Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 23 - Light: Geometric Optics - General Problems - Page 677: 81

Answer

$n_p\geq1.60$

Work Step by Step

We need to find the index of refraction of the prism $n_p$. According to Snell's law, $$\overbrace{n_{air}}^{=1}\;\sin45^\circ=n_p\sin\theta_2$$ Thus, $$n_p=\dfrac{\sin45^\circ}{\sin\theta_2}\tag 1$$ And to find $n_p$, we need to find $\theta_2$ the angle of refraction, as you see in the figure below. We need to apply Snell's law to the second surface of the prism. $$n_p\sin\theta_3=\overbrace{n_{air}}^{=1}\;\;\overbrace{\sin\theta_{4,max}}^{=1}$$ We know that the maximum refracted angle is $90^\circ$ which occurs before the total internally reflected occurs. It is occur when the out light arrow comes out parallel to the right side of our prisme. Thus, $$n_p\sin\theta_3\geq 1\tag 2$$ From the geometry of the figure below; $$\theta_2=90^\circ -\phi_1$$ and $$\theta_3=90^\circ -\phi_2$$ Adding these two angles together; $$\theta_2+\theta_3=180^\circ-(\phi_1+\phi_2)\tag 3$$ and from the upper triangle, $65^\circ+\phi_1+\phi_2=180^\circ$ and hence, $\phi_1+\phi_2=180^\circ-65^\circ=\bf 115^{\circ}$ Plugging into (3); $$\theta_2+\theta_3=180^\circ- 115^\circ=\bf 65^\circ$$ Therefore, $$\theta_3=65^\circ-\theta_2$$ Plugging into (2); $$n_p\sin(65^\circ-\theta_2)\geq1 $$ Thus, $$n_p\geq\dfrac{1}{\sin(65^\circ-\theta_2)}$$ Let's ignore the greater and work for equal only. So plug this result into (1); $$\dfrac{1}{\sin(65^\circ-\theta_2)}=\dfrac{\sin45^\circ}{\sin\theta_2}$$ Thus, $$\sin45^\circ \sin(65^\circ-\theta_2)=\sin\theta_2$$ $$\sin45^\circ [\sin 65^\circ\cos\theta_2 -\sin\theta_2\cos65^\circ]=\sin\theta_2$$ $$\sin45^\circ \sin 65^\circ\cos\theta_2 -\sin45^\circ\sin\theta_2\cos65^\circ -\sin\theta_2=0$$ $$\sin45^\circ \sin 65^\circ\cos\theta_2 -\sin\theta_2(\sin45^\circ\cos65^\circ +1)=0$$ $$\sin45^\circ \sin 65^\circ\cos\theta_2 =\sin\theta_2(\sin45^\circ\cos65^\circ +1) $$ $$\dfrac{\sin45^\circ \sin 65^\circ}{ (\sin45^\circ\cos65^\circ +1) }=\dfrac{\sin\theta_2}{\cos\theta_2}=\tan\theta_2$$ Therefore, $$\theta_2=\tan^{-1}\left[\dfrac{\sin45^\circ \sin 65^\circ }{ (\sin45^\circ\cos65^\circ +1) }\right]=\bf 26.3^\circ$$ Plugging into (1); $$n_p=\dfrac{\sin45^\circ}{\sin 26.3^\circ} =1.60$$ Therefore, and from all the above, the index of refraction of the prism is greater than or equal to 1.6. $$n_p\geq\color{red}{\bf 1.60}$$
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