Chapter 23, Problem 18P

To determine

The expression for the angle of deviation $\delta$ in term of $\beta$ and $n$ and to show that $\delta$ is proportional to the angle $\beta$ of the prism.

Answer to Problem 18P

The expression for the angle of deviation is $\delta =\beta \left(n-1\right)$, and the value of $\delta$ is proportional to the angle $\beta$ of the prism.

Explanation of Solution

The schematic ray diagram involving the prism is shown in Figure 1. The refractive index of the medium outside the prism is $n_1$, and that of the material of the prism is $n_2$. The angles $\alpha$, $\beta$,$\gamma$, $\delta$, ${\theta }_1$, ${\theta }_2$, ${\theta }_3$, and ${\theta }_4$ are marked in the Figure 1.

From the Figure 1, the following relations can be deduced.

$\beta +2\alpha =180°$ (I)

${\theta }_1=90°-\alpha$ (II)

Solve equation (I) for $\alpha$.

$\begin{array}{c}\alpha =\frac{180°-\beta }{2}\\ =90°-\frac{\beta }{2}\end{array}$ (III)

Use equation (III) in (II).

$\begin{array}{c}{\theta }_1=90°-\left(90°-\frac{\beta }{2}\right)\\ =\frac{\beta }{2}\end{array}$ (IV)

Apply Snell’s law to the boundary where light enters the prism.

$n_{\text{1}}\sin {\theta }_1=n_2\sin {\theta }_2$ (V)

Solve equation (V) for $\sin {\theta }_2$.

$\sin {\theta }_2=\frac{n_{\text{1}}}{n_2}\sin {\theta }_1$ (VI)

Use equation (IV) in (VI).

$\sin {\theta }_2=\frac{n_{\text{1}}}{n_2}\sin \frac{\beta }{2}$ (VII)

Use the approximation $\sin \theta \approx \theta$ in equation (VII) since $\beta$ is very small.

$\sin {\theta }_2\approx \frac{n_{\text{1}}}{n_2}\frac{\beta }{2}$ (VIII)

Use the approximation $\sin \theta \approx \theta$ in equation (VIII) since ${\theta }_2$ is also small.

${\theta }_2\approx \frac{n_{\text{1}}}{n_2}\frac{\beta }{2}$ (IX)

Use $n_1=1$ (refractive index of air), and $n_2=n$ to simplify equation (IX).

${\theta }_2=\frac{\beta }{2n}$ (X)

In the triangle formed by the ray crossing the prism and the two surfaces of the prism, the angles are $\left(90°-{\theta }_2\right)$, $\beta$, and $\left(90°-{\theta }_3\right)$. As interior angles of a triangle, the sum of these angles must be $180°$.

$\left(90°-{\theta }_2\right)+\beta +\left(90°-{\theta }_3\right)=180°$ (XI)

Solve equation (XI) for ${\theta }_3$.

${\theta }_3=\beta -{\theta }_2$ (XII)

Use equation (X) in (XII).

$\begin{array}{c}{\theta }_3=\beta -\frac{\beta }{2n}\\ =\beta \left(1-\frac{1}{2n}\right)\end{array}$ (XIII)

Apply Snell’s law to the boundary where light leaves the prism.

$n_1\sin \theta =n_{\text{2}}\sin {\theta }_3$ (XIV)

Use $n_1=1$ (refractive index of air), and $n_2=n$ to simplify equation (XIV).

$\sin {\theta }_4=n\sin {\theta }_3$ (XV)

Use equation (XIII) in (XV).

$\sin {\theta }_4=n\sin \left[\beta \left(1-\frac{1}{2n}\right)\right]$ (XVI)

Since $\beta \left(1-\frac{1}{2n}\right)<\beta$, it is also small. Moreover, ${\theta }_4$ is small as well. Thus the approximation $\sin \theta \approx \theta$ can be applied to equation (XVI).

$\begin{array}{c}{\theta }_4\approx n\beta \left(1-\frac{1}{2n}\right)\\ =\beta \left(n-\frac{1}{2}\right)\end{array}$ (XVII)

Write the expression for the angle of deviation $\delta$ from the figure.

$\delta ={\theta }_4-{\theta }_1$ (XVIII)

Use equation (XVII) and (IV) in (XVIII).

$\begin{array}{c}\delta =\beta \left(n-\frac{1}{2}\right)-\frac{\beta }{2}\\ =\beta \left(n-\frac{1}{2}-\frac{1}{2}\right)\\ =\beta \left(n-1\right)\end{array}$

This indicates that, the angle of deviation $\delta$ is proportional to the angle $\beta$ of the prism.

Conclusion:

Therefore, the expression for the angle of deviation is $\delta =\beta \left(n-1\right)$, and the value of $\delta$ is proportional to the angle $\beta$ of the prism.