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 1004: 46c

Answer

The index of refraction of material 1 is $~~1.88$

Work Step by Step

We can write Snell's law: $n_w~sin~\theta_2 = n_i~sin~\theta_1$ In this equation: $n_w$ is the index of refraction of water $n_i$ is the index of refraction of the material We can use the data pair $(\theta_1, \theta_2) = (45^{\circ}, 90^{\circ})$ We can use Snell's law to find the index of refraction of the material: $n_w~sin~\theta_2 = n_i~sin~\theta_1$ $n_i = \frac{n_w~sin~\theta_2}{sin~\theta_1}$ $n_i = \frac{1.33~sin~90^{\circ}}{sin~45^{\circ}}$ $n_i = 1.88$ The index of refraction of material 1 is $~~1.88$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.