A detector at `x=0` receives waves from three sources each of amplitude `A` and frequencies `f + 2,f` and `f-2`.
If `I_(0) prop A^(2)`, then the value of maximum intensity,is

Options

A. `2I_(0)`
B. `3 I_(0)`
C. `4 I_(0)`
D. `9 I_(0)`

Explanation

To solve the problem, we need to determine the maximum intensity at the detector located at \( x = 0 \) when it receives waves from three sources with given amplitudes and frequencies.

### Step-by-Step Solution:

1. **Identify the Frequencies and Amplitudes:**
- The three sources have frequencies \( f + 2 \), \( f \), and \( f - 2 \).
- Each source has an amplitude \( A \).

2. **Understand the Superposition of Waves:**
- When waves from different sources interfere, the resultant amplitude can be calculated using the principle of superposition.
- The intensity \( I \) of a wave is proportional to the square of its amplitude: \( I \propto A^2 \).

3. **Calculate the Resultant Amplitude:**
- The resultant amplitude \( A_{r} \) when three waves interfere can be calculated as follows:
\[ A_{r} = A_1 + A_2 + A_3 \]
- Since all sources have the same amplitude \( A \), we have:
\[ A_{r} = A + A + A = 3A \]

4. **Calculate the Maximum Intensity:**
- The maximum intensity \( I_{max} \) is given by:
\[ I_{max} \propto (A_{r})^2 = (3A)^2 = 9A^2 \]

5. **Relate Maximum Intensity to Initial Intensity:**
- The initial intensity \( I_{0} \) is proportional to \( A^2 \):
\[ I_{0} \propto A^2 \]
- Therefore, we can express \( I_{max} \) in terms of \( I_{0} \):
\[ I_{max} = 9 \times I_{0} \]

6. **Conclusion:**
- The maximum intensity at the detector is \( 9I_{0} \).

### Final Answer:
The value of maximum intensity is \( 9I_{0} \).

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