Chapter 35, Problem 35.71AP

To determine

The relation between the emergent angle and $n$, $R$ and $L$.

Answer to Problem 35.71AP

The relations between $\theta$ and $n$, $L$ and $R$ are,

$\theta ={\sin }^{-1}\left(\left(\frac{L}{R^2}\right)\left[\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2}\right]\right)$ or $\theta ={\sin }^{-1}\left(n\sin \left({\sin }^{-1}\left(\frac{L}{R}\right)-{\sin }^{-1}\left(\frac{L}{R}\right)\right)\right)$

Explanation of Solution

Given info: The refractive index of the material is $n$ . Τhe radius of the curvature of the material is $R$. The light parallel to the base of the material strikes it from a height $L$. The emergent ray makes an angle $\theta$ with the material.

Let the angle of incidence is $\gamma$ as shown in figure below.

Figure (I)

From the Figure (1) from triangle $OPQ$,

$\begin{array}{c}\sin \gamma =\frac{L}{R}\\ \gamma ={\sin }^{-1}\left(\frac{L}{R}\right)\end{array}$

From the Figure (1) from triangle $OPQ$,

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

From Snell’s Law

$n_1\sin \gamma =n_2\sin \varphi$ (1)

Here,

$n_1$ is the refractive index of the vacuum.

$n_2$ is the refractive index of material.

$\gamma$ is the angle of incidence.

$\varphi$ is the angle of refraction at point P.

Substitute $1.00$ for $n_1$, $n$ for $n_2$ in equation (1).

$\begin{array}{l}\left(1.00\right)\sin \gamma =n\sin \varphi \\ \sin \varphi =\frac{\sin \gamma }{n}\end{array}$

Substitute $\frac{L}{R}$ for $\sin \gamma$ in the above equation.

$\begin{array}{c}\sin \varphi =\frac{L}{nR}\\ \varphi ={\sin }^{-1}\left(\frac{L}{nR}\right)\end{array}$

The value of $\cos \varphi$ is ,

$\cos \varphi =\sqrt{1-{\sin }^2\varphi }$

Substitute $\frac{L}{nR}$ for $\sin \varphi$ in the above equation.

$\cos \varphi =\frac{\sqrt{n^2{R}^2-L^2}}{nR}$

From figure (1) in triangle $\text{OPS}$,

$\begin{array}{c}\varphi +\left(\alpha +90.0°\right)+\left(90.0°-\gamma \right)=180\\ \alpha =\gamma -\varphi \end{array}$

Apply Snell’s law at the point S.

$n_2\sin \theta =n_1\sin \alpha$

Here,

$n_2$ is the refractive index of vacuum.

$n_1$ is the refractive index of the material.

Substitute $1.00$ for $n_2$, $n$ for $n_1$, $\gamma -\varphi$ for $\alpha$ in the above equation.

$\begin{array}{c}1.00\sin \theta =n\sin \left(\gamma -\varphi \right)\\ \sin \theta =n\sin \left(\gamma -\varphi \right)\end{array}$ (2)

From trigonometry, the formula of $\sin \left(\gamma -\varphi \right)$ is,

$\sin \left(\gamma -\varphi \right)=\sin \gamma \cos \varphi -\cos \gamma \sin \varphi$

Substitute $\sin \gamma \cos \varphi -\cos \gamma \sin \varphi$ for $\sin \left(\gamma -\varphi \right)$ in equation (2).

$\sin \theta =n\left(\sin \gamma \cos \varphi -\cos \gamma \sin \varphi \right)$

Substitute $\frac{L}{R}$ for $\sin \gamma$, $\frac{\left(\sqrt{R^2-L^2}\right)}{R}$ for $\cos \gamma$, $\frac{L}{nR}$ for $\sin \varphi$ $\frac{\sqrt{n^2{R}^2-L^2}}{nR}$ for $\cos \varphi$ in the above equation.

$\begin{array}{c}\sin \theta =n\left(\left(\frac{L}{R}\right)\left(\frac{\sqrt{n^2{R}^2-L^2}}{nR}\right)-\frac{\left(\sqrt{R^2-L^2}\right)}{R}\left(\frac{L}{nR}\right)\right)\\ \theta ={\sin }^{-1}\left(\left(\sqrt{n^2{R}^2-L^2}\right)-\left(\sqrt{R^2-L^2}\right)\right)\\ ={\sin }^{-1}\left(\left(\frac{L}{R^2}\right)\left[\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2}\right]\right)\end{array}$ (3)

The another solution for the equation (2) is,

$\sin \theta =n\sin \left(\gamma -\varphi \right)$

Substitute ${\sin }^{-1}\left(\frac{L}{R}\right)$ for $\gamma$ and ${\sin }^{-1}\left(\frac{L}{nR}\right)$ for $\varphi$ in the above equation.

$\begin{array}{c}\sin \theta =n\sin \left(\left({\sin }^{-1}\left(\frac{L}{R}\right)\right)-\left({\sin }^{-1}\left(\frac{L}{nR}\right)\right)\right)\\ \theta ={\sin }^{-1}\left(n\sin \left({\sin }^{-1}\left(\frac{L}{R}\right)-{\sin }^{-1}\left(\frac{L}{R}\right)\right)\right)\end{array}$ (4)

Thus the relations between $\theta$ and $n$, $L$ and $R$ are

$\theta ={\sin }^{-1}\left(\left(\frac{L}{R^2}\right)\left[\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2}\right]\right)$ and $\theta ={\sin }^{-1}\left(n\sin \left({\sin }^{-1}\left(\frac{L}{R}\right)-{\sin }^{-1}\left(\frac{L}{R}\right)\right)\right)$

Conclusion:

Therefore, the relations between $\theta$ and $n$, $L$ and $R$ are

$\theta ={\sin }^{-1}\left(\left(\frac{L}{R^2}\right)\left[\sqrt{n^2{R}^2-L^2}-\sqrt{R^2-L^2}\right]\right)$ or $\theta ={\sin }^{-1}\left(n\sin \left({\sin }^{-1}\left(\frac{L}{R}\right)-{\sin }^{-1}\left(\frac{L}{R}\right)\right)\right)$