Explanation Given data: The refractive index of sodium fluoride is 1.33. Formula used: Consider the formula for critical angle. θ c = sin − 1 ( n medium 2 n medium 1 ) (1) Here, n medium 2 is the refractive index of medium 2, and n medium 1 is the refractive index of medium 1. Modify FIGURE P18.16 as shown in Figure 1. As one medium is sodium fluoride (NaF) and the other medium is air. As the light rays moves from a denser medium to a rarer medium so there will be a critical angle θ c . Modify equation (1) for air and NaF medium. θ c = sin − 1 ( n air n NaF ) Here, n air is refractive index of air medium, which has the standard value as 1, and n NaF is refractive index of sodium fluoride medium, which is 1.33. Substitute 1 for n air , and 1.33 for n NaF , θ c = sin − 1 ( 1 1.33 ) = 48.75 ° From the figure we know that; sum of all the angle in a right angled triangle is 180 ° , θ 1 + θ c + 90 ° = 180 ° θ 1 = 90 ° − θ c Substitute 48

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ISBN: 9780321989246

Author: Knight

Publisher: PEARSON CO

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Chapter 18, Problem 16P

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The angle θ2 at which the ray emerges from the sodium fluoride surface to the air.

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Chapter 18, Problem 15P

Chapter 18, Problem 17P

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The angle θ 2 at which the ray emerges from the sodium fluoride surface to the air.

Solution Summary: The author calculates the angle at which the ray of light emerges from the sodium fluoride surface to the air.

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The angle θ2 at which the ray emerges from the sodium fluoride surface to the air.

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Chapter 18 Solutions