Answered

\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}

An incident ray strikes a boundary at a [tex]\(72.5^\circ\)[/tex] angle, resulting in an angle of refraction of [tex]\(39.6^\circ\)[/tex]. The index of refraction of the first medium is 1.0003.

What is medium 1? [tex]\(\square\)[/tex]

What is medium 2? [tex]\(\square\)[/tex]



Answer :

GinnyAnswer
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