Answer
$N=23100$
Work Step by Step
We know that:
$N=\frac{\lambda_{avg}}{m\Delta \lambda}$
We plug in the known values to obtain:
$N=\frac{\frac{415.496nm+415.487nm}{2}}{2(415.496nm-415.487nm)}=23100$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 36 - Diffraction - Problems - Page 1113: 56a
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
