Chapter 7.3, Problem 113AYU

To determine

Find refractive index of two mediums.

Answer to Problem 113AYU

  $\boxed{1.25to1.34}$

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.

  

Ptolemy, who lived in the city of Alexandria in Egypt during the second century AD, gave the measured values in the following table foe the angle of incidence ${\theta }_1$ and the angel of ${\theta }_2$ for a light beam passing from air into water. Do these values agree with Snell’s Law? If so, what index of refraction results?

  $\begin{array}{cccc}{\theta }_1& {\theta }_2& {\theta }_1& {\theta }_2\\ 10°& 8°& 50°& 35°0\text{'}\\ 20°& 15°30\text{'}& 60°& 40°30\text{'}\\ 30°& 22°30\text{'}& 70°& 45°30\text{'}\\ 40°& 29°0\text{'}& 80°& 50°0\text{'}\end{array}$

Calculation: The Snell’s law states that $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{v_1}{v_2}$ , where ${\theta }_1$ is the angle of incidence, ${\theta }_2$ is the angle of refraction, $v_1$ is the speed of light in air , and $v_2$ is the speed of light in water.

We know that the correct value of the index of refraction of water is $1.33$ .

In order to determine whether the experiment done satisfies the Snell’s law, we have to check whether the indices of refraction obtained are about $1.33$ .

We have ${\theta }_1=10°and{\theta }_2=8°$ .

Then, $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 10}{\sin 8}=1.25$ .

We have ${\theta }_1=20°and{\theta }_2=15°30\text{'}$

Then, $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 10}{\sin 8}=1.28$

We have ${\theta }_1=30°and{\theta }_2=22°30\text{'}$ .

Then, $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 30}{\sin 22°30\text{'}}=1.31$

We have ${\theta }_1=40°and{\theta }_2=29°$

Then, $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 40}{\sin 29}=1.33$

We have ${\theta }_1=50°and{\theta }_2=35°$ .

Then, $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 50}{\sin 35}=1.34$

We have ${\theta }_1=60°and{\theta }_2=40°30\text{'}$ .

We have $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 60}{\sin 40°30\text{'}}=1.33$

We have ${\theta }_1=70°and{\theta }_2=45°30\text{'}$

Then, $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 70}{\sin 45°35\text{'}}=1.32$

We have ${\theta }_1=80°and{\theta }_2=50°$

Then, $\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 80}{\sin 50}=1.29$

Here, we can see that the indices of refraction are all close to $1.33$ .

Hence, we can say that the Snell’s law is satisfied and the value of the indices of refraction varies from $1.25to1.34$ .