Chapter 7.3, Problem 111AYU

To determine

Determine the angle of refraction.

Answer to Problem 111AYU

  $\boxed{28.9°}$

Explanation of Solution

Given information:

The following discussion of Snell’s Law of Refraction is needed for problems. Light, sound, and other waves travel at different speeds, depending on the media (air, water, wood, and so on) through which they pass. Suppose that light travels from a point A in one medium, where its speed is $v_1$ , to a point B in another medium, where its speed is $v_2$ . Refer to the figure, where the angle ${\theta }_1$ is called the angle of incidence and the angle ${\theta }_2$ is the angle of refraction. Snell’s Law, which can be proved, using calculus, states that

  $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{v_1}{v_2}$

The ratio $\frac{v_1}{v_2}$ is called the index of refraction. Some values are given in the table shown to the right.

  

The index of refraction of light in passing from a vacuum into water is $1.33$ . If the angle of incidence is $40°$ , determine the angle of refraction.

Calculation: The index of refraction of light passing from vacuum to water is $1.33$ . Thus, we can say that $\frac{v_1}{v_2}=1.33$ , where $v_1$ is the speed of light in vacuum and $v_2$ is the speed of light in water.

According to the Snell’s law, we have $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{v_1}{v_2}$ , where ${\theta }_1$ is the angle of incidence and ${\theta }_2$ is the angle of refraction.

Substitute $40$ for ${\theta }_1$ and $1.33$ for $\frac{v_1}{v_2}$ in $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{v_1}{v_2}$ to get ${\theta }_2$ .

  $\frac{\sin 40}{\sin {\theta }_2}=1.33$

Multiply both sides of the equation by $\sin {\theta }_2$ .

  $\frac{\sin 40}{\sin {\theta }_2}(\sin {\theta }_2)=1.33(\sin {\theta }_2)$

  $\sin 40=1.33\left(\sin {\theta }_2\right)$

Divide both sides of the equation by $1.33$ .

  $\begin{array}{l}\frac{\sin 40}{1.33}=\frac{1.33(\sin {\theta }_2)}{1.33}\\ 0.4833=\sin {\theta }_2\end{array}$

Take the inverse sine of both sides of the equation.

  $\begin{array}{l}{\sin }^{-1}\left(0.4833\right)={\sin }^{-1}\left(\sin {\theta }_2\right)\\ 28.9°={\theta }_2\end{array}$

Hence, we have the angle of refraction ${\theta }_2$ as $28.9°$ .