In case of a virtual and erect image, the magnification of a mirror is

Options

A. positive
B. negative
C. unity
D. infinity

Answer

To solve the question regarding the magnification of a mirror when a virtual and erect image is formed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- A virtual image is one that cannot be projected onto a screen; it appears to be behind the mirror.
- An erect image is one that is oriented in the same direction as the object (not inverted).

2. **Recall the Magnification Formula**:
- The magnification (m) of a mirror is given by the formula:
\[ m = -\frac{V}{U} \]
where \( V \) is the image distance and \( U \) is the object distance.

3. **Determine the Sign Conventions**:
- For mirrors, the object distance \( U \) is taken as negative (since the object is placed in front of the mirror).
- The image distance \( V \) for a virtual image is positive (as it appears to be behind the mirror).

4. **Substituting Values**:
- Let’s denote the object distance as \( U = -x \) (where \( x \) is a positive value).
- Since the image is virtual, we denote the image distance as \( V = +y \) (where \( y \) is also a positive value).
- Now substituting these into the magnification formula:
\[ m = -\frac{y}{-x} = \frac{y}{x} \]

5. **Conclusion about Magnification**:
- Since both \( y \) and \( x \) are positive integers, the magnification \( m \) will also be positive:
\[ m = \frac{y}{x} > 0 \]
- This indicates that the image is virtual and erect, confirming that the magnification is positive.

### Final Answer:
The magnification of a mirror when a virtual and erect image is formed is **positive**.

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