A pin of length `1cm` lies along the principle axis of a converging lens, the centre being at a distance of `5,5cm` form the lens. The focal length of the lens is `3 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 1 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_o) = 1 cm
- Distance of the pin's center from the lens (u) = -5.5 cm (the negative sign indicates the object is on the same side as the incoming light)
- Focal length of the lens (f) = 3 cm (positive for a converging lens)

### Step 2: Determine the distances of the ends of the pin from the lens
Since the pin is 1 cm long and its center is at -5.5 cm from the lens:
- Distance of the top end (P) from the lens (u_P) = -5.5 cm + 0.5 cm = -5 cm
- Distance of the bottom end (Q) from the lens (u_Q) = -5.5 cm - 0.5 cm = -6 cm

### Step 3: Use the lens formula to find the image distances
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
We can rearrange this to find \(v\):
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

#### For the top end (P):
Using \(u_P = -5\) cm:
\[ \frac{1}{v_P} = \frac{1}{3} + \frac{1}{-5} \]
Finding a common denominator (15):
\[ \frac{1}{v_P} = \frac{5}{15} - \frac{3}{15} = \frac{2}{15} \]
Thus,
\[ v_P = \frac{15}{2} = 7.5 \text{ cm} \]

#### For the bottom end (Q):
Using \(u_Q = -6\) cm:
\[ \frac{1}{v_Q} = \frac{1}{3} + \frac{1}{-6} \]
Finding a common denominator (6):
\[ \frac{1}{v_Q} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} \]
Thus,
\[ v_Q = 6 \text{ cm} \]

### Step 4: Calculate the size of the image
The size of the image can be calculated using the magnification formula:
\[ \text{Magnification} (m) = \frac{h_i}{h_o} = -\frac{v}{u} \]
Where \(h_i\) is the height of the image and \(h_o\) is the height of the object.

#### For the top end (P):
Using \(v_P = 7.5\) cm and \(u_P = -5\) cm:
\[ m_P = -\frac{7.5}{-5} = 1.5 \]
Thus,
\[ h_{i_P} = m_P \cdot h_o = 1.5 \cdot 1 = 1.5 \text{ cm} \]

#### For the bottom end (Q):
Using \(v_Q = 6\) cm and \(u_Q = -6\) cm:
\[ m_Q = -\frac{6}{-6} = 1 \]
Thus,
\[ h_{i_Q} = m_Q \cdot h_o = 1 \cdot 1 = 1 \text{ cm} \]

### Step 5: Find the size of the image
The total height of the image will be the difference in heights of the image formed by the top and bottom ends:
\[ \text{Size of the image} = h_{i_P} - h_{i_Q} = 1.5 \text{ cm} - 1 \text{ cm} = 0.5 \text{ cm} \]

### Final Answer
The size of the image is **0.5 cm**.

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