Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 22 - Wave Optics - Exercises and Problems - Page 650: 29

Answer

$0.2894\;\rm mm$

Work Step by Step

We have here two wavlengths $\lambda_1=589.6$ nm and $\lambda_2=589.0$ nm where $\lambda_2\lt \lambda_1$. We need to find the distance mirror 2 would be moved so that one wavelength has produced one more new maximum than the other wavelength. So, $$\Delta L_2=\dfrac{m\lambda_1}{2}=\dfrac{(m+1)\lambda_2}{2}$$ Thus, $$m\lambda_1=m\lambda_2+\lambda_2$$ $$m(\lambda_1- \lambda_2)=\lambda_2$$ Hence, $$m=\dfrac{\lambda_2}{\lambda_1- \lambda_2}\tag 1$$ So the distance the mirror moved is given by $$\Delta L_2=\dfrac{ m \lambda_1}{2}$$ Plugging from (1); $$\Delta L_2=\dfrac{\lambda_2}{\lambda_1- \lambda_2}\dfrac{ \lambda_1}{2}$$ Plugging the known; $$\Delta L_2=\dfrac{589.0}{589.6- 589.0}\dfrac{ 589.6}{2}=\bf 289395\;\rm nm$$ $$\Delta L_2= \color{red}{\bf 2.894\times 10^4}\;\rm m$$
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.