A square wire of side `3.0 cm` is placed `25 cm` away from a concave mirror of focal length `10 cm`. What is the area enclosed by the image of the wire ? The centre of the wire is on the axis of the mirror, with its two sides normal to the axis.

Options

A. `4 cm ^(2)`
B. `6 cm^(2)`
C. `16cm ^(2)`
D. `36 cm ^(2)`

Answer

To solve the problem step by step, we will use the mirror formula and the concept of magnification.

### Step 1: Identify the given values
- Side of the square wire (object height, h) = 3.0 cm
- Object distance (u) = -25 cm (negative because the object is in front of the mirror)
- Focal length (f) = -10 cm (negative for a concave mirror)

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-25} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-10} + \frac{1}{25} \]

### Step 3: Calculate the image distance (v)
Finding a common denominator (which is 50):
\[ \frac{1}{-10} = \frac{-5}{50}, \quad \frac{1}{25} = \frac{2}{50} \]
Thus,
\[ \frac{1}{v} = \frac{-5}{50} + \frac{2}{50} = \frac{-3}{50} \]
Taking the reciprocal gives:
\[ v = -\frac{50}{3} \text{ cm} \]

### Step 4: Calculate the magnification (m)
The magnification is given by:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
Substituting the values:
\[ m = -\left(-\frac{50}{3}\right) \div (-25) = \frac{50/3}{25} = \frac{50}{75} = \frac{2}{3} \]

### Step 5: Calculate the image height (h')
Using the magnification formula:
\[ h' = m \cdot h = \frac{2}{3} \cdot 3 \text{ cm} = 2 \text{ cm} \]

### Step 6: Calculate the area of the image
The area enclosed by the image of the wire is given by:
\[ \text{Area} = (h')^2 = (2 \text{ cm})^2 = 4 \text{ cm}^2 \]

### Final Answer
The area enclosed by the image of the wire is **4 cm²**.

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