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: 105

Answer

(a) increase (b) $15.1^{\circ}$

Work Step by Step

(a) We know that if the frequency of light is decreased then the wavelength increases and therefore the angle of the first dark fringe increases. (b) We know that $\lambda=\frac{c}{f}$ $\lambda=\frac{3.0\times 10^8m/s}{5.22\times 10^{14}Hz}=57.5\times 10^{-9}m$ Now $\theta=\frac{m\lambda}{W}$ We plug in the known values to obtain: $\theta=\frac{1\times 575\times 10^{-9}m}{2.20\times 10^{-6}m}$ $\theta=15.1^{\circ}$
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