A boy 2 m tall stands 40 cm in front of a mirror. He sees an erect image, 1 m high. Which of the following is/are correct about the mirror :

Options

A. Concave
B. Convex
C. f = 40 cm
D. f = 50 cm

Answer

To solve the problem, we need to determine the type of mirror based on the given information about the boy and his image. Let's break it down step by step.

### Step 1: Identify the given information
- Height of the boy (object height, \( h_o \)) = 2 m = 200 cm
- Distance of the boy from the mirror (object distance, \( u \)) = 40 cm (since the object is in front of the mirror, \( u \) will be negative, so \( u = -40 \) cm)
- Height of the image (\( h_i \)) = 1 m = 100 cm

### Step 2: Use the magnification formula
The magnification (\( m \)) of a mirror is given by the formula:
\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \]
Substituting the known values:
\[ m = \frac{100 \text{ cm}}{200 \text{ cm}} = \frac{1}{2} \]
This means the image is half the height of the object.

### Step 3: Relate magnification to object and image distances
From the magnification formula, we can express it as:
\[ \frac{1}{2} = -\frac{v}{-40} \]
This simplifies to:
\[ \frac{1}{2} = \frac{v}{40} \]
Cross-multiplying gives:
\[ v = 20 \text{ cm} \]

### Step 4: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Substituting the values of \( u \) and \( v \):
\[ \frac{1}{f} = \frac{1}{-40} + \frac{1}{20} \]
Finding a common denominator (which is 40):
\[ \frac{1}{f} = -\frac{1}{40} + \frac{2}{40} = \frac{1}{40} \]
Thus, we have:
\[ f = 40 \text{ cm} \]

### Step 5: Determine the type of mirror
Since the focal length \( f \) is positive, we conclude that the mirror is a **convex mirror**.

### Conclusion
Based on the calculations, we find that:
1. The image is erect and diminished.
2. The focal length is positive, indicating that the mirror is a convex mirror.