University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 33 - The Nature and Propagation of Light - Problems - Exercises - Page 1106: 33.22

Answer

$0.41^{\circ}$.

Work Step by Step

Use Snell’s Law, where the angles are measured from the normal to the interface. First for the red light. $$n_a sin\theta_a= n_b sin\theta_b$$ $$\theta_b=sin^{-1}(\frac{n_a}{n_b}sin\theta_1) $$ $$=sin^{-1}(\frac{1.00}{2.41}sin53.5^{\circ}) =19.48^{\circ}$$ Next for the violet light. $$n_a sin\theta_a= n_b sin\theta_b$$ $$\theta_b=sin^{-1}(\frac{n_a}{n_b}sin\theta_1) $$ $$=sin^{-1}(\frac{1.00}{2.46}sin53.5^{\circ}) =19.07^{\circ}$$ From these two values, calculate the final angle between the two initially parallel rays. $$\Delta \theta=19.48^{\circ}-19.07^{\circ}=0.41^{\circ}$$
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