To determine the indices of refraction and the corresponding media for the incident and refracted rays, we can follow these steps:
### Step 1: Identify the Index of Refraction for the First Medium
The index of refraction of the first medium, given in the problem, is 1.0003. Referring to the provided table:
[tex]\[
\begin{tabular}{|c|c|}
\hline Material & Index of Refraction \\
\hline Air & 1.0003 \\
\hline Water & 1.33 \\
\hline Quartz & 1.46 \\
\hline Lucite & 1.50 \\
\hline Zircon & 1.92 \\
\hline Diamond & 2.42 \\
\hline
\end{tabular}
\][/tex]
We see that an index of refraction of 1.0003 corresponds to "Air."
Hence, Medium 1 is Air.
### Step 2: Apply Snell's Law
We need to use Snell's Law to find the index of refraction for the second medium. Snell's Law is given by the formula:
[tex]\[
n_1 \cdot \sin(\theta_1) = n_2 \cdot \sin(\theta_2)
\][/tex]
Where:
- [tex]\( n_1 \)[/tex] is the index of refraction of the first medium (1.0003).
- [tex]\(\theta_1\)[/tex] is the angle of incidence (72.5 degrees).
- [tex]\( n_2 \)[/tex] is the index of refraction of the second medium.
- [tex]\(\theta_2\)[/tex] is the angle of refraction (39.6 degrees).
The desired index of refraction [tex]\( n_2 \)[/tex] is found by rearranging this equation:
[tex]\[
n_2 = \frac{n_1 \cdot \sin(\theta_1)}{\sin(\theta_2)}
\][/tex]
### Step 3: Matching the Index of Refraction to Known Materials
Once we calculate [tex]\( n_2 \)[/tex] and round it to the nearest value provided in the table for practical purposes, we get:
[tex]\[
n_2 \approx 1.50
\][/tex]
From the table:
[tex]\[
\begin{tabular}{|c|c|}
\hline Material & Index of Refraction \\
\hline Air & 1.0003 \\
\hline Water & 1.33 \\
\hline Quartz & 1.46 \\
\hline Lucite & 1.50 \\
\hline Zircon & 1.92 \\
\hline Diamond & 2.42 \\
\hline
\end{tabular}
\][/tex]
An index of refraction of 1.50 corresponds to "Lucite."
Hence, Medium 2 is Lucite.
### Final Answer
- Medium 1: Air
- Medium 2: Lucite
