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: 89

Answer

$0.569\;\rm m$

Work Step by Step

First of all, we need to find the angle of the first order of the two wavelengths. We know, for bright fringes in a diffraction grating, that $$m\lambda=d\sin\theta $$ $$\theta=\sin ^{-1}\left(\dfrac{m\lambda}{d}\right)\tag 1 $$ And we know that the distance from the center fringe to the first-order bright fringe is given by $$x=L\tan\theta $$ Plugging from (1); $$x=L\tan\left[\sin ^{-1}\left(\dfrac{m\lambda}{d}\right)\right] $$ So, for the first order for the first wavelength; $$x_1=L\tan\left[\sin ^{-1}\left(\dfrac{ \lambda_1}{d}\right)\right] $$ and for the first order for the second wavelength; $$x_2=L\tan\left[\sin ^{-1}\left(\dfrac{ \lambda_2}{d}\right)\right] $$ Therefore, the linear distance between the first-order bright fringes of these two wavelengths on the screen is given by $$\Delta x=x_2-x_1$$ $$\Delta x=L\tan\left[\sin ^{-1}\left(\dfrac{ \lambda_2}{d}\right)\right]-L\tan\left[\sin ^{-1}\left(\dfrac{ \lambda_1}{d}\right)\right] $$ Plugging the known; $$\Delta x=2.5\tan\left[\sin ^{-1}\left(\dfrac{ 6.8\times10^{-7}}{\dfrac{1}{7200}\times10^{-2}}\right)\right]-2.5\tan\left[\sin ^{-1}\left(\dfrac{ 4.4\times10^{-7}}{\dfrac{1}{7200}\times10^{-2}}\right)\right] $$ $$\Delta x=\color{red}{\bf 0.569}\;\rm cm$$
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