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: 92

Answer

$\approx12^\circ$

Work Step by Step

We know, for bright fringes in a diffraction grating, that $$m\lambda=d\sin\theta$$ and for the first order, $m=1$ $$ \lambda=d\sin\theta$$ So, the angle of the first order is given by $$\theta=\sin^{-1}\left[\dfrac{\lambda}{d}\right] $$ Noting that $d=\dfrac{1}{N}\times10^{-2}\;\rm m$ $$\theta=\sin^{-1}\left[\dfrac{N\lambda}{10^{-2}}\right] $$ Therefore, the angular separation is given by $$\Delta \theta=\theta_2-\theta_1$$ $$\Delta \theta=\bigg|\sin^{-1}\left[\dfrac{N\lambda_2}{10^{-2}}\right] -\sin^{-1}\left[\dfrac{N\lambda_1}{10^{-2}}\right] \bigg|$$ Plugging the given; $$\Delta \theta=\bigg|\sin^{-1}\left[\dfrac{7700\times 410\times10^{-9}}{10^{-2}}\right] -\sin^{-1}\left[\dfrac{7700\times 656\times10^{-9}}{10^{-2}}\right] \bigg|= $$ $$\Delta \theta=\color{red}{\bf11.94^\circ}$$
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