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.

Options

A. moves away from the lens with an uniform speed `5 ms^(-1)`
B. moves away from the lens with an 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 how the image behaves as an object approaches a convergent lens with a uniform speed, we can follow these steps:

### Step 1: Understand the Lens Formula
The lens formula relates the object distance (u), the image distance (v), and the focal length (f) of the lens:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( u \) is the object distance (negative in lens convention),
- \( v \) is the image distance (positive for real images),
- \( f \) is the focal length (positive for converging lenses).

### Step 2: Define the Initial Conditions
Assume the focal length \( f \) of the lens is given (for example, 20 cm). The object starts at a distance greater than the focal length and approaches the lens at a speed of \( 5 \, \text{m/s} \).

### Step 3: Calculate Image Distance at Various Object Distances
As the object approaches the lens, we can calculate the image distance \( v \) for several values of \( u \) (object distance). For example:
- If \( u = -50 \, \text{cm} \):
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-50} \implies v = 33.3 \, \text{cm} \]
- If \( u = -45 \, \text{cm} \):
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-45} \implies v = 36 \, \text{cm} \]
- If \( u = -40 \, \text{cm} \):
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-40} \implies v = 40 \, \text{cm} \]

### Step 4: Analyze the Variation of Image Distance
As the object moves closer to the lens, we observe that the image distance \( v \) changes non-linearly. The image distance increases at a varying rate as the object approaches the focus.

### Step 5: Determine the Nature of Image Motion
Since the object is moving towards the lens at a constant speed, the change in image distance \( v \) is not uniform. The image distance increases with a non-uniform acceleration because the rate of change of \( v \) with respect to \( u \) is not constant.

### Conclusion
Thus, the image moves away from the lens with a non-uniform acceleration as the object approaches the focus.