Question

In about A.D. 150, Claudius Ptolemy gave the following measured values for the angle of incidence${\mathit{\theta }}_{\mathbf{1}}$ and the angle of refraction ${\mathit{\theta }}_{\mathbf{2}}$for a light beam passing from air to water:

Assuming these data are consistent with the law of refraction, use them to find the index of refraction of water. These data are interesting as perhaps the oldest recorded physical measurements.

Short answer

The Refractive index for water is.$1.3$

Step-by-step solution

The given data

The index of the refraction for vacuum is.$\text{n}=1$

The table showing values of the angle of incidence${\theta }_1$ and the angle of refraction${\theta }_2$ is given.

Understanding the concept of Snell’s law

Snell’s law gives the relation between the angle of incidence and the angle of refraction. We have to use Snell’s law to find the refractive index of the water.

Formula:

The Snell’s law of refraction,$n_1\sin {\theta }_1=n_2\sin {\theta }_2$(1)

Calculation of the refractive index of water 

Here, ${\text{n}}_{\text{1}}$is the refractive index of thevacuum and${\text{n}}_{\text{2}}$is the refractive index of the water.

For, angle of incidence${\theta }_1=10°$ and angle of emergence ,${\theta }_2=8°$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{l}sin(10°)=n_{2}sin(8°)\\ n_2=\frac{sin(10°)}{sin(8°)}\\ n_2=1.245\end{array}$

Forthe angle of incidence${\theta }_1=20°$and angle of emergence,${\theta }_2=15°{30}^{\text{'}}$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(20°)=n_{2}sin(15°{30}^{\text{'}})\\ n_2=\frac{\text{sin}(20°)}{\text{sin}(15°{30}^{\text{'}})}\\ n_2=1.29\end{array}$

Forthe angle of incidence${\theta }_1=30°$and angle of emergence,${\theta }_2=22°{30}^{\text{'}}$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(30°)=n_{2}sin(22°{30}^{\text{'}})\\ n_2=\frac{sin(30°)}{sin(22°{30}^{\text{'}})}\\ n_2=1.31\end{array}$

Forthe angle of incidence${\theta }_1=40°$and angle of emergence,${\theta }_2=29°$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(40°)=n_{2}sin(29°)\\ n_2=\frac{sin(40°)}{sin(29°)}\\ n_2=1.32\end{array}$

Forthe angle of incidence${\theta }_1=50°$and angle of emergence,data-custom-editor="chemistry" ${\theta }_2=35°$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}\text{sin}(50°)=n_2\text{sin}(35°)\\ n_2=\frac{\text{sin}(50°)}{\text{sin}(35°)}\\ n_2=1.33\end{array}$

Forthe angle of incidence${\theta }_1=60°$and angle of emergence,${\theta }_2=40°{30}^{\text{'}}$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(60°)=n_{2}sin(40°{30}^{\text{'}})\\ n_2=\frac{sin(60°)}{sin40°30\text{'}}\\ =1.3\end{array}$

Forthe angle of incidence${\theta }_1=70°$and angle of emergence,${\theta }_2=45°{30}^{\text{'}}$the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{c}sin(70°)=n_{2}sin(45°{30}^{\text{'}})\\ n_2=\frac{sin(70°)}{sin(45°{30}^{\text{'}})}\\ n_2=1.34\end{array}$

Fortheangle of incidence ${\theta }_1=80°$and angle of emergence,${\theta }_2=50°$ the refractive index of water can be given using the data in equation (1) as follows:

$\begin{array}{l}sin(80°)=n_{2}sin(50°)\\ n_2=\frac{sin(80°)}{sin(50°)}\\ n_2=1.29\end{array}$

Thus, from all these results, we can say that the refractive index for the water is.$1.3$