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 - General Problems - Page 711: 78

Answer

$544\;\rm nm$

Work Step by Step

As we see below, we have two reflected rays with one $\pi$-phase change. This means that the path length difference for destructive interference is given by $$2t=m\lambda_n=\dfrac{m\lambda}{n}\tag 1$$ and for constructive interference the path length difference is given by $$2t=\left(m+\frac{1}{2}\right)\lambda_n=\dfrac{\left(m+\frac{1}{2}\right)\lambda}{n}\tag 2$$ And since we have a maximum reflected light for two wavelengths, we will use equation (2) to find the thickness of the film. - For $\lambda_1=688\;\rm nm$, $$t=\dfrac{\left(m_1+\frac{1}{2}\right)\lambda_1}{2n}\tag 3$$ - For $\lambda_2=491.4\;\rm nm$, $$t=\dfrac{\left(m_2+\frac{1}{2}\right)\lambda_2}{2n}\tag 4$$ The left side of the two last formulas (3) and (4) is the same, so doe the right side. Thus; $$ \dfrac{\left(m_2+\frac{1}{2}\right)\lambda_2}{2n}=\dfrac{\left(m_1+\frac{1}{2}\right)\lambda_1}{2n} $$ $$ \left(m_2+\frac{1}{2}\right)\lambda_2 = \left(m_1+\frac{1}{2}\right)\lambda_1 $$ $$ \dfrac{\lambda_1}{\lambda_2} =\dfrac{\left(m_2+\frac{1}{2}\right)}{\left(m_1+\frac{1}{2}\right) } $$ Thus, $$ \dfrac{\lambda_1}{\lambda_2} =\dfrac{\left(m_1+1+\frac{1}{2}\right)}{\left(m_1+\frac{1}{2}\right) } $$ Plugging the known $$ \dfrac{688}{491.4}= \dfrac{\left(m_1 +\frac{3}{2}\right)}{\left(m_1+\frac{1}{2}\right) } $$ $$ \dfrac{688}{491.4}m_1+\dfrac{688}{491.4} \cdot \frac{1}{2}= \left(m_1 +\frac{3}{2}\right) $$ $$ \dfrac{688}{491.4} m_1 -m_1= \frac{3}{2}-\dfrac{688}{982.8} $$ Thus, $m_1=1.9999\approx 2$. Plugging that into (3) $$t=\dfrac{\left(2+\frac{1}{2}\right)\cdot 688}{2\cdot 1.58} =\color{red}{\bf 544}\;\rm nm$$
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