`{:("Column I","Column II"),((A)" An object is placed at a distance equal to focal length from pole before convex mirror",(p)" Magnification is "(infty)),((B)" An object is placed at focus before a concave mirror",(q)" Magnification is (0.5)"),((C) "An object is placed at the centre of curvature before a concave mirror" , (r) "Magnification is (1//3)") , ((D) " An object is placed at a distance equal to radius of curvature before a convex mirror" , (s) "Magnification is (-1)"):}`

Explanation

To solve the problem, we will analyze each case in the question and calculate the magnification for the given object placements in relation to the convex and concave mirrors.

### Step-by-Step Solution:

1. **Understanding the Magnification Formula**:
The magnification (M) for mirrors is given by the formula:
\[ M = \frac{h'}{h} = \frac{f}{f - u} \]
where:
- \( h' \) is the height of the image,
- \( h \) is the height of the object,
- \( f \) is the focal length,
- \( u \) is the object distance (taken as negative in mirror conventions).

2. **Case A**: An object is placed at a distance equal to the focal length from the pole before a convex mirror.
- For a convex mirror, the focal length \( f \) is positive.
- Given \( u = -f \).
- Substitute into the magnification formula:
\[ M = \frac{f}{f - (-f)} = \frac{f}{f + f} = \frac{f}{2f} = \frac{1}{2} = 0.5 \]
- **Match**: A matches with q (Magnification is 0.5).

3. **Case B**: An object is placed at the focus before a concave mirror.
- For a concave mirror, the focal length \( f \) is negative.
- Given \( u = -f \).
- Substitute into the magnification formula:
\[ M = \frac{f}{f - (-f)} = \frac{f}{f + f} = \frac{f}{0} = \infty \]
- **Match**: B matches with p (Magnification is ∞).

4. **Case C**: An object is placed at the center of curvature before a concave mirror.
- The center of curvature is at \( u = -2f \).
- Substitute into the magnification formula:
\[ M = \frac{f}{f - (-2f)} = \frac{f}{f + 2f} = \frac{f}{3f} = \frac{1}{3} \]
- **Match**: C matches with r (Magnification is 1/3).

5. **Case D**: An object is placed at a distance equal to the radius of curvature before a convex mirror.
- The radius of curvature \( R = 2f \) for a convex mirror, hence \( u = -2f \).
- Substitute into the magnification formula:
\[ M = \frac{f}{f - (-2f)} = \frac{f}{f + 2f} = \frac{f}{3f} = \frac{1}{3} \]
- **Match**: D matches with s (Magnification is -1).

### Final Matches:
- A → q (Magnification is 0.5)
- B → p (Magnification is ∞)
- C → r (Magnification is 1/3)
- D → s (Magnification is -1)