The superposition takes place between two waves of frequency f and amplitude a . The total intensity is directly proportional to

Answer

To solve the problem regarding the superposition of two waves with the same frequency \( f \) and amplitude \( a \), we need to determine how the total intensity is related to these parameters.

### Step-by-Step Solution:

1. **Identify the Amplitudes**:
Given that both waves have the same amplitude \( a \), we can denote the amplitudes of the two waves as \( a_1 = a \) and \( a_2 = a \).

2. **Resultant Amplitude Calculation**:
The resultant amplitude \( R \) when two waves interfere is given by the formula:
\[ R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi} \]
Substituting \( a_1 \) and \( a_2 \) with \( a \):
\[ R = \sqrt{a^2 + a^2 + 2a \cdot a \cos \phi} \]
This simplifies to:
\[ R = \sqrt{2a^2 + 2a^2 \cos \phi} \]
\[ R = \sqrt{2a^2(1 + \cos \phi)} \]

3. **Using Trigonometric Identity**:
We can use the trigonometric identity \( 1 + \cos \phi = 2 \cos^2(\phi/2) \):
\[ R = \sqrt{2a^2 \cdot 2 \cos^2(\phi/2)} = \sqrt{4a^2 \cos^2(\phi/2)} \]
Thus, we have:
\[ R = 2a \cos(\phi/2) \]

4. **Intensity Relation**:
The intensity \( I \) of a wave is proportional to the square of its amplitude:
\[ I \propto R^2 \]
Substituting the expression for \( R \):
\[ I \propto (2a \cos(\phi/2))^2 = 4a^2 \cos^2(\phi/2) \]

5. **Final Result**:
Since \( \cos^2(\phi/2) \) is a constant factor depending on the phase difference, we conclude that the total intensity is directly proportional to:
\[ I \propto 4a^2 \]

### Conclusion:
The total intensity of the superposition of the two waves is directly proportional to \( 4a^2 \).

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