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 676: 58

Answer

7.2 cm past the second lens.

Work Step by Step

The first lens is converging, with f = 20.0 cm. An object placed at infinity forms an image at the focal point of the converging lens (see equation 23–8). $$d_{i1}=f_1=20.0cm$$ This image is now the object for the second, diverging lens. This image is 6.0 cm beyond the second lens, so the object distance is negative. $$d_{o2}=-6.0cm$$ Use the lens equation, 23–8, for the second lens. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ $$d_i=\frac{fd_o}{d_o-f}=\frac{(-36.5cm)(-6.0cm)}{-6.0cm-(-36.5cm)}=7.2cm$$ The final image is real, and is 7.2 cm past the second lens.
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