A ray of light is incident on the plane mirror at rest. The mirr starts turning at a uniform angular acceleration of `pi rad s^(-2)` . The reflected ray at the end of `1/4` s must have turned through

Options

A. `90^@`
B. `45^@`
C. `22.5^@`
D. `11.25^@`

Answer

To solve the problem, we need to determine how much the reflected ray has turned after the mirror has been rotating for \( \frac{1}{4} \) seconds with a uniform angular acceleration of \( \pi \, \text{rad/s}^2 \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Angular acceleration (\( \alpha \)) = \( \pi \, \text{rad/s}^2 \)
- Time (\( t \)) = \( \frac{1}{4} \, \text{s} \)

2. **Use the Angular Displacement Formula:**
The angular displacement (\( \theta \)) of the mirror can be calculated using the formula:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
where \( \omega_0 \) is the initial angular velocity. Since the mirror starts from rest, \( \omega_0 = 0 \).

3. **Substituting the Values:**
\[ \theta = 0 \cdot \frac{1}{4} + \frac{1}{2} \cdot \pi \cdot \left(\frac{1}{4}\right)^2 \]
\[ = \frac{1}{2} \cdot \pi \cdot \frac{1}{16} \]
\[ = \frac{\pi}{32} \, \text{radians} \]

4. **Determine the Angle of Reflection:**
According to the law of reflection, the angle of incidence (\( i \)) equals the angle of reflection (\( r \)). Since the mirror has turned by \( \frac{\pi}{32} \) radians, the reflected ray will turn through twice this angle:
\[ \text{Total turn of the reflected ray} = 2 \cdot \theta = 2 \cdot \frac{\pi}{32} = \frac{\pi}{16} \, \text{radians} \]

5. **Final Answer:**
The reflected ray at the end of \( \frac{1}{4} \) seconds must have turned through \( \frac{\pi}{16} \) radians.