A 5 mm high pin is placed at a distance of 15 cm from a convex lens of focal length 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.

Explanation

To solve the problem step by step, we will break it down into parts (a), (b), and (c) as requested.

### Given Data:
- Height of the pin (object) \( h_o = 5 \, \text{mm} = 0.5 \, \text{cm} \)
- Distance of the pin from lens 1 \( u_1 = -15 \, \text{cm} \) (negative as per sign convention)
- Focal length of lens 1 \( f_1 = 10 \, \text{cm} \)
- Distance between lens 1 and lens 2 = 40 cm
- Focal length of lens 2 \( f_2 = 5 \, \text{cm} \)

### Part (a): Position of the Final Image

1. **Find the image formed by lens 1:**
We use the lens formula:
\[ \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \]
Substituting the values:
\[ \frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10} \]
Rearranging gives:
\[ \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15} \]
Finding a common denominator (30):
\[ \frac{1}{v_1} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \]
Therefore:
\[ v_1 = 30 \, \text{cm} \]
This means the image formed by lens 1 is located 30 cm to the right of lens 1.

2. **Determine the object distance for lens 2:**
The distance from lens 1 to lens 2 is 40 cm, so the object distance for lens 2 \( u_2 \) is:
\[ u_2 = 40 \, \text{cm} - 30 \, \text{cm} = 10 \, \text{cm} \]
Since the image formed by lens 1 acts as an object for lens 2, we take:
\[ u_2 = -10 \, \text{cm} \, (\text{negative as per sign convention}) \]

3. **Find the image formed by lens 2:**
Using the lens formula for lens 2:
\[ \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \]
Substituting the values:
\[ \frac{1}{v_2} - \frac{1}{-10} = \frac{1}{5} \]
Rearranging gives:
\[ \frac{1}{v_2} = \frac{1}{5} - \frac{1}{10} \]
Finding a common denominator (10):
\[ \frac{1}{v_2} = \frac{2}{10} - \frac{1}{10} = \frac{1}{10} \]
Therefore:
\[ v_2 = 10 \, \text{cm} \]
This means the final image is located 10 cm to the right of lens 2.

4. **Calculate the position of the final image from the pin:**
The total distance from the pin to the final image is:
\[ 40 \, \text{cm} + 10 \, \text{cm} = 50 \, \text{cm} \]

### Part (b): Nature of the Final Image
Since the final image is formed by lens 2 and we will calculate magnification to determine its nature.

### Part (c): Size of the Final Image

1. **Calculate the magnification for lens 1:**
\[ m_1 = \frac{v_1}{u_1} = \frac{30}{-15} = -2 \]
The negative sign indicates that the image is inverted.

2. **Calculate the height of the image formed by lens 1:**
\[ h_1 = m_1 \cdot h_o = -2 \cdot 0.5 \, \text{cm} = -1 \, \text{cm} \]
(The height is negative, indicating it is inverted.)

3. **Calculate the magnification for lens 2:**
\[ m_2 = \frac{v_2}{u_2} = \frac{10}{-10} = -1 \]
The negative sign indicates that the image is inverted.

4. **Calculate the height of the final image:**
\[ h_2 = m_2 \cdot h_1 = -1 \cdot (-1 \, \text{cm}) = 1 \, \text{cm} \]

### Summary of Results:
- (a) Position of the final image: **50 cm from the pin**
- (b) Nature of the final image: **Erect**
- (c) Size of the final image: **1 cm**