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Problem 18.16 Review 1 Constants Periodic Table Part A Figure 1) shows a ray of light entering an equilateral prism, with all

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Answer #1

orma bo 66 G 0 f 6

\theta = 40o

Angle of incidence = \theta1

\theta + \theta1 = 90

\theta1 = 50o

Angle of refraction = \theta2

From the figure at the interface point,

\theta + \theta1 + 90 + 60 + \theta2 + 90 = 360

40 + 50 + 90 + 60 + \theta2 + 90 = 360

\theta2 = 30o

Index of refraction of air = n1 = 1.0003

Index of refraction of prism = n2

By Snell's Law,

n1Sin\theta1 = n2Sin\theta2

(1.0003)Sin(50) = n2Sin(30)

n2 = 1.532

Index of refraction of the prism = 1.532

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