Explanation Given data: The exit angle θ is given in the Figure P18.56. Formula used: Write the expression for Snell’s law. n air sin θ air = n glass sin θ glass (1) Here, n air is the refractive index of air and has the standard value of 1.00. n glass is the refractive index of water and has the standard value of 1.52. θ air is the angle made by the light ray with respect to the normal. θ glass is the angle of refraction in the glass. Redraw Figure P18.56 as shown in Figure 1. From figure 1, θ 1 = θ 2 + θ 3 θ 2 + θ 3 = 45 ° (2) Substitute 1.00 for n air , 1.52 for n glass and 45 ° for θ 1 in equation (1), 1.00 ( sin 45 ° ) = 1.52 ( sin θ 2 ) sin θ 2 = sin 45 ° 1.52 sin θ 2 = 0.7071 1.52 sin θ 2 = 0.4652 θ 2 = sin − 1 ( 0

Student Workbook for College Physics: A Strategic Approach Volume 1 (Chs. 1-16)

3rd Edition

ISBN: 9780321908865

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 18, Problem 56GP

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Exit angle (θ) from the glass prism.

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Chapter 18, Problem 55GP

Chapter 18, Problem 57GP

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Chapter 18 Solutions

Student Workbook for College Physics: A Strategic Approach Volume 1 (Chs. 1-16)

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Exit angle ( θ ) from the glass prism.

Solution Summary: The formula used to determine the exit angle theta is given in Figure P18.56.

Exit angle (θ) from the glass prism.

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Chapter 18 Solutions