It is found that what waves of same intensity from two coherent sources superpose at a certain point, then the resultant intensity is equal to the intensity of one wave only. This means that the phase difference between the two waves at that point is

Options

A. zero
B. `pi/3`
C. `(2pi)/3`
D. `pi`

Explanation

To solve the problem, we need to determine the phase difference between two coherent waves when their resultant intensity is equal to the intensity of one of the waves. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have two coherent waves coming from two sources.
- Both waves have the same intensity, denoted as \( I_0 \).
- The resultant intensity \( I_R \) at a certain point is equal to the intensity of one wave, which is \( I_0 \).

2. **Using the Formula for Resultant Intensity**:
- The formula for the resultant intensity \( I_R \) when two coherent waves interfere is given by:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
- Since both waves have the same intensity \( I_0 \), we can substitute \( I_1 = I_0 \) and \( I_2 = I_0 \):
\[ I_R = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi \]
\[ I_R = 2I_0 + 2I_0 \cos \phi \]

3. **Setting the Resultant Intensity Equal to One Wave's Intensity**:
- According to the problem, \( I_R = I_0 \). Therefore, we can set up the equation:
\[ 2I_0 + 2I_0 \cos \phi = I_0 \]

4. **Simplifying the Equation**:
- Rearranging the equation gives:
\[ 2I_0 \cos \phi = I_0 - 2I_0 \]
\[ 2I_0 \cos \phi = -I_0 \]
- Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 2 \cos \phi = -1 \]

5. **Solving for Cosine**:
- Dividing both sides by 2:
\[ \cos \phi = -\frac{1}{2} \]

6. **Finding the Phase Difference**:
- The cosine of the angle is \( -\frac{1}{2} \) at specific angles. The general solutions for \( \phi \) where \( \cos \phi = -\frac{1}{2} \) are:
\[ \phi = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad \phi = \frac{4\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]
- The principal value we are interested in is \( \phi = \frac{2\pi}{3} \).

7. **Conclusion**:
- Therefore, the phase difference \( \phi \) between the two waves at that point is:
\[ \phi = \frac{2\pi}{3} \]

### Final Answer:
The phase difference between the two waves at that point is \( \frac{2\pi}{3} \).

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