University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 33 - The Nature and Propagation of Light - Problems - Exercises - Page 1106: 33.28

Answer

a. $ \phi=63.4^{\circ} $ b. $ \phi=71.6^{\circ} $.

Work Step by Step

The first polarizer filters out half the incident unpolarized light. $$I_1=\frac{1}{2}I_o$$ The fraction filtered out by the second polarizer depends on the angle between the axes of the two polarizers. At the end: $$I_2=I_1cos^2\phi=\frac{1}{2}I_o cos^2 \phi=0.100I_0 $$ $$ cos \phi=\sqrt{2/10} $$ $$ \phi=63.4^{\circ} $$ b. Now the first filter passes all of the incident light. $$I_1=I_o$$ The fraction filtered out by the second polarizer depends on the angle between the axes of the two polarizers. At the end: $$I_2=I_1cos^2\phi=I_o cos^2 \phi=0.100I_0 $$ $$ cos \phi=\sqrt{1/10} $$ $$ \phi=71.6^{\circ} $$
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