Physics: Principles with Applications (7th Edition)

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

Chapter 24 - The Wave Nature of Light - Problems - Page 708: 22

Answer

629 nm

Work Step by Step

The width of the entire central peak is two times this angular width of the central peak to the first minimum. $$\Delta \theta =28.0^{\circ}= 2 \theta_1$$ $$\theta_1=14.0^{\circ}$$ Use equation 24–3a to calculate the angular width from the middle of the central peak to the first minimum. $$sin\theta_1=\frac{\lambda}{D}$$ $$\lambda = D sin\theta_1=(2.60\times10^{-6}m)sin14^{\circ}$$ $$\lambda = 6.29\times10^{-7}m=629nm$$
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.