Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 33 - Electromagnetic Waves - Problems - Page 1008: 83c

Answer

There is no refracted light.

Work Step by Step

We can use Snell's law to find $\theta_r$, the angle of refraction of red light: $n_2~sin~\theta_r = n_1~sin~\theta_1$ $sin~\theta_r = \frac{n_1~sin~\theta_1}{n_2}$ $sin~\theta_r = \frac{1.456~sin~44.00^{\circ}}{1.00}$ $sin~\theta_r = 1.011$ Since there is no angle $\theta_r$ such that $sin~\theta_r = 1.011$, the red end of the spectrum is not refracted. We can use Snell's law to find $\theta_b$, the angle of refraction of blue light: $n_2~sin~\theta_b = n_1~sin~\theta_1$ $sin~\theta_b = \frac{n_1~sin~\theta_1}{n_2}$ $sin~\theta_b = \frac{1.470~sin~44.00^{\circ}}{1.00}$ $sin~\theta_b = 1.021$ Since there is no angle $\theta_b$ such that $sin~\theta_b = 1.021$, the blue end of the spectrum is not refracted. There is no refracted light.
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