Two point sources `S_(1)` and `S_(2)` are 24 cm apart. What should a convex lens of focal length 9 cm be placed between them so that the images of both sources formed at the same place ?

Options

A. 6 cm from `S_(1)`
B. 15 cm from `S_(1)`
C. 10 cm from `S_(1)`
D. 12 cm from `S_(1)`

Answer

To solve the problem of determining the position of a convex lens such that the images of two point sources \( S_1 \) and \( S_2 \) are formed at the same place, we can follow these steps:

### Step 1: Understand the Configuration
We have two point sources \( S_1 \) and \( S_2 \) that are 24 cm apart. We need to place a convex lens of focal length 9 cm between them. Let's denote the distance from \( S_1 \) to the lens as \( x \), and consequently, the distance from \( S_2 \) to the lens will be \( 24 - x \).

### Step 2: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
where \( v \) is the image distance, \( u \) is the object distance (taken as negative for real objects), and \( f \) is the focal length of the lens.

### Step 3: Set Up the Equations for Both Sources
1. For the source \( S_1 \):
- The object distance \( u_1 = -x \) (since it is on the same side as the incoming light).
- The image distance \( v_1 \) will be positive if the image is real.
- Using the lens formula:
\[ \frac{1}{v_1} - \frac{1}{-x} = \frac{1}{9} \]
Rearranging gives:
\[ \frac{1}{v_1} + \frac{1}{x} = \frac{1}{9} \quad \text{(1)} \]

2. For the source \( S_2 \):
- The object distance \( u_2 = -(24 - x) \).
- The image distance \( v_2 \) will also be positive if the image is real.
- Using the lens formula:
\[ \frac{1}{v_2} - \frac{1}{-(24 - x)} = \frac{1}{9} \]
Rearranging gives:
\[ \frac{1}{v_2} + \frac{1}{(24 - x)} = \frac{1}{9} \quad \text{(2)} \]

### Step 4: Set the Image Distances Equal
Since we want the images of both sources to be formed at the same place, we set \( v_1 = v_2 \). Thus, we can equate the two equations derived from the lens formula.

### Step 5: Solve the Equations
From equation (1):
\[ \frac{1}{v_1} = \frac{1}{9} - \frac{1}{x} \]
From equation (2):
\[ \frac{1}{v_2} = \frac{1}{9} - \frac{1}{(24 - x)} \]
Setting these equal gives:
\[ \frac{1}{9} - \frac{1}{x} = \frac{1}{9} - \frac{1}{(24 - x)} \]
This simplifies to:
\[ \frac{1}{x} = \frac{1}{(24 - x)} \]
Cross-multiplying gives:
\[ 24 - x = x \]
Solving for \( x \):
\[ 24 = 2x \implies x = 12 \text{ cm} \]

### Conclusion
The convex lens should be placed 12 cm from the source \( S_1 \) to ensure that the images of both sources are formed at the same place.