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: 79

Answer

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Work Step by Step

We know, in diffraction gratings, that the angle $\theta$ increases when the wavlelngth increases. $$d\sin\theta=m\lambda$$ And to see, or to proof, that the second and the third order of spectra of white light produced by a diffraction grating is overlapping, we need to find the angle of the second and the third spectra. Thus, $$\sin\theta=\dfrac{m\lambda}{d}$$ Now we need to find the ratio of the second order and third order. $$\dfrac{\sin\theta_2}{\sin\theta_3}=\dfrac{\dfrac{2\lambda_2}{d}}{\dfrac{3\lambda_3}{d}}=\dfrac{2\lambda_2}{3\lambda_3}$$ Plugging the known, $$\dfrac{\sin\theta_2}{\sin\theta_3} =\dfrac{2\cdot 700 }{3\cdot 400}=1.17$$ This means that, $$ \sin\theta_2 =1.17\sin\theta_3$$ This means that the maximum angle of the second order is greater than the minimum angle of the third order; so that we have an overlap. To find which wavelengths overlap, we need to assume that the two angles are equal, so that $\dfrac{\sin\theta_2}{\sin\theta_3}=1$. Thus, $$1=\dfrac{2\lambda_2}{3\lambda_3}$$ And hence, $$\lambda_3=\dfrac{2\lambda_{2,maximum}}{3}=\dfrac{2\cdot 700}{3}=\bf 466.67\;\rm nm$$ $$\lambda_2=\dfrac{3\lambda_{3,minimum}}{2}=\dfrac{3\cdot 400}{2}=\bf 600\;\rm nm$$ It is obvious now that the two wavelengths of 600 nm and 700 nm overlapping with the two wavelengths of 400 nm and 467 nm.
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