An object approaches a convergent lens from the left of the lens with a uniform speed `5 m//s` and stops at the focus. The image.q

Options

A. moves away from the lens with uniform speed `5 m//s`.
B. moves away from the lens with uniform acceleration.
C. moves away from the lens with a non-uniform acceleration.
D. moves towards the lens with a non-uniform acceleration.

Explanation

To solve the problem of the object approaching a convergent lens and determining the motion of the image, we can follow these steps:

### Step 1: Understand the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where \( f \) is the focal length of the lens, \( v \) is the image distance, and \( u \) is the object distance (with the convention that \( u \) is negative for real objects).

### Step 2: Analyze the Object's Motion
The object approaches the lens with a uniform speed of \( 5 \, m/s \). As it moves towards the lens, the object distance \( u \) decreases uniformly.

### Step 3: Determine the Image Motion
As the object approaches the lens, the image distance \( v \) changes according to the lens formula. Initially, when the object is at a large distance (approaching infinity), the image forms at the focus \( f \). As the object moves closer, the image distance \( v \) will change.

### Step 4: Graphical Representation
The relationship between \( u \) and \( v \) can be represented graphically. The graph of \( u \) vs. \( v \) is hyperbolic, indicating that as \( u \) decreases, \( v \) increases.

### Step 5: Rate of Change of Distances
Since the object is moving towards the lens at a constant speed, the rate of change of \( u \) is constant. However, the rate of change of \( v \) is not constant. As the object gets closer to the lens, the change in \( v \) becomes larger for the same change in \( u \).

### Step 6: Determine the Type of Acceleration
- The image moves away from the lens as the object approaches it.
- The change in image distance \( v \) is not uniform because the rate of change of \( v \) increases as \( u \) decreases.
- Thus, the image moves away from the lens with **non-uniform acceleration**.

### Conclusion
The correct answer is that the image moves away from the lens with a non-uniform acceleration.