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

Options

A. (a) Phase difference of `3pi`
B. (b) Phase difference of `5pi//2`
C. (c) Path difference of `3lambda`
D. (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) is given by the formula:
\[ \text{Path difference} = \frac{(2n - 1) \lambda}{2} \]
where \( n \) is the order of the minimum (1st, 2nd, 3rd, etc.) and \( \lambda \) is the wavelength of the light used.

2. **Identifying the Order of the Minimum**:
For the third minimum, we set \( n = 3 \).

3. **Substituting the Value of \( n \)**:
Now, substituting \( n = 3 \) into the formula for path difference:
\[ \text{Path difference} = \frac{(2 \cdot 3 - 1) \lambda}{2} \]

4. **Calculating the Path Difference**:
Simplifying the expression:
\[ \text{Path difference} = \frac{(6 - 1) \lambda}{2} = \frac{5\lambda}{2} \]

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

### Final Answer:
The path difference for the third minimum is \( \frac{5\lambda}{2} \).