In Young's double slit experiment, when two light waves form third minimum, they have

Options

A. phase difference of `3pi`
B. phase difference of `(5pi)/2`
C. path difference of `3lambda`
D. path difference of `(5lambda)/2`

Explanation

To solve the problem regarding the path difference in Young's double slit experiment when two light waves form the third minimum, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Condition for Minima**:
In Young's double slit experiment, the condition for destructive interference (minima) occurs when the path difference (\( \Delta x \)) between the two waves is an odd multiple of half the wavelength (\( \lambda \)). This can be expressed mathematically as:
\[ \Delta x = \left( n - \frac{1}{2} \right) \lambda \]
where \( n \) is a positive integer (1, 2, 3, ...).

2. **Identifying the Third Minimum**:
For the third minimum, we set \( n = 3 \). Thus, we can substitute \( n \) into the formula:
\[ \Delta x = \left( 3 - \frac{1}{2} \right) \lambda \]

3. **Calculating the Path Difference**:
Simplifying the expression:
\[ \Delta x = \left( 2.5 \right) \lambda = \frac{5}{2} \lambda \]

4. **Conclusion**:
Therefore, the path difference when two light waves form the third minimum in Young's double slit experiment is:
\[ \Delta x = \frac{5\lambda}{2} \]

### Final Answer:
The path difference when two light waves form the third minimum is \( \frac{5\lambda}{2} \).

---