Glencoe Physics: Principles and Problems, Student Edition
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN: 9780078807213
Author: Paul W. Zitzewitz
Publisher: Glencoe/McGraw-Hill
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Chapter 17, Problem 40A

(a)

To determine

The angle at which the light ray is reflected from the other mirror.

(a)

Expert Solution
Check Mark

Answer to Problem 40A

The angle at which the light ray is reflected from the other mirror is 60° .

Explanation of Solution

Given:

The given diagram is shown below:

  Glencoe Physics: Principles and Problems, Student Edition, Chapter 17, Problem 40A , additional homework tip  1

The angle between the two mirror is 90°.

The angle of incident is 30°.

Calculation:

According to the law of reflection, the angle made by the reflected ray measured from the normal to a reflecting surface is equal to the angle made by the incident ray measured from the same normal.

Using this information, the sketch of incident and reflected rays of the two mirrors is shown below:

  Glencoe Physics: Principles and Problems, Student Edition, Chapter 17, Problem 40A , additional homework tip  2

Now, let’s find the angle of incidence of the second mirror. In the figure, one can easily see that the reflected ray of the first mirror is the incident ray of the other.

Consider the quadrilateral formed by the sides of the mirrors and the two normal lines. The angle made by the two normal lines will be equal to 90° .

  Glencoe Physics: Principles and Problems, Student Edition, Chapter 17, Problem 40A , additional homework tip  3

Now, consider the triangle that is formed by two normal lines and the reflected ray of the first mirror. Since the sum of angle of the triangle is 180° , we can say that the angle of the incident ray of the second mirror with the normal is 60° . By the law of reflection, the reflected ray will also make an angle of 60° with its normal.

Conclusion:

Hence, the angle at which the light ray is reflected from the other mirror is 60° .

(b)

To determine

To draw: The diagram in which the reflected light rays to show that this mirror system acts as a retro reflector.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The given diagram is shown below:

  Glencoe Physics: Principles and Problems, Student Edition, Chapter 17, Problem 40A , additional homework tip  4

The angle between the two mirror is 90°.

The angle of incident is 30°.

Calculation:

In a retro reflector, the reflected ray follows the same direction and the path as the incident ray. Put two arrows each in the opposite on the same ray to show the incident and the reflected rays in the mirror system that acts as the retro reflector.

The diagram is shown below:

  Glencoe Physics: Principles and Problems, Student Edition, Chapter 17, Problem 40A , additional homework tip  5

Chapter 17 Solutions

Glencoe Physics: Principles and Problems, Student Edition

Ch. 17.1 - Prob. 11SSCCh. 17.1 - Prob. 12SSCCh. 17.2 - Prob. 13PPCh. 17.2 - Prob. 14PPCh. 17.2 - Prob. 15PPCh. 17.2 - Prob. 16PPCh. 17.2 - Prob. 17PPCh. 17.2 - Prob. 18PPCh. 17.2 - Prob. 19PPCh. 17.2 - Prob. 20PPCh. 17.2 - Prob. 21PPCh. 17.2 - Prob. 22PPCh. 17.2 - Prob. 23SSCCh. 17.2 - Prob. 24SSCCh. 17.2 - Prob. 25SSCCh. 17.2 - Prob. 26SSCCh. 17.2 - Prob. 27SSCCh. 17.2 - Prob. 28SSCCh. 17.2 - Prob. 29SSCCh. 17.2 - Prob. 30SSCCh. 17 - Prob. 31ACh. 17 - Prob. 32ACh. 17 - Prob. 33ACh. 17 - Prob. 34ACh. 17 - Prob. 35ACh. 17 - Prob. 36ACh. 17 - Prob. 37ACh. 17 - Prob. 38ACh. 17 - Prob. 39ACh. 17 - Prob. 40ACh. 17 - Prob. 41ACh. 17 - Prob. 42ACh. 17 - Prob. 43ACh. 17 - Prob. 44ACh. 17 - Prob. 45ACh. 17 - Prob. 46ACh. 17 - Prob. 47ACh. 17 - Prob. 48ACh. 17 - Prob. 49ACh. 17 - Prob. 50ACh. 17 - Prob. 51ACh. 17 - Prob. 52ACh. 17 - Prob. 53ACh. 17 - Prob. 54ACh. 17 - Prob. 55ACh. 17 - Prob. 56ACh. 17 - Prob. 57ACh. 17 - Prob. 58ACh. 17 - Prob. 59ACh. 17 - Prob. 60ACh. 17 - Prob. 61ACh. 17 - Prob. 62ACh. 17 - Prob. 63ACh. 17 - Prob. 64ACh. 17 - Prob. 65ACh. 17 - Prob. 66ACh. 17 - Prob. 67ACh. 17 - Prob. 68ACh. 17 - Prob. 69ACh. 17 - Prob. 70ACh. 17 - Prob. 71ACh. 17 - Prob. 72ACh. 17 - Prob. 73ACh. 17 - Prob. 74ACh. 17 - Prob. 75ACh. 17 - Prob. 76ACh. 17 - Prob. 77ACh. 17 - Prob. 78ACh. 17 - Prob. 79ACh. 17 - Prob. 80ACh. 17 - Prob. 81ACh. 17 - Prob. 82ACh. 17 - Prob. 83ACh. 17 - Prob. 84ACh. 17 - Prob. 85ACh. 17 - Prob. 86ACh. 17 - Prob. 87ACh. 17 - Prob. 88ACh. 17 - Prob. 89ACh. 17 - Prob. 90ACh. 17 - Prob. 91ACh. 17 - Prob. 92ACh. 17 - Prob. 93ACh. 17 - Prob. 94ACh. 17 - Prob. 95ACh. 17 - Prob. 96ACh. 17 - Prob. 97ACh. 17 - Prob. 98ACh. 17 - Prob. 99ACh. 17 - Prob. 100ACh. 17 - Prob. 101ACh. 17 - Prob. 102ACh. 17 - Prob. 103ACh. 17 - Prob. 104ACh. 17 - Prob. 1STPCh. 17 - Prob. 2STPCh. 17 - Prob. 3STPCh. 17 - Prob. 4STPCh. 17 - Prob. 5STPCh. 17 - Prob. 6STPCh. 17 - Prob. 7STPCh. 17 - Prob. 8STPCh. 17 - Prob. 9STPCh. 17 - Prob. 10STP

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