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: 34

Answer

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Work Step by Step

For the longest wavelength, $\lambda_{max}=700$ nm, we will solve for the maximum angle of $90^\circ$. $$d\sin\theta= m\lambda_{max}$$ Solving for $m$; $$m=\dfrac{d\sin\theta}{\lambda_{max}}$$ Recall that the slit width $d$ is given by $1/7400$ cm, $$m=\dfrac{\dfrac{1}{7400}\times10^{-2}\cdot \sin90^\circ}{ 700\times 10^{-9}}=\bf 1.93$$ Since $m$ must be an integer number, thus $m=1$, which means that we can see only one full order, the red light, while the second order is almost fully visible as well. ---- For the smallest wavelength, $\lambda_{min}=400$ nm, we will solve for the maximum angle of $90^\circ$. $$d\sin\theta= m\lambda_{min}$$ Solving for $m$; $$m=\dfrac{d\sin\theta}{\lambda_{min}}$$ Recall that the slit width $d$ is given by $1/7400$ cm, $$m=\dfrac{\dfrac{1}{7400}\times10^{-2}\cdot \sin90^\circ}{ 400\times 10^{-9}}=\bf 3.38$$ Since $m$ must be an integer number, thus $m=3$, which means that we can see part of the third order.
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