Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1009: 103

Answer

(a) longer than (b) $0.63\mu m$

Work Step by Step

(a) We know that $\lambda\frac{1}{m}$. This shows that if $m$ is increased from $10$ to $7$, then the wavelength will be increased and hence the new wavelength will be greater than $440nm$. (b) We have $\theta=tan^{-1}(\frac{y}{L})$ $\theta=\tan^{-1}(\frac{0.12m}{2.3m})=3.0^{\circ}$ Now $\lambda=\frac{d}{m}sin\theta$ We plug in the known values to obtain: $\lambda=\frac{8.5\times 10^{-5}m}{7}sin 3.0^{\circ}=0.63\mu m$
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.