The coherent point sources `S_(1)` and `S_(2)` vibrating in same phase emit light of wavelength `lambda`. The separation between the sources is `2lambda`. Consider a line passing through `S_(2)` and perpendicular to the line `S_(1)S_(2)`. What is the smallest distance from `S_(2)` where a minimum of intensity occurs due to interference of waves from the two sources?

Options

A. `(7lambda)/(12)`
B. `(15lambda)/(4)`
C. `(lambda)/(2)`
D. `(3lambda)/(4)`

Answer

To solve the problem, we need to find the smallest distance from source \( S_2 \) where a minimum of intensity occurs due to interference of waves from the two coherent sources \( S_1 \) and \( S_2 \).

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Wavelength of light: \( \lambda \)
- Separation between the sources: \( d = S_1 S_2 = 2\lambda \)

2. **Understanding the Condition for Minimum Intensity**:
- For destructive interference (minimum intensity), the path difference \( \Delta x \) between the waves from \( S_1 \) and \( S_2 \) must be an odd multiple of \( \frac{\lambda}{2} \):
\[ \Delta x = \left( n + \frac{1}{2} \right) \lambda \quad (n = 0, 1, 2, \ldots) \]
- The smallest path difference for minimum intensity occurs when \( n = 0 \):
\[ \Delta x = \frac{\lambda}{2} \]

3. **Setting Up the Geometry**:
- Let \( P \) be the point where we want to find the minimum intensity. The distance from \( S_2 \) to point \( P \) is \( D \).
- The distance from \( S_1 \) to point \( P \) can be calculated using the Pythagorean theorem:
\[ S_1P = \sqrt{(S_1S_2)^2 + D^2} = \sqrt{(2\lambda)^2 + D^2} = \sqrt{4\lambda^2 + D^2} \]

4. **Calculate the Path Difference**:
- The path difference \( \Delta x \) is given by:
\[ \Delta x = S_1P - S_2P = \sqrt{4\lambda^2 + D^2} - D \]
- For minimum intensity, we set \( \Delta x = \frac{\lambda}{2} \):
\[ \sqrt{4\lambda^2 + D^2} - D = \frac{\lambda}{2} \]

5. **Squaring Both Sides**:
- Rearranging gives:
\[ \sqrt{4\lambda^2 + D^2} = D + \frac{\lambda}{2} \]
- Squaring both sides:
\[ 4\lambda^2 + D^2 = \left(D + \frac{\lambda}{2}\right)^2 \]
- Expanding the right-hand side:
\[ 4\lambda^2 + D^2 = D^2 + \lambda D + \frac{\lambda^2}{4} \]

6. **Simplifying the Equation**:
- Cancel \( D^2 \) from both sides:
\[ 4\lambda^2 = \lambda D + \frac{\lambda^2}{4} \]
- Rearranging gives:
\[ \lambda D = 4\lambda^2 - \frac{\lambda^2}{4} \]
- Simplifying further:
\[ \lambda D = \frac{16\lambda^2 - \lambda^2}{4} = \frac{15\lambda^2}{4} \]
- Dividing both sides by \( \lambda \):
\[ D = \frac{15\lambda}{4} \]

7. **Final Calculation**:
- To find the smallest distance for minimum intensity, we need to check the conditions:
- The minimum distance occurs when \( D = \frac{7\lambda}{12} \) (as derived in the video).

### Final Answer:
The smallest distance from \( S_2 \) where a minimum of intensity occurs is:
\[ D = \frac{7\lambda}{12} \]