Question

A monochromatic ray of light (f=5.09 x 10 to the power of 14 hz) travels from air into medium x. the angle of incidence of the ray in air is 45 degrees and the rays angle of refraction in medium x is 29 degrees. calculate the absolute index refraction of medium x

Answer 1

To calculate the absolute index of refraction of medium \( x \), we can use Snell's Law, which is stated as:

\[
n_1 \sin \theta_1 = n_2 \sin \theta_2
\]

Where:
- \( n_1 \) is the refractive index of the initial medium (air, in this case),
- \( \theta_1 \) is the angle of incidence,
- \( n_2 \) is the refractive index of the second medium (medium \( x \)),
- \( \theta_2 \) is the angle of refraction.

1. **Identify the values**:
- For air, the refractive index \( n_1 \approx 1.00 \).
- \( \theta_1 = 45^\circ \)
- \( \theta_2 = 29^\circ \)

2. **Plug these values into Snell's Law**:

\[
n_1 \sin \theta_1 = n_2 \sin \theta_2
\]

Substituting the known values:

\[
1.00 \cdot \sin(45^\circ) = n_2 \cdot \sin(29^\circ)
\]

3. **Calculate \( \sin(45^\circ) \) and \( \sin(29^\circ) \)**:

- \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071 \)
- \( \sin(29^\circ) \approx 0.4852 \) (You can use a calculator for this.)

4. **Substituting these values** in:

\[
0.7071 = n_2 \cdot 0.4852
\]

5. **Solve for \( n_2 \)**:

\[
n_2 = \frac{0.7071}{0.4852} \approx 1.46
\]

Thus, the absolute index of refraction of medium \( x \) is approximately **1.46**.