Chapter 25, Problem 61P

To determine

An expression for $\theta$ in terms of $n,R$, and $L$.

Answer to Problem 61P

An expression for $\theta$ in terms of $n,R$, and $L$ is $\underset{\_}{\theta ={\sin }^{-1}\left[\frac{L}{R^2}\left(\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2}\right)\right]}$ or $\underset{\_}{\theta ={\sin }^{-1}\left[n\sin \left({\sin }^{-1}\frac{L}{R}-{\sin }^{-1}\frac{L}{nR}\right)\right]}$.

Explanation of Solution

The path of the ray in the quarter circle is shown in the Figure.

Consider the triangle $OPQ$ from the Figure, and write the expression for $\sin \gamma$ .

    $\sin \gamma =\frac{L}{R}$        (I)

Here, $\gamma$ is the angle of incidence, $L$ is the distance between the parallel incoming ray to the base of the quarter, and $R$ is the radius of the quarter.

Using Pythagorean theorem, write the expression for $\cos \gamma$.

    $\cos \gamma =\frac{\sqrt{R^2-L^2}}{R}$        (II)

Apply Snell’s law at the point $P$.

    $1.00\sin \gamma =n\sin \varphi$        (III)

Here, $\varphi$ is the angle of refraction.

Solve the equation (III) for $\sin \varphi$, and use equation (I).

    $\begin{array}{c}\sin \varphi =\frac{\sin \gamma }{n}\\ =\frac{L}{nR}\end{array}$        (IV)

Using trigonometric relation, write the expression for $\cos \varphi$, and use equation (IV).

    $\begin{array}{c}\cos \varphi =\sqrt{1-{\sin }^2\varphi }\\ =\frac{\sqrt{n^2{R}^2-L^2}}{nR}\end{array}$        (V)

Consider the triangle $OPS$, the sum of interior angle should be zero.

    $\varphi +\left(\alpha +90.0°\right)+\left(90.0°-\gamma \right)=180°$        (VI)

Here, $\alpha$ is the angle of incidence at the point $S$.

Solve the equation (VI).

    $\alpha =\gamma -\varphi$        (VII)

Apply Snell’s law at the point $S$, and use equation (VII).

    $\begin{array}{c}1.00\sin \theta =n\sin \alpha \\ \sin \theta =n\sin \left(\gamma -\varphi \right)\\ =n\left[\sin \gamma \cos \varphi -\cos \gamma \sin \varphi \right]\end{array}$        (VIII)

Use equation (I), (II), (IV), and (V) in (VIII), and solve for $\theta$.

    $\begin{array}{c}\sin \theta =n\left[\left(\frac{L}{R}\right)\frac{\sqrt{n^2{R}^2-L^2}}{nR}-\frac{\sqrt{R^2-L^2}}{R}\left(\frac{L}{nR}\right)\right]\\ =\frac{L}{R^2}\left(\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2}\right)\\ \theta ={\sin }^{-1}\left[\frac{L}{R^2}\left(\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2}\right)\right]\end{array}$        (IX)

Solve the equation (I) for $\gamma$.

    $\gamma ={\sin }^{-1}\left(\frac{L}{R}\right)$        (X)

Solve the equation (IV).

    $\varphi ={\sin }^{-1}\left(\frac{L}{nR}\right)$        (XI)

Use equation (X) and (XI) in the equation $\sin \theta =n\sin \left(\gamma -\varphi \right)$.

    $\begin{array}{c}\sin \theta =n\sin \left({\sin }^{-1}\frac{L}{R}-{\sin }^{-1}\frac{L}{nR}\right)\\ \theta ={\sin }^{-1}\left[n\sin \left({\sin }^{-1}\frac{L}{R}-{\sin }^{-1}\frac{L}{nR}\right)\right]\end{array}$        (XII)

Conclusion:

Therefore, an expression for $\theta$ in terms of $n,R$, and $L$ is $\underset{\_}{\theta ={\sin }^{-1}\left[\frac{L}{R^2}\left(\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2}\right)\right]}$ or $\underset{\_}{\theta ={\sin }^{-1}\left[n\sin \left({\sin }^{-1}\frac{L}{R}-{\sin }^{-1}\frac{L}{nR}\right)\right]}$.