A rectangular wire of length 2.0 cm, breadth 1.5 cm is placed 25 cm in front of a concave mirror of focal length 10 cm with its centre on the axis of the mirror and its plane normal to the axis. The area enclosed by the image of the wire is

Options

A. `7.5cm^(2)`
B. `6cm^(2)`
C. `4cm^(2)`
D. `1.33cm^(2)`

Explanation

To solve the problem, we need to find the area enclosed by the image of a rectangular wire placed in front of a concave mirror. Let's break down the solution step by step.

### Step 1: Identify the given data
- Length of the wire (object) \( L_o = 2.0 \, \text{cm} \)
- Breadth of the wire (object) \( B_o = 1.5 \, \text{cm} \)
- Distance of the object from the mirror \( u = -25 \, \text{cm} \) (negative as per the sign convention)
- Focal length of the concave mirror \( f = -10 \, \text{cm} \) (negative for concave mirrors)

### Step 2: Use the mirror formula to find the image distance \( v \)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v} = \frac{1}{-10} - \frac{1}{-25} \]
Calculating the right side:
\[ \frac{1}{v} = -\frac{1}{10} + \frac{1}{25} = -\frac{5}{50} + \frac{2}{50} = -\frac{3}{50} \]
Thus,
\[ v = -\frac{50}{3} \, \text{cm} \approx -16.67 \, \text{cm} \]

### Step 3: Calculate the magnification \( m \)
The magnification \( m \) for mirrors is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\left(-\frac{50}{3}\right) \div (-25) = \frac{50}{3 \times 25} = \frac{50}{75} = \frac{2}{3} \]

### Step 4: Find the dimensions of the image
Using the magnification to find the image dimensions:
- Image length \( L_i = m \cdot L_o = \frac{2}{3} \cdot 2.0 \, \text{cm} = \frac{4}{3} \, \text{cm} \approx 1.33 \, \text{cm} \)
- Image breadth \( B_i = m \cdot B_o = \frac{2}{3} \cdot 1.5 \, \text{cm} = 1.0 \, \text{cm} \)

### Step 5: Calculate the area of the image
The area \( A \) of the image is given by:
\[ A = L_i \cdot B_i = \left(\frac{4}{3} \, \text{cm}\right) \cdot (1.0 \, \text{cm}) = \frac{4}{3} \, \text{cm}^2 \approx 1.33 \, \text{cm}^2 \]

### Final Answer
The area enclosed by the image of the wire is approximately \( \frac{4}{3} \, \text{cm}^2 \) or \( 1.33 \, \text{cm}^2 \).

---