Chapter 23, Problem 17P

To determine

The angle of deviation of the ray.

Answer to Problem 17P

The angle of deviation of the ray is $16.5°$ .

Explanation of Solution

The diagram is shown in figure 1.

To find $\alpha$ , use geometry.

The sum of the angles of the triangle must be $180°$ .

$\beta +2\alpha =180°$

Rewrite the above equation for $\alpha$ .

$\alpha =\frac{180°-\beta }{2}$

Given that the value of $\beta$ is $30.0°$ . Substitute $30.0°$ for $\beta$ in the above equation to find $\alpha$ .

$\begin{array}{c}\alpha =\frac{180°-30.0°}{2}\\ =75.0°\end{array}$

$n_1$ is a normal. The sum of the angles $\alpha$ and ${\theta }_1$ must be $90°$ .

${\theta }_1=90°-\alpha$

Here, ${\theta }_1$ is the angle of incidence

Substitute $75.0°$ for $\alpha$ in the above equation to find ${\theta }_1$ .

$\begin{array}{c}{\theta }_1=90°-75.0°\\ =15.0°\end{array}$

The index of refraction of air is $1.000$ and that of crown glass is $1.517$ .

Write the Snell’s law.

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, $n_1$ is the index of refraction of air and $n_2$ is the index of refraction of crown glass

Rewrite the above equation for ${\theta }_2$ .

${\theta }_2={\sin }^{-1}\left(\frac{n_1}{n_2}\sin {\theta }_1\right)$

Substitute $1.000$ for $n_1$ , $1.517$ for $n_2$ and $15.0°$ for ${\theta }_1$ in the above equation to find ${\theta }_2$ .

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{1.000}{1.517}\sin 15.0°\right)\\ =9.823°\end{array}$

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 $\left(90°-{\theta }_3\right)$ . The sum of these angles must be equal to $180°$ .

$\begin{array}{c}\left(90°-{\theta }_2\right)+\beta +\left(90°-{\theta }_3\right)=180°\\ {\theta }_2+{\theta }_3=\beta \end{array}$ (I)

Rewrite t equation (I) for ${\theta }_3$ .

${\theta }_3=\beta -{\theta }_2$

Substitute $30.0°$ for $\beta$ and $9.823°$ for ${\theta }_2$ in the above equation to find ${\theta }_3$ .

$\begin{array}{c}{\theta }_3=30.0°-9.823°\\ =20.18°\end{array}$

To find the angle ${\theta }_4$ , write the Snell’s law.

$n_2\sin {\theta }_3=n_1\sin {\theta }_4$

Rewrite the above equation for ${\theta }_4$ .

${\theta }_4={\sin }^{-1}\left(\frac{n_2}{n_1}\sin {\theta }_3\right)$

Substitute $1.517$ for $n_2$ , $1.00$ for $n_1$ and $20.18°$ for ${\theta }_3$ in the above equation to find ${\theta }_4$ .

$\begin{array}{c}{\theta }_4={\sin }^{-1}\left(\frac{1.517}{1.00}\sin 20.18°\right)\\ =31.55°\end{array}$

On entering into the prism, the ray is deviated down by the angle ${\theta }_1-{\theta }_2$ . As it leaves the prism, it changes direction downward through the angle ${\theta }_4-{\theta }_3$ .

Write the equation for the net deviation.

$\begin{array}{c}\delta ={\theta }_1-{\theta }_2+{\theta }_4-{\theta }_3\\ ={\theta }_1+{\theta }_4-\left({\theta }_2+{\theta }_3\right)\end{array}$

Here, $\delta$ is the angle of deviation

Put equation (I) in the above equation.

$\delta ={\theta }_1+{\theta }_4-\beta$ (II)

Conclusion:

Substitute $15.0°$ for ${\theta }_1$, $31.55°$ for ${\theta }_4$ and $30.0°$ for $\beta$ in equation (II) to find $\delta$ .

$\begin{array}{c}\delta =15.0°+31.55°-30.0°\\ =16.5°\end{array}$

Therefore, the angle of deviation of the ray is $16.5°$ .