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 709: 50

Answer

$9800\;\rm nm$

Work Step by Step

As we see in the figure below, the second ray (ray 2) reflects with $\pi$-phase change. And since we have 35 bright rings and 35 dark rings without counting the central dark one, so we have actually an odd number of dark fringes. The author here asks about the thickness of the lens, which is actually the thickness of the air film at the right (or at the left) edge of the lens between the lens and the flat glass, as you see below, Now we have a destructive interference at the edge where $m=35$ (since the central dark one is at $m=0$), and we have two reflected rays with one phase change, so the thickness is given by $$ t=m\dfrac{\lambda}{2}$$ Plugging the known; $$ t=35\cdot \dfrac{560}{2}=\color{red}{\bf 9800}\;\rm nm$$
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