A real object is placed infront of a convex mirror (focal length f). It moves towards the mirror, the image also moves. If `V_(i)=` speed of image and `V_(0)=` speed of the object and u is the distance of object from mirror along principal axis, then

Options

A. `V_(i) lt V_(0)` if `|u| lt |f|`
B. `V_(i) gt V_(0)` if `|u| gt |f|`
C. `V_(i) lt V_(0)` if `|u|gt|f|`
D. `V_(i)=V_(0)` if `|u|=|f|`

Explanation

To solve the problem, we need to analyze the behavior of the image formed by a convex mirror when a real object moves towards it. We will use the mirror formula and the concept of magnification to derive the relationship between the speeds of the object and the image.

### Step-by-Step Solution:

1. **Understanding the Mirror Formula**:
The mirror formula for a convex mirror is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
where:
- \( f \) is the focal length of the convex mirror (positive for convex mirrors),
- \( v \) is the image distance (negative for virtual images formed by convex mirrors),
- \( u \) is the object distance (negative as per the sign convention).

2. **Differentiating the Mirror Formula**:
To find the relationship between the speeds of the object and the image, we differentiate the mirror formula with respect to time \( t \):
\[ 0 = -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} \]
Here, \( \frac{du}{dt} \) is the speed of the object \( V_0 \) (which is negative since the object is moving towards the mirror), and \( \frac{dv}{dt} \) is the speed of the image \( V_i \) (which is also negative since the image moves towards the mirror).

3. **Rearranging the Equation**:
Rearranging the differentiated equation gives:
\[ \frac{dv}{dt} = -\frac{u^2}{v^2} \frac{du}{dt} \]
Substituting \( V_0 = \frac{du}{dt} \) and \( V_i = \frac{dv}{dt} \):
\[ V_i = -\frac{u^2}{v^2} V_0 \]

4. **Analyzing the Relationship**:
Since \( u \) is the distance of the object from the mirror (which is negative) and \( v \) is the image distance (also negative), the magnitudes can be considered. The image distance \( v \) is always less than the object distance \( u \) for a convex mirror, leading to:
\[ |V_i| < |V_0| \]
This implies that the speed of the image \( V_i \) is less than the speed of the object \( V_0 \).

5. **Conclusion**:
Therefore, we conclude that the speed of the image is always less than the speed of the object when the object is moving towards the convex mirror.

### Final Answer:
The correct options are:
- The speed of the image \( V_i \) is less than the speed of the object \( V_0 \).