An object of height 2 cm is placed at a distance 20cm in front of a concave mirror of focal length 12 cm. Find the position, size and nature of the image

Answer

To solve the problem step-by-step, we will follow these procedures:

### Step 1: Identify the given values
- Height of the object (h₀) = 2 cm
- Object distance (u) = -20 cm (negative because the object is in front of the mirror)
- Focal length (f) = -12 cm (negative for concave mirrors)

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- f = focal length
- v = image distance
- u = object distance

### Step 3: Substitute the known values into the mirror formula
Substituting the values we have:
\[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-20} \]

### Step 4: Simplify the equation
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{-12} + \frac{1}{20} \]
Finding a common denominator (which is 60):
\[ \frac{1}{v} = \frac{-5 + 3}{60} = \frac{-2}{60} \]
Thus:
\[ \frac{1}{v} = -\frac{1}{30} \]

### Step 5: Calculate the image distance (v)
Taking the reciprocal gives:
\[ v = -30 \text{ cm} \]

### Step 6: Calculate the magnification (M)
The magnification formula is:
\[ M = -\frac{v}{u} \]
Substituting the values:
\[ M = -\frac{-30}{-20} = \frac{30}{20} = -1.5 \]

### Step 7: Calculate the height of the image (hᵢ)
Using the magnification formula:
\[ M = \frac{h_i}{h_0} \]
Substituting the known values:
\[ -1.5 = \frac{h_i}{2} \]
Thus:
\[ h_i = -1.5 \times 2 = -3 \text{ cm} \]

### Step 8: Determine the nature of the image
- The negative sign of hᵢ indicates that the image is inverted.
- Since v is negative, the image is real.
- The magnitude of hᵢ (3 cm) indicates that the image is magnified.

### Final Result
- **Position of the image (v)**: -30 cm (real image)
- **Size of the image (hᵢ)**: -3 cm (inverted and magnified)
- **Nature of the image**: Real, inverted, and magnified.

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