Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 35, Problem 75AP

(a)

To determine

The angle θ2 of the emerging beam.

(a)

Expert Solution
Check Mark

Answer to Problem 75AP

The angle θ2 of the emerging beam is 53.1°.

Explanation of Solution

Apply Snell’s law for the first layer.

    n1sinθ1=n2sinθ2'                                                                                                      (I)

Here, n1 is the refractive index of the first layer, θ1 is the incident angle, n2 is the refractive index of the second layer and θ2' is the emerging angle for the second layer.

Apply Snell’s law for the second layer.

    n2sinθ2'=n3sinθ3                                                                                                    (II)

Here, n2 is the refractive index of the second layer and θ2' is the incident angle, n3 is the refractive index of the third layer and θ3 is the emerging angle.

Apply Snell’s law for the third layer.

    n3sinθ3=nsinθ2                                                                                                    (III)

Here, n3 is the refractive index of the third layer, θ3 is the incident angle, n is the refractive index of air and θ2 is the emerging angle.

Conclusion:

Substitute 1.60 for n1, 30° for θ1 and 1.40 for n2 in equation (I) to calculate θ2'.

    (1.60)sin(30°)=(1.40)sinθ2'sinθ2'=1.601.40sin30°θ2'=sin1(0.5714)=34.8°

Substitute 1.40 for n2, 34.8° for θ2' and 1.20 for n3 in equation (II)  to calculate θ3.

    (1.40)sin(34.8°)=(1.20)sinθ3sinθ3=1.401.20sin34.8°θ3=sin1(0.6628)=41.74°

Substitute 1.20 for n3, 41.74° for θ3 and 1 for n in equation (III) to calculate θ2.

    (1.20)sin(41.74°)=(1)sinθ2sinθ2=1.201sin(41.74°)θ2=sin1(0.7989)=53.1°

Therefore, the angle θ2 of the emerging beam is 53.1°.

(b)

To determine

The value of incident angle θ1 to have total internal reflection at the surface between the medium with n=1.20 and the medium with n=1.00.

(b)

Expert Solution
Check Mark

Answer to Problem 75AP

The value of incident angle θ1 to have total internal reflection at the surface between the medium with n=1.20 and the medium with n=1.00 should be greater than 38.7°.

Explanation of Solution

Apply Snell’s law for the third layer.

    n3sinθc=nsinθ2                                                                                                    (IV)

Here, θc is the critical angle.

Apply Snell’s law for the second layer.

    n3sinθi=n2sinθr                                                                                                     (V)

Here, θr is the refracted angle.

Apply Snell’s law for the first layer.

    n2sinθi'=n1sinθr'                                                                                                   (VI)

Here, θi' is the incident angle and θr' is the refracted angle.

Conclusion:

Substitute 1 for n, 1.20 for n3 and 90° for θ2 in equation (IV) to calculate θc.

    (1.20)sinθc=(1)sin90°sinθc=11.20sin90°θc=sin1(0.8333)=56.44°

Substitute 1.40 for n3, 1.20 for n2 and 56.44° for θr in equation (V) to calculate θi.

    (1.40)sinθi=(1.20)sin56.44°sinθi=1.201.40sin56.44°θi=sin1(0.7142)=45.58°

Substitute 1.60 for n2 and 1.40 for n1 and 45.68° for θr' in equation (VI) to calculate θi'.

    (1.60)sinθi'=(1.40)sin45.68°sinθi'=1.401.60sin45.68°θi'=sin1(0.626)=38.7°

Therefore, the value of incident angle θ1 to have total internal reflection at the surface between the medium with n=1.20 and the medium with n=1.00 should be greater than 38.7°.

