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

Answer

To solve the problem step by step, we will analyze the situation involving two identical sound sources \( S_1 \) and \( S