Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 34 - Ray Optics - Exercises and Problems - Page 992: 50

Answer

$x = 1.9~m$

Work Step by Step

We can find the angle from the normal that the light ray from the near side of the meter stick makes with the water surface: $tan~\theta_n = \frac{1.0~m}{3.0~m}$ $\theta_n = tan^{-1}~(\frac{1.0~m}{3.0~m})$ $\theta_n = 18.435^{\circ}$ We can find the angle this light ray makes with the normal when it emerges from the water: $n_a~sin~\theta_1 = n_w~sin~\theta_n$ $sin~\theta_1 = \frac{n_w~sin~\theta_n}{n_a}$ $\theta_1 = sin^{-1}~(\frac{n_w~sin~\theta_n}{n_a})$ $\theta_1 = sin^{-1}~(\frac{1.33~sin~18.435^{\circ}}{1.00})$ $\theta_1 = 24.87^{\circ}$ We can find the apparent position of the near side of the meter stick: $\frac{x_n}{3.0~m} = tan~24.87^{\circ}$ $x_n = (3.0~m)~(tan~24.87^{\circ})$ $x_n = 1.39~m$ We can find the angle from the normal that the light ray from the far side of the meter stick makes with the water surface: $tan~\theta_f = \frac{2.0~m}{3.0~m}$ $\theta_f = tan^{-1}~(\frac{2.0~m}{3.0~m})$ $\theta_f = 33.695^{\circ}$ We can find the angle this light ray makes with the normal when it emerges from the water: $n_a~sin~\theta_2 = n_w~sin~\theta_f$ $sin~\theta_2 = \frac{n_w~sin~\theta_f}{n_a}$ $\theta_2 = sin^{-1}~(\frac{n_w~sin~\theta_f}{n_a})$ $\theta_2 = sin^{-1}~(\frac{1.33~sin~33.695^{\circ}}{1.00})$ $\theta_2 = 47.54^{\circ}$ We can find the apparent position of the far side of the meter stick: $\frac{x_f}{3.0~m} = tan~47.54^{\circ}$ $x_f = (3.0~m)~(tan~47.54^{\circ})$ $x_f = 3.28~m$ We can find the apparent length of the meter stick: $x = x_f-x_n$ $x = (3.28~m)-(1.39~m)$ $x = 1.9~m$
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.