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Chapter 35 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

Ch. 35 - Prob. 6OQCh. 35 - Prob. 7OQCh. 35 - Prob. 8OQCh. 35 - Prob. 9OQCh. 35 - Prob. 10OQCh. 35 - Prob. 11OQCh. 35 - Prob. 12OQCh. 35 - Prob. 13OQCh. 35 - Prob. 14OQCh. 35 - Prob. 15OQCh. 35 - Prob. 1CQCh. 35 - Prob. 2CQCh. 35 - Prob. 3CQCh. 35 - Prob. 4CQCh. 35 - Prob. 5CQCh. 35 - Prob. 6CQCh. 35 - Prob. 7CQCh. 35 - Prob. 8CQCh. 35 - Prob. 9CQCh. 35 - Prob. 10CQCh. 35 - Prob. 11CQCh. 35 - (a) Under what conditions is a mirage formed?...Ch. 35 - Prob. 13CQCh. 35 - Prob. 14CQCh. 35 - Prob. 15CQCh. 35 - Prob. 16CQCh. 35 - Prob. 17CQCh. 35 - Prob. 1PCh. 35 - Prob. 2PCh. 35 - In an experiment to measure the speed of light...Ch. 35 - As a result of his observations, Ole Roemer...Ch. 35 - Prob. 5PCh. 35 - Prob. 6PCh. 35 - Prob. 7PCh. 35 - Prob. 8PCh. 35 - Prob. 9PCh. 35 - Prob. 10PCh. 35 - Prob. 11PCh. 35 - A ray of light strikes a flat block of glass (n =...Ch. 35 - Prob. 13PCh. 35 - Prob. 14PCh. 35 - Prob. 15PCh. 35 - Prob. 16PCh. 35 - Prob. 17PCh. 35 - Prob. 18PCh. 35 - When you look through a window, by what time...Ch. 35 - Two flat, rectangular mirrors, both perpendicular...Ch. 35 - Prob. 21PCh. 35 - Prob. 22PCh. 35 - Prob. 23PCh. 35 - Prob. 24PCh. 35 - Prob. 25PCh. 35 - Prob. 26PCh. 35 - Prob. 27PCh. 35 - Prob. 28PCh. 35 - Prob. 29PCh. 35 - Prob. 30PCh. 35 - Prob. 31PCh. 35 - Prob. 32PCh. 35 - Prob. 33PCh. 35 - A submarine is 300 m horizontally from the shore...Ch. 35 - Prob. 35PCh. 35 - Prob. 36PCh. 35 - Prob. 37PCh. 35 - Prob. 39PCh. 35 - Prob. 40PCh. 35 - Prob. 41PCh. 35 - Prob. 42PCh. 35 - Prob. 43PCh. 35 - Prob. 44PCh. 35 - Assume a transparent rod of diameter d = 2.00 m...Ch. 35 - Consider a light ray traveling between air and a...Ch. 35 - Prob. 47PCh. 35 - Prob. 48PCh. 35 - Prob. 49PCh. 35 - Prob. 50PCh. 35 - Prob. 51APCh. 35 - Prob. 52APCh. 35 - Prob. 53APCh. 35 - Prob. 54APCh. 35 - Prob. 55APCh. 35 - Prob. 56APCh. 35 - Prob. 57APCh. 35 - Prob. 58APCh. 35 - Prob. 59APCh. 35 - A light ray enters the atmosphere of a planet and...Ch. 35 - Prob. 61APCh. 35 - Prob. 62APCh. 35 - Prob. 63APCh. 35 - Prob. 64APCh. 35 - Prob. 65APCh. 35 - Prob. 66APCh. 35 - Prob. 67APCh. 35 - Prob. 68APCh. 35 - Prob. 69APCh. 35 - Prob. 70APCh. 35 - Prob. 71APCh. 35 - Prob. 72APCh. 35 - Prob. 73APCh. 35 - Prob. 74APCh. 35 - Prob. 75APCh. 35 - Prob. 76APCh. 35 - Prob. 77APCh. 35 - Prob. 78APCh. 35 - Prob. 79APCh. 35 - Prob. 80APCh. 35 - Prob. 81CPCh. 35 - Prob. 82CPCh. 35 - Prob. 83CPCh. 35 - Prob. 84CPCh. 35 - Prob. 85CPCh. 35 - Prob. 86CPCh. 35 - Prob. 87CP
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