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 707: 5

Answer

$17^{\circ}$ and $64^{\circ}$

Work Step by Step

For destructive interference to occur, the path difference is a half-wavelength, three half-wavelengths, five half-wavelengths, etc. This is stated in equation 24–2b, where m = 0, 1, 2, 3... $$d sin \theta = (m+\frac{1}{2})\lambda $$ Solve for the possible angles. $$sin \theta = \frac{(m+\frac{1}{2})\lambda }{d}$$ $$\theta = sin^{-1}(\frac{(m+\frac{1}{2})\lambda }{d})$$ Find the first few angles, relative to the straight-through direction, at which destructive interference occurs. Let m = 0, then 1, then 2, etc. $$\theta = sin^{-1}(\frac{(0+\frac{1}{2})(4.5cm) }{7.5cm})=17^{\circ}$$ $$\theta = sin^{-1}(\frac{(1+\frac{1}{2})(4.5cm) }{7.5cm})=64^{\circ}$$ There are no solutions for m=3, or higher, because the maximum of the sine function is 1.
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.