The distance between two point sources of light is `24cm.` Find out where would you place a converging lens of focal length `9cm,` so that the images of both the sources are formed at the same point.

Explanation

To solve the problem of finding the position of a converging lens so that the images of two point sources of light are formed at the same point, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Distance between the two point sources (S1 and S2) = 24 cm
- Focal length of the converging lens (f) = 9 cm

2. **Define the Variables:**
- Let the distance from the first source (S1) to the lens be \( x \) cm.
- Therefore, the distance from the second source (S2) to the lens will be \( 24 - x \) cm.

3. **Set Up the Lens Formula:**
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) = focal length of the lens
- \( v \) = image distance from the lens
- \( u \) = object distance from the lens

4. **Apply the Lens Formula for Both Sources:**
- For source S1:
\[ \frac{1}{9} = \frac{1}{-y} - \frac{1}{-x} \quad \text{(Equation 1)} \]
- For source S2:
\[ \frac{1}{9} = \frac{1}{y} - \frac{1}{(24 - x)} \quad \text{(Equation 2)} \]

5. **Rearranging the Equations:**
- From Equation 1:
\[ \frac{1}{y} - \frac{1}{x} = \frac{1}{9} \]
- From Equation 2:
\[ \frac{1}{y} + \frac{1}{(24 - x)} = \frac{1}{9} \]

6. **Add the Two Equations:**
Adding both equations gives:
\[ \left(\frac{1}{y} - \frac{1}{x}\right) + \left(\frac{1}{y} + \frac{1}{(24 - x)}\right) = \frac{1}{9} + \frac{1}{9} \]
Simplifying this:
\[ 2 \cdot \frac{1}{y} = \frac{2}{9} + \frac{1}{x} - \frac{1}{(24 - x)} \]

7. **Combine and Solve:**
Rearranging gives:
\[ \frac{1}{x} + \frac{1}{(24 - x)} = \frac{2}{9} \]
Cross-multiplying:
\[ 24 - x + x = \frac{2 \cdot 24 \cdot 9}{9} \Rightarrow 24 = \frac{48}{9} \Rightarrow 24x = 48 - 2x \]
Rearranging leads to:
\[ 9x^2 - 24x + 108 = 0 \]

8. **Solve the Quadratic Equation:**
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 9 \cdot 108}}{2 \cdot 9} \]
This results in two possible values for \( x \): 6 cm and 18 cm.

9. **Conclusion:**
The converging lens can be placed either 6 cm from source S1 or 18 cm from source S1.