An object is approaching a fixed plane mirror with velocity `5ms^(-1)` making an angle of `45^(@)` with the normal. The speed of image w.r.t. the mirror is

Options

A. `5ms^(-1)`
B. `5//sqrt(2)ms^(-1)`
C. `5sqrt(2)ms^(-1)`
D. `10ms^(-1)`

Explanation

To solve the problem of finding the speed of the image with respect to the mirror when an object approaches a fixed plane mirror at an angle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Velocity of the object, \( v_o = 5 \, \text{m/s} \)
- Angle with the normal, \( \theta = 45^\circ \)

2. **Resolve the Velocity into Components:**
- The object's velocity can be resolved into two components:
- **Perpendicular Component (towards the mirror):**
\[ v_{o\perp} = v_o \cos(\theta) = 5 \cos(45^\circ) = 5 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \, \text{m/s} \]
- **Parallel Component (along the mirror):**
\[ v_{o\parallel} = v_o \sin(\theta) = 5 \sin(45^\circ) = 5 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \, \text{m/s} \]

3. **Calculate the Speed of the Image:**
- The image in a plane mirror moves with the same speed as the object but in the opposite direction. Therefore, the speed of the image \( v_i \) with respect to the mirror is:
- The perpendicular component doubles because the image moves away from the mirror as the object approaches:
\[ v_{i\perp} = 2 v_{o\perp} = 2 \times \frac{5}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{m/s} \]
- The parallel component remains the same:
\[ v_{i\parallel} = v_{o\parallel} = \frac{5}{\sqrt{2}} \, \text{m/s} \]

4. **Calculate the Resultant Speed of the Image:**
- The resultant speed of the image with respect to the mirror can be calculated using the Pythagorean theorem:
\[ v_i = \sqrt{(v_{i\perp})^2 + (v_{i\parallel})^2} \]
\[ v_i = \sqrt{\left(\frac{10}{\sqrt{2}}\right)^2 + \left(\frac{5}{\sqrt{2}}\right)^2} \]
\[ v_i = \sqrt{\frac{100}{2} + \frac{25}{2}} = \sqrt{\frac{125}{2}} = \sqrt{62.5} \approx 7.91 \, \text{m/s} \]

### Final Answer:
The speed of the image with respect to the mirror is approximately \( 7.91 \, \text{m/s} \).