Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 23 - Light: Geometric Optics - Problems - Page 674: 17

Answer

The image is 2.0 cm behind the surface of the ball, virtual, and upright.

Work Step by Step

Use the mirror equation, 23–2. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ Solve for the image distance. The object distance is 25.0 cm. The focal length is half the radius, and is negative because the ball acts as a convex mirror. The radius is 4.4 cm, so $f=-2.2 cm$. $$d_i=\frac{d_of}{d_o-f}=\frac{(25.0cm)(-2.2cm)}{25.0cm-(-2.2cm)}$$ $$=-2.02cm$$ The image is 2.0 cm behind the surface of the ball, and virtual. Calculate the magnification. $$m=-\frac{d_i}{d_o}=-\frac{-2.022cm}{25.0cm}=+0.081$$ The magnification is positive. The image is upright.
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.