Two coherent point sources `S_1 and S_2` emit light of wavelength `lambda`. The separation between sources is `2lambda`. Consider a line passing through `S_2` and perpendicular to the line `S_(1)S_(2)`. What is the smallest distance on this line from `S_2` where a minimum of intensity occurs?

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two coherent point sources \( S_1 \) and \( S_2 \) separated by a distance \( d = 2\lambda \).
- A line is drawn through \( S_2 \) perpendicular to the line \( S_1 S_2 \).

2. **Path Difference for Minima**:
- For destructive interference (minimum intensity), the path difference between the waves from \( S_1 \) and \( S_2 \) must be an odd multiple of half the wavelength:
\[ \Delta = S_1P - S_2P = (2n - 1) \frac{\lambda}{2} \]
- Here, \( n \) is an integer (0, 1, 2, ...).

3. **Calculating Distances**:
- Let \( d \) be the distance from \( S_2 \) to the point \( P \) on the perpendicular line.
- The distance \( S_1P \) can be calculated using the Pythagorean theorem:
\[ S_1P = \sqrt{(2\lambda)^2 + d^2} = \sqrt{4\lambda^2 + d^2} \]
- The distance \( S_2P \) is simply \( d \).

4. **Setting Up the Equation**:
- For the first minimum (taking \( n = 1 \)):
\[ S_1P - S_2P = \frac{\lambda}{2} \]
- Substituting the distances:
\[ \sqrt{4\lambda^2 + d^2} - d = \frac{\lambda}{2} \]

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 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:
\[ 4\lambda^2 - \frac{\lambda^2}{4} = \lambda d \]
- Finding a common denominator:
\[ \frac{16\lambda^2}{4} - \frac{\lambda^2}{4} = \lambda d \]
- Simplifying:
\[ \frac{15\lambda^2}{4} = \lambda d \]

7. **Solving for \( d \)**:
- Dividing both sides by \( \lambda \):
\[ d = \frac{15\lambda}{4} \]

### Final Answer:
The smallest distance on the line from \( S_2 \) where a minimum of intensity occurs is:
\[ d = \frac{15\lambda}{4} \]