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 708: 20

Answer

$68.12^\circ$, $64.7^\circ$

Work Step by Step

We know, from Snell's law, that $$n_1\sin\theta_1=n_2\sin\theta_2$$ On the left side, $n_1$ is air for air and $n_2$ is for flint glass. We have two cases here for the two wavelengths, so we need to solve each one on its own. Noting that both wavelengths are having the same incident angle on the left side of the prism. In the first case, for $\lambda_1=455\;\rm nm$; $$\overbrace{n_{air}}^{=1}\sin\theta_1=n_{\lambda_1, glass}\sin\theta_{\lambda_1}$$ Plugging the known; see figure 24-14; $$ \sin45^\circ=1.64\sin\theta_{\lambda_1}$$ Thus, $$\theta_{\lambda_1}=\sin^{-1}\left[\dfrac{\sin45^\circ}{1.64}\right]=\bf 25.54^\circ$$ From the geometry of the first figure below, we can see that $$\phi_1+\phi_2=120^\circ$$ Thus, $$(90^\circ -\theta_2)+(90^\circ -\theta_3)=120^\circ$$ We found $\theta_2$ above which is the refraction angle, So, solving for $\theta_3$; $$ 90^\circ -25.54^\circ + 90^\circ -\theta_3 =120^\circ$$ $$\theta_3=\bf 34.46^\circ$$ Using Snell's law again to find $\theta_4$ which is the needed one. Note that $n_1$ here is for the glass and $n_2$ is for the air. $$1.64\sin\theta_3=\sin\theta_4$$ Thus, $$\theta_4=\sin^{-1}\left[1.64\sin 34.46^\circ\right]=\color{red}{\bf 68.12^\circ}$$ --- In the second case, for $\lambda_2=642\;\rm nm$; $$ \sin45^\circ=1.615\sin\theta_{\lambda_2}$$ And by the same approach; $$\theta_{\lambda_2}=\sin^{-1}\left[\dfrac{\sin45^\circ}{1.615}\right]=\bf 25.96^\circ$$ From the geometry of the second figure below, we can see that $$\phi_1+\phi_2=120^\circ$$ Thus, $$(90^\circ -\theta_2)+(90^\circ -\theta_3)=120^\circ$$ We found $\theta_2$ above which is the refraction angle, So, solving for $\theta_3$; $$ 90^\circ -25.96^\circ + 90^\circ -\theta_3 =120^\circ$$ $$\theta_3=\bf 34.04^\circ$$ Using Snell's law again to find $\theta_4$ which is the needed one. Note that $n_1$ here is for the glass and $n_2$ is for the air. $$1.615\sin\theta_3=\sin\theta_4$$ Thus, $$\theta_4=\sin^{-1}\left[1.615\sin 34.46^\circ\right]=\color{red}{\bf 64.7^\circ}$$
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