Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 37, Problem 19PQ
To determine

The distance between the image of your roommate’s face and you when she stands behind you in such a way that her face is 18in behind your face.

Expert Solution & Answer
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Answer to Problem 19PQ

The distance between the image of your roommate’s face and you is 8.5ft.

Explanation of Solution

Write the expression for magnification of the plane mirror.

    M=did0                                                                                                           (I)

Here, d0 is the object distance, and di is the image distance.

The magnification of the plane mirror will be constant for both person and his roommate.

    Mp=Mf

Here, Mp is the magnification of the plane mirror for the person, and Mf is the magnification of the mirror for his roommate.

Write the mirror equation for the person and his roommate.

    1(d0)p+1(di)p=1(d0)f+1(di)f                                                                      (II)

Write the expression for the distance between the images of the person roommate from the person.

    d=(d0)p(di)f                                                                                    (III)

Here, (d0)p is the object distance for person, and (di)f is the image distance for his roommate, and d is the image distance between roommate and person.

Substitute (di)p(d0)p for Mp, and (di)f(d0)f for Mf in the equation (I).

    (di)p(d0)p=(di)f(d0)f(di)p(d0)p=(di)f(d0)f

Conclusion:

Substitute 3.5ft for (d0)p and 3.5ft+1.8in for (d0)f in the above equation to calculate (di)p.

    (di)p3.5ft=(di)f3.5ft+1.8in×(1ft12in)(di)p=3.5ft(3.5ft+1.5ft)(di)f=(3.5ft5)(di)f

Substitute 3.5ft for (d0)p, (3.5ft5)(di)f for (di)p, and 5ft for (d0)f in the equation (II) to calculate (di)f.

    13.5ft+1(3.5ft5)(di)f=15ft+1(di)f13.5ft+53.5ft×(di)f=15ft+1(di)f(di)f=5ft

Here, the negative sign indicate the distance of the image from the mirror.

Substitute 3.5ft for (d0)p, and 5ft for (di)f in the equation (III) to calculate d.

    d=3.5ft(5ft)=8.5ft

Therefore, the image of the roommate from the person is 8.5ft.

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Chapter 37 Solutions

Physics for Scientists and Engineers: Foundations and Connections

Ch. 37 - Prob. 4PQCh. 37 - Prob. 5PQCh. 37 - Prob. 6PQCh. 37 - Prob. 7PQCh. 37 - Prob. 8PQCh. 37 - Prob. 9PQCh. 37 - Prob. 10PQCh. 37 - Prob. 11PQCh. 37 - Prob. 12PQCh. 37 - Prob. 13PQCh. 37 - Prob. 14PQCh. 37 - Light rays strike a plane mirror at an angle of...Ch. 37 - Prob. 16PQCh. 37 - Prob. 17PQCh. 37 - Prob. 18PQCh. 37 - Prob. 19PQCh. 37 - Prob. 20PQCh. 37 - Prob. 21PQCh. 37 - Prob. 22PQCh. 37 - Prob. 23PQCh. 37 - Prob. 24PQCh. 37 - Prob. 25PQCh. 37 - Prob. 26PQCh. 37 - Prob. 27PQCh. 37 - Prob. 28PQCh. 37 - A convex mirror with a radius of curvature of 25.0...Ch. 37 - The magnitude of the radius of curvature of a...Ch. 37 - Prob. 31PQCh. 37 - The image formed by a convex spherical mirror with...Ch. 37 - An object is placed 25.0 cm from the surface of a...Ch. 37 - Prob. 34PQCh. 37 - Prob. 35PQCh. 37 - Prob. 36PQCh. 37 - Prob. 37PQCh. 37 - Prob. 38PQCh. 37 - Prob. 39PQCh. 37 - Prob. 40PQCh. 37 - Prob. 41PQCh. 37 - Prob. 42PQCh. 37 - Prob. 43PQCh. 37 - Prob. 44PQCh. 37 - Prob. 45PQCh. 37 - Prob. 46PQCh. 37 - Prob. 47PQCh. 37 - Prob. 48PQCh. 37 - Prob. 49PQCh. 37 - Prob. 50PQCh. 37 - Prob. 51PQCh. 37 - Prob. 52PQCh. 37 - Prob. 53PQCh. 37 - Prob. 54PQCh. 37 - Prob. 55PQCh. 37 - Prob. 56PQCh. 37 - You see the image of a sign through a camera...Ch. 37 - Prob. 58PQCh. 37 - Prob. 59PQCh. 37 - Prob. 60PQCh. 37 - An object is placed midway between two concave...Ch. 37 - Prob. 62PQCh. 37 - Prob. 63PQCh. 37 - Prob. 64PQCh. 37 - Prob. 65PQCh. 37 - Prob. 66PQCh. 37 - Observe your reflection in the back of a spoon....Ch. 37 - Prob. 68PQCh. 37 - A small convex mirror and a large concave mirror...Ch. 37 - Prob. 70PQCh. 37 - Prob. 71PQCh. 37 - Prob. 72PQCh. 37 - Prob. 73PQ
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