A pin of length 2.0 cm lies along the principal axis of a converging lens, the centre being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.

Explanation

To find the size of the image formed by a converging lens when a pin of length 2.0 cm is placed along the principal axis, we can follow these steps:

### Step 1: Identify the given data
- Length of the pin (object height, h) = 2.0 cm
- Distance of the center of the pin from the lens (object distance, u) = 11 cm
- Focal length of the lens (f) = 6 cm

### Step 2: Determine the object distances for the ends of the pin
Since the pin is 2.0 cm long and its center is 11 cm from the lens, we can find the distances of the ends of the pin from the lens:
- Distance of the top end (A) from the lens:
\[ u_A = 11 \, \text{cm} - 1 \, \text{cm} = 10 \, \text{cm} \]
- Distance of the bottom end (B) from the lens:
\[ u_B = 11 \, \text{cm} + 1 \, \text{cm} = 12 \, \text{cm} \]

### Step 3: Apply the lens formula for each end of the pin
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) = focal length
- \( v \) = image distance
- \( u \) = object distance (with sign convention)

#### For end A:
- Object distance \( u_A = -10 \, \text{cm} \) (negative as per sign convention)
- Focal length \( f = +6 \, \text{cm} \)

Using the lens formula:
\[ \frac{1}{6} = \frac{1}{v_A} - \frac{1}{-10} \]
\[ \frac{1}{v_A} = \frac{1}{6} - \frac{1}{10} \]
Finding a common denominator (30):
\[ \frac{1}{v_A} = \frac{5}{30} - \frac{3}{30} = \frac{2}{30} = \frac{1}{15} \]
Thus,
\[ v_A = 15 \, \text{cm} \]

#### For end B:
- Object distance \( u_B = -12 \, \text{cm} \)

Using the lens formula:
\[ \frac{1}{6} = \frac{1}{v_B} - \frac{1}{-12} \]
\[ \frac{1}{v_B} = \frac{1}{6} - \frac{1}{12} \]
Finding a common denominator (12):
\[ \frac{1}{v_B} = \frac{2}{12} - \frac{1}{12} = \frac{1}{12} \]
Thus,
\[ v_B = 12 \, \text{cm} \]

### Step 4: Calculate the size of the image
The image distances for the ends of the pin are:
- \( v_A = 15 \, \text{cm} \) (image of A)
- \( v_B = 12 \, \text{cm} \) (image of B)

The size of the image (height of the image) can be calculated as:
\[ \text{Length of image} = v_A - v_B = 15 \, \text{cm} - 12 \, \text{cm} = 3 \, \text{cm} \]

### Final Answer
The size of the image of the pin is **3.0 cm**.