To determine when the image of an extended object placed perpendicular to the principal axis of a mirror is erect, we can analyze the magnification formula and the nature of the image formed by different types of mirrors.
### Step-by-Step Solution:
1. **Understanding Magnification**:
The magnification (m) of a mirror is given by the formula:
\[
m = \frac{h'}{h} = -\frac{v}{u}
\]
where:
- \( h' \) is the height of the image,
- \( h \) is the height of the object,
- \( v \) is the image distance,
- \( u \) is the object distance.
2. **Condition for Erect Image**:
For the image to be erect, the magnification must be positive:
\[
m > 0
\]
This means that both \( h' \) and \( h \) must have the same sign (both positive or both negative).
3. **Analyzing Different Cases**:
We will analyze the four options provided in the question:
- **Case 1**: Both object and image are real.
- For a real object, \( u < 0 \) (object distance is negative).
- For a real image, \( v < 0 \) (image distance is negative).
- Therefore, \( m = -\frac{v}{u} \) will be negative (since both \( v \) and \( u \) are negative).
- **Conclusion**: The image is not erect.
- **Case 2**: Both object and image are virtual.
- For a virtual object, \( u > 0 \) (object distance is positive).
- For a virtual image, \( v > 0 \) (image distance is positive).
- Therefore, \( m = -\frac{v}{u} \) will be negative (since both \( v \) and \( u \) are positive).
- **Conclusion**: The image is not erect.
- **Case 3**: The object is real, but the image is virtual.
- For a real object, \( u < 0 \).
- For a virtual image, \( v > 0 \).
- Therefore, \( m = -\frac{v}{u} \) will be positive (since \( v \) is positive and \( u \) is negative).
- **Conclusion**: The image is erect.
- **Case 4**: The object is virtual, but the image is real.
- For a virtual object, \( u > 0 \).
- For a real image, \( v < 0 \).
- Therefore, \( m = -\frac{v}{u} \) will be positive (since \( v \) is negative and \( u \) is positive).
- **Conclusion**: The image is erect.
4. **Final Conclusion**:
The image of an extended object placed perpendicular to the principal axis of a mirror will be erect in the following cases:
- Case 3: The object is real, and the image is virtual.
- Case 4: The object is virtual, and the image is real.
### Summary:
The image will be erect if:
- The object is real and the image is virtual (Case 3).
- The object is virtual and the image is real (Case 4).
