) and the magnification would be 1\. Find the image distance and magnification for this mirror.

Short Answer

Expert verified
Answer: The image distance is approximately 4.29 m, and the magnification is approximately -0.39.

Step by step solution

01

Identify the given information

We are given: 1. Radius of curvature (\(R\)) for the convex mirror: \(14.0 \mathrm{m}\) 2. Object distance (\(o\)): \(11.0 \mathrm{m}\)
02

Find the focal length of the convex mirror

The focal length (\(f\)) of a mirror can be obtained from its radius of curvature (\(R\)) using the following formula: $$ f = | \frac{R}{2} | $$ Since it is a convex mirror, the focal length is positive, so no need for absolute value: $$ f=\frac{14.0}{2}=7 \mathrm{m} $$
03

Apply the mirror formula

The mirror formula is given by: $$ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} $$ Where \(f\) is the focal length, \(o\) is the object distance, and \(i\) is the image distance. Rearrange the formula to find the image distance (\(i\)): $$ \frac{1}{i} = \frac{1}{f} - \frac{1}{o} $$ Substitute the values of \(f\) and \(o\): $$ \frac{1}{i} = \frac{1}{7} - \frac{1}{11} $$
04

Calculate the image distance

Now, solve for \(i\): $$ i = \frac{1}{\frac{1}{7} - \frac{1}{11}} = 4.2941 \mathrm{m} $$ So, the image distance is approximately \(4.29 \mathrm{m}\).
05

Find the magnification of the convex mirror

The magnification (\(M\)) of a mirror can be obtained using the formula: $$ M = - \frac{i}{o} $$ Substitute the values of \(i\) and \(o\): $$ M = - \frac{4.2941}{11} = -0.39037 $$ So the magnification is approximately \(-0.39\).
06

Interpret the magnification

Since the magnification is negative, it indicates that the image is inverted. The magnitude of the magnification being less than 1 indicates that the image is smaller compared to the object which means "Objects in mirror are closer than they appear." In conclusion, the image distance is approximately \(4.29 \mathrm{m}\), and the magnification is approximately \(-0.39\).

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Most popular questions from this chapter

Standing by a pool filled with water, under what condition will you see a reflection of the scenery on the opposite side through total internal reflection of the light from the scenery? a) Your eyes are level with the water. b) You observe the pool at an angle of \(41.8^{\circ}\) c) Under no condition. d) You observe the pool at an angle of \(48.2^{\circ}\)

For specular reflection of a light ray, the angle of incidence a) must be equal to the angle of reflection. b) is always less than the angle of reflection. c) is always greater than the angle of reflection. d) is equal to \(90^{\circ}\) - the angle of reflection. e) may be greater than, less than, or equal to the angle of reflection.

Reflection and refraction, like all classical features of light and other electromagnetic waves, are governed by the Maxwell equations. The Maxwell equations are time-reversal invariant, which means that any solution of the equations reversed in time is also a solution. a) Suppose some configuration of electric charge density \(\rho,\) current density \(\vec{j},\) electric field \(\vec{E},\) and magnetic field \(\vec{B}\) is a solution of the Maxwell equations. What is the corresponding time-reversed solution? b) How, then, do "one-way mirrors" work?

What kinds of images, virtual or real, are formed by a converging mirror when the object is placed a distance away from the mirror that is a) beyond the center of curvature of the mirror, b) between the center of curvature and half the center of curvature, and c) closer than half of the center of curvature.

A \(5.00-\mathrm{cm}\) object is placed \(30.0 \mathrm{~cm}\) away from a convex mirror with a focal length of \(-10.0 \mathrm{~cm}\). Determine the size, orientation, and position of the image.
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