Physics: Principles with Applications (7th Edition)

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

Chapter 25 - Optical Instruments - General Problems - Page 743: 84

Answer

0.269 nm

Work Step by Step

Apply equation 25–10 with m = 2. $$m\lambda=2dsin\phi$$ $$d= \frac{m \lambda}{2sin\phi}= \frac{(2)(0.0973nm)}{2(sin21.2^{\circ})}$$ $$d=0.269nm$$
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