A point object is placed at equal distance 3 f in front of a concave mirror, a convex mirror and plane mirror separately (event-1). Now, the distance is decreased to 1.5 f from all the three mirror (event-2). Magnitude of focal length of convex mirror and concave mirror is f . Then choose the correct options.

Options

A. Maximum distance of object in event-1 from the mirror is from plane mirror
B. Minimum distanc of object in event-1 from the mirrro is from convex mirror.
C. Maximum distance of object in event-2 from the mirror is from concave mirror
D. Minimum distance of object in event-2 from the mirror is from plane mirrror

Explanation

To solve the problem step by step, we will analyze the situation for each type of mirror (concave, convex, and plane) for both events (event 1 and event 2).

### Step 1: Analyze Event 1

**Concave Mirror:**
1. The object is placed at a distance \( u = -3f \) (the negative sign indicates that the object is in front of the mirror).
2. Using the mirror formula:
\[ \frac{1}{v} = -\frac{1}{f} + \frac{1}{u} \]
Substitute \( u = -3f \):
\[ \frac{1}{v} = -\frac{1}{f} - \frac{1}{3f} \]
\[ \frac{1}{v} = -\frac{3}{3f} - \frac{1}{3f} = -\frac{4}{3f} \]
Therefore,
\[ v = -\frac{3f}{4} \]

**Convex Mirror:**
1. The object is also at \( u = -3f \).
2. Using the mirror formula:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substitute \( u = -3f \):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{3f} \]
\[ \frac{1}{v} = \frac{3}{3f} - \frac{1}{3f} = \frac{2}{3f} \]
Therefore,
\[ v = \frac{3f}{2} \]

**Plane Mirror:**
1. The object is at \( u = -3f \).
2. For a plane mirror, the image distance \( v \) is equal to the object distance:
\[ v = -u = 3f \]

### Step 2: Summary of Event 1 Results
- **Concave Mirror:** \( v = -\frac{3f}{4} \)
- **Convex Mirror:** \( v = \frac{3f}{2} \)
- **Plane Mirror:** \( v = 3f \)

### Step 3: Analyze Event 2

**Concave Mirror:**
1. The object is placed at \( u = -1.5f \).
2. Using the mirror formula:
\[ \frac{1}{v} = -\frac{1}{f} + \frac{1}{u} \]
Substitute \( u = -1.5f \):
\[ \frac{1}{v} = -\frac{1}{f} - \frac{2}{3f} \]
\[ \frac{1}{v} = -\frac{3}{3f} - \frac{2}{3f} = -\frac{5}{3f} \]
Therefore,
\[ v = -\frac{3f}{5} \]

**Convex Mirror:**
1. The object is at \( u = -1.5f \).
2. Using the mirror formula:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substitute \( u = -1.5f \):
\[ \frac{1}{v} = \frac{1}{f} - \frac{2}{3f} \]
\[ \frac{1}{v} = \frac{3}{3f} - \frac{2}{3f} = \frac{1}{3f} \]
Therefore,
\[ v = 3f \]

**Plane Mirror:**
1. The object is at \( u = -1.5f \).
2. For a plane mirror:
\[ v = -u = 1.5f \]

### Step 4: Summary of Event 2 Results
- **Concave Mirror:** \( v = -\frac{3f}{5} \)
- **Convex Mirror:** \( v = 3f \)
- **Plane Mirror:** \( v = 1.5f \)

### Final Results
- In **Event 1**, the maximum distance from the mirror is from the plane mirror and the minimum distance is from the convex mirror.
- In **Event 2**, the maximum distance from the mirror is from the convex mirror.