When an object moves towards a convex lens, the size of the image

Options

A. decreases
B. increases
C. first decreases then increases
D. remains the same

Answer

To solve the question regarding how the size of the image changes when an object moves towards a convex lens, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding 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.

2. **Initial Position of the Object**:
When the object is at infinity (\( u \to \infty \)), the image is formed at the focal point (\( v = f \)). At this position, the size of the image is very small.

3. **Moving the Object Closer**:
As the object moves closer to the lens (from infinity to the center of curvature \( C \)), we need to analyze several positions:
- **At \( u = C \)**: The image is formed at \( C' \) (the center of curvature on the other side). Here, the magnification \( m \) is 1, meaning the size of the image is equal to the size of the object.
- **At \( u = F \)**: When the object is at the focal point, the rays of light diverge and the image is formed at infinity. The magnification becomes very large.

4. **Analyzing Magnification**:
The magnification \( m \) is given by:
\[ m = \frac{v}{u} \]
- When \( u \) is very large (object at infinity), \( m \) is very small.
- When \( u \) decreases to \( C \), \( m \) increases to 1.
- When \( u \) approaches \( F \), \( m \) approaches infinity.

5. **Conclusion**:
As the object moves from infinity to the focal point, the size of the image first increases from a very small size, becomes equal to the size of the object at \( C \), and then increases significantly as it approaches \( F \). Therefore, the size of the image is continuously increasing.

### Final Answer:
The size of the image increases continuously as the object moves towards the convex lens.

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