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 710: 54

Answer

$325\;\rm nm$

Work Step by Step

As we see in the figure below, the first ray experiences a $\pi$ phase change while the second one does not. So, we have here one phase change from the two reflected rays and we got a constructive interference for two wavelengths 390 nm, and 650 nm. To find the minimum thickness that allows for these both colors to appear, we need to find the common thickness that works for both. In our case, the thickness is given by $$t=(m+\frac{1}{2})\dfrac{\lambda_n}{2}$$ $$\boxed{t=(m+\frac{1}{2})\dfrac{\lambda}{2n}}$$ For $\lambda=390$ nm: when $m=0$; $$t=(0+\frac{1}{2})\dfrac{390}{2\cdot 1.5}=\bf 65\;\rm nm$$ when $m=1$; $$t=(1+\frac{1}{2})\dfrac{390}{2\cdot 1.5}=\bf 195\;\rm nm$$ when $m=2$; $$t=(2+\frac{1}{2})\dfrac{390}{2\cdot 1.5}=\color{red}{\bf 325}\;\rm nm$$ when $m=3$; $$t=(3+\frac{1}{2})\dfrac{390}{2\cdot 1.5}=\bf 455\;\rm nm$$ --- For $\lambda=650$ nm: when $m=0$; $$t=(0+\frac{1}{2})\dfrac{650}{2\cdot 1.5}=\bf 108\;\rm nm$$ when $m=1$; $$t=(1+\frac{1}{2})\dfrac{650}{2\cdot 1.5}=\color{red}{\bf 325}\;\rm nm$$ when $m=2$; $$t=(2+\frac{1}{2})\dfrac{650}{2\cdot 1.5}=\bf 541\;\rm nm$$ when $m=3$; $$t=(3+\frac{1}{2})\dfrac{650}{2\cdot 1.5}=\bf 758\;\rm nm$$ Therefore, the minimum thickness is 325 nm, as you see above.
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