Sounds from two identical sources `S_(1)` and `S_(2)` reach a point `P`. When the sounds reach directly, and in the same phase, the intensity at `P` is `I_(0)`. The power of `S_(1)` is now reduced by `64%` and the phase difference between `S_(1)` and `S_(2)` is varied continuously. The maximum and minimum intensities recorded at `P` are now `I_("max")` and `I_("min")`

Options

A. `I_(max) = 0.64 I_(0)`
B. `I_(min) = 0.36 I_(0)`
C. `(I_(max))/(I_(min)) = 16`
D. `(I_(max))/(I_(min)) = (1.64)/(0.36)`

Explanation

To solve the problem, we need to determine the maximum and minimum intensities at point P when the power of source S1 is reduced by 64% and the phase difference between S1 and S2 is varied continuously.

### Step-by-Step Solution:

1. **Understanding Initial Conditions:**
- When both sources S1 and S2 are in phase, the intensity at point P is given as \( I_0 \).
- Since both sources are identical, we can denote their amplitudes as \( A \).

2. **Calculating Initial Intensity:**
- The intensity \( I_0 \) when both sources are in phase is given by:
\[ I_0 = k (A + A)^2 = k (2A)^2 = 4kA^2 \]
- Here, \( k \) is a constant that relates intensity to amplitude.

3. **Power Reduction of Source S1:**
- The power of S1 is reduced by 64%. Therefore, the remaining power is 36% of the original.
- Since power is proportional to the square of the amplitude, the new amplitude \( A' \) of S1 can be calculated as:
\[ A' = A \sqrt{0.36} = 0.6A \]

4. **New Amplitudes:**
- The amplitude of S2 remains \( A \).
- The new amplitudes are:
- For S1: \( A' = 0.6A \)
- For S2: \( A = A \)

5. **Calculating Maximum Intensity:**
- The maximum intensity occurs when the waves from S1 and S2 are in phase. The resultant amplitude \( A_{max} \) is:
\[ A_{max} = A + A' = A + 0.6A = 1.6A \]
- The maximum intensity \( I_{max} \) is then:
\[ I_{max} = k (A_{max})^2 = k (1.6A)^2 = 2.56kA^2 \]

6. **Calculating Minimum Intensity:**
- The minimum intensity occurs when the waves from S1 and S2 are out of phase. The resultant amplitude \( A_{min} \) is:
\[ A_{min} = A - A' = A - 0.6A = 0.4A \]
- The minimum intensity \( I_{min} \) is then:
\[ I_{min} = k (A_{min})^2 = k (0.4A)^2 = 0.16kA^2 \]

7. **Finding the Ratios:**
- We can express \( I_{max} \) and \( I_{min} \) in terms of \( I_0 \):
\[ I_0 = 4kA^2 \]
- Thus,
\[ I_{max} = 2.56kA^2 = \frac{2.56}{4} I_0 = 0.64 I_0 \]
\[ I_{min} = 0.16kA^2 = \frac{0.16}{4} I_0 = 0.04 I_0 \]

### Final Results:
- The maximum intensity \( I_{max} = 0.64 I_0 \)
- The minimum intensity \( I_{min} = 0.04 I_0 \)