A square wire of side 1cm is placed perpendicular to the principal axis of a concave mirror of focal length 15cm at a distance of 20 cm. The area enclosed by the image of the wire is

Options

A. `4" cm"^(2)`
B. `6" cm"^(2)`
C. `2"cm"^(2)`
D. `9 cm^(2) `

Explanation

To solve the problem of finding the area enclosed by the image of a square wire placed in front of a concave mirror, we can follow these steps:

### Step 1: Identify the given values
- Side length of the square wire (object height, h_o) = 1 cm
- Focal length of the concave mirror (f) = -15 cm (negative because it is a concave mirror)
- Distance of the object from the mirror (u) = -20 cm (negative as per sign convention)

### 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:
\[ \frac{1}{v} = \frac{1}{-15} - \frac{1}{-20} \]
Finding a common denominator (60):
\[ \frac{1}{v} = -\frac{4}{60} + \frac{3}{60} = -\frac{1}{60} \]
Thus,
\[ v = -60 \text{ cm} \]

### Step 3: Calculate the magnification (m)
The magnification formula is:
\[ m = -\frac{v}{u} = \frac{h_i}{h_o} \]
Substituting the known values:
\[ m = -\frac{-60}{-20} = 3 \]

### Step 4: Calculate the height of the image (h_i)
Using the magnification:
\[ h_i = m \cdot h_o = 3 \cdot 1 \text{ cm} = 3 \text{ cm} \]

### Step 5: Calculate the area of the image
Since the image is a square, the area (A) is given by:
\[ A = (h_i)^2 = (3 \text{ cm})^2 = 9 \text{ cm}^2 \]

### Final Answer
The area enclosed by the image of the wire is **9 cm²**.

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