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 of finding the position of the object so that the final image formed by the combination of a diverging lens and a converging lens is at infinity, we can follow these steps:

### Step 1: Understand the Lens Configuration
We have two lenses:
1. A diverging lens (concave) with a focal length \( f_1 = -20 \) cm (negative because it is diverging).
2. A converging lens (convex) with a focal length \( f_2 = 30 \) cm.

The distance between the two lenses is \( d = 15 \) cm.

### Step 2: Determine the Condition for Image at Infinity
For the final image to be at infinity, the rays coming out of the converging lens must be parallel. This occurs when the object for the converging lens is placed at its focal point. Therefore, the object distance \( u_2 \) for the converging lens must be equal to the focal length \( f_2 \):
\[ u_2 = -f_2 = -30 \text{ cm} \]

### Step 3: Relate Object Distance of Converging Lens to Image Distance of Diverging Lens
The object for the converging lens is the image formed by the diverging lens. The image distance \( v_1 \) for the diverging lens can be calculated using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{u_1} = \frac{1}{v_1} + \frac{1}{f_1} \]
Where:
- \( u_1 \) is the object distance for the diverging lens.
- \( v_1 \) is the image distance for the diverging lens.

### Step 4: Set Up the Relationship
Since the distance between the two lenses is 15 cm, we can relate \( v_1 \) and \( u_2 \):
\[ v_1 = d + u_2 \]
Substituting \( u_2 = -30 \) cm gives:
\[ v_1 = 15 - 30 = -15 \text{ cm} \]

### Step 5: Substitute into the Lens Formula for Diverging Lens
Now we substitute \( v_1 \) and \( f_1 \) into the lens formula:
\[ \frac{1}{u_1} = \frac{1}{-15} + \frac{1}{-20} \]

### Step 6: Calculate \( u_1 \)
Finding a common denominator (60):
\[ \frac{1}{u_1} = -\frac{4}{60} - \frac{3}{60} = -\frac{7}{60} \]
Thus,
\[ u_1 = -\frac{60}{7} \approx -8.57 \text{ cm} \]

### Step 7: Conclusion
The object should be placed approximately 8.57 cm in front of the diverging lens (to the left of it, since the distance is negative).