5.
A ray of light is incident on a mirror and the light travels in the direction 𝑥 + 3 𝑦 and
after reflection, travels in the direction 𝑥. What is the angle of incidence with respect to
the normal to the mirror?
a. 30 degrees
b. 60 degrees
c. 90 degrees
d. 120 degrees
Ray of Light Incident on Mirror: Direction \( \mathbf{x + 3y} \) to \( \mathbf{x} \), Angle of Incidence Solved
Problem Analysis
Direction vectors represent the incident ray \( \mathbf{i} = \langle 1, 3 \rangle \) and reflected ray \( \mathbf{r} = \langle 1, 0 \rangle \). Normalize these: \( \hat{\mathbf{i}} = \langle 0.316, 0.949 \rangle \), \( \hat{\mathbf{r}} = \langle 1, 0 \rangle \). The law of reflection states \( \mathbf{r} = \mathbf{i} – 2 (\mathbf{i} \cdot \mathbf{n}) \mathbf{n} \), where \( \mathbf{n} \) is the unit normal.
Rearranging gives \( \mathbf{n} \) proportional to \( \hat{\mathbf{i}} – \hat{\mathbf{r}} = \langle -0.684, 0.949 \rangle \), so \( \hat{\mathbf{n}} = \langle -0.585, 0.811 \rangle \). The angle of incidence \( \theta_i = \cos^{-1} | \hat{\mathbf{i}} \cdot \hat{\mathbf{n}} | = \cos^{-1}(0.5) = 60^\circ \).
Option Evaluation
- a. 30°: Matches angle between reflected ray and normal (\( \cos^{-1}(0.866) \approx 30^\circ \)), but incidence uses incident-normal angle. Incorrect.
- b. 60°: Equals \( \cos^{-1} | \hat{\mathbf{i}} \cdot \hat{\mathbf{n}} | \). Law confirms \( i = r \). Correct.
- c. 90°: Would mean grazing incidence (\( \mathbf{i} \) parallel to mirror), but vectors show acute angle. Incorrect.
- d. 120°: Exceeds 90°, impossible for incidence (measured 0°-90° from normal). Incorrect.
| Option | Calculation | Valid? |
|---|---|---|
| 30° | Reflected-normal angle | No |
| 60° | Incident-normal angle | Yes |
| 90° | Grazing case | No |
| 120° | Obtuse incidence | No |
Vector Reflection Verification
Computed \( \mathbf{r}’ = \hat{\mathbf{i}} – 2 (\hat{\mathbf{i}} \cdot \hat{\mathbf{n}}) \hat{\mathbf{n}} \) matches \( \hat{\mathbf{r}} \) exactly. Tangent component unchanged; normal reverses. Confirms 60°.
