A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image formed at infinity ?

Explanation

To solve the problem, we need to determine where to place an object on the principal axis so that the final image formed by the combination of a diverging lens and a converging lens is at infinity.

### Step-by-Step Solution:

1. **Identify the Focal Lengths**:
- The focal length of the diverging lens (f1) is -20 cm (negative because it is a diverging lens).
- The focal length of the converging lens (f2) is +30 cm.

2. **Distance Between the Lenses**:
- The distance (d) between the two lenses is given as 15 cm.

3. **Use the Lens Formula**:
The lens formula for a combination of two lenses is given by:
\[ \frac{1}{f_{total}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]

4. **Calculate the Total Focal Length**:
Substituting the values into the formula:
\[ \frac{1}{f_{total}} = \frac{1}{-20} + \frac{1}{30} - \frac{15}{(-20)(30)} \]

First, calculate each term:
- \(\frac{1}{-20} = -0.05\)
- \(\frac{1}{30} = 0.0333\)
- \(-\frac{15}{(-20)(30)} = \frac{15}{600} = 0.025\)

Now, combine these:
\[ \frac{1}{f_{total}} = -0.05 + 0.0333 + 0.025 = 0.0083 \]
Therefore,
\[ f_{total} \approx 120 \text{ cm} \]

5. **Determine the Object Distance for the Diverging Lens**:
To form an image at infinity, the object must be placed at the focal point of the diverging lens. The object distance (u1) from the diverging lens can be calculated using the lens formula:
\[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1} \]
Here, \(v_1\) (the image distance from the diverging lens) should be at the focal point of the converging lens, which is 30 cm away from the converging lens. Thus, the image formed by the diverging lens must be at a distance of \(15 + 30 = 45\) cm from the diverging lens.

Rearranging the lens formula gives:
\[ \frac{1}{-20} = \frac{1}{45} - \frac{1}{u_1} \]
Solving for \(u_1\):
\[ -0.05 = \frac{1}{45} - \frac{1}{u_1} \]
\[ -0.05 + 0.0222 = -\frac{1}{u_1} \]
\[ -0.0278 = -\frac{1}{u_1} \]
Therefore,
\[ u_1 \approx 36 \text{ cm} \]

6. **Calculate the Object Distance from the Converging Lens**:
The object distance from the converging lens (u2) can be calculated as:
\[ u_2 = u_1 + d = 36 + 15 = 51 \text{ cm} \]

### Final Answer:
The object should be placed approximately **36 cm from the diverging lens** or **51 cm from the converging lens** to ensure that the final image is formed at infinity.