Question

A material having an index of refraction $n$ is surrounded by vacuum and is in the shape of a quarter circle of radius $R$ (Fig. $P35.71$ ). A light ray parallel to the base of the material is incident from the left at a distance $L$ above the base and emerges from the material at the angle $\theta$. Determine an expression for $\theta$ in terms of $n,R,$ and $L$

Answer 1

Observe in ANS. FIG. P35.71 that the angle of incidence at point $P$ is $\gamma$, and using triangle $OPQ:$
Also,
$\begin{array}{c}\sin \gamma =\frac{L}{R}\\ \cos \gamma =\sqrt{1-{\sin }^2\gamma }=\frac{\sqrt{R^2-L^2}}{R}\end{array}$
Apply Snell's law at point $P$ :
$1.00\sin \gamma =n\sin \varphi$
Thus, $\sin \varphi =\frac{\sin \gamma }{n}=\frac{L}{nR}$ and $\cos \varphi =\sqrt{1-{\sin }^2\varphi }=\frac{\sqrt{n^2{R}^2-L^2}}{nR}$
From triangle $OPS,\varphi +(\alpha +{90.0}^{\circ })+({90.0}^{\circ }-\gamma )={180}^{\circ },$ or the angle ofincidence at point $S$ is $\alpha =\gamma -\varphi .$ Then, applying Snell's law at point $S$
$\begin{array}{c}\text{gives}1.00\sin \theta =n\sin \alpha =n\sin (\gamma -\varphi )\\ \text{or}\sin \theta =n\sin (\gamma -\varphi )\end{array}$
thus,
$\begin{array}{c}=n[\sin \gamma \cos \varphi -\cos \gamma \sin \varphi ]\\ =n[\left(\frac{L}{R}\right)\frac{\sqrt{n^2{R}^2-L^2}}{nR}-\frac{\sqrt{R^2-L^2}}{R}\left(\frac{L}{nR}\right)]\\ =\frac{L}{R^2}(\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2})\\ \theta ={\sin }^{-1}\left[\frac{L}{R^2}(\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2})\right]\end{array}$
or, using from above $\sin \gamma =\frac{L}{R}\to \gamma ={\sin }^{-1}\frac{L}{R}$ and $\varphi ={\sin }^{-1}\frac{L}{nR}$
$\begin{array}{c}\sin \theta =n\sin (\gamma -\varphi )=n\sin ({\sin }^{-1}\frac{L}{R}-{\sin }^{-1}\frac{L}{nR})\\ \theta =[{\sin }^{-1}\left[n\sin ({\sin }^{-1}\frac{L}{R}-{\sin }^{-1}\frac{L}{nR})\right]\end{array}$