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: 17

Answer

567 nm.

Work Step by Step

The number of wavelengths, N, present in a distance t is equal to the distance t, divided by the appropriate wavelength $\lambda$. $$N=\frac{t}{\lambda}$$ We are told that the center point changes from constructive to destructive interference upon insertion of the plastic. The wavelength of the colored light is shorter in plastic than it is in air, so the phase shift must be the result of an extra half wavelength present in the distance t, when the plastic is inserted. Use the above equation. $$N_{plastic}-N_{air}=\frac{1}{2}$$ $$ \frac{t}{\lambda_{plastic}}-\frac{t}{\lambda}=\frac{1}{2}$$ $$ \frac{tn_{plastic}}{\lambda}-\frac{t}{\lambda}=\frac{1}{2}$$ $$ t=\frac{\lambda}{2(n_{plastic}-1)}=\frac{680nm}{2(1.60-1)}=567nm$$
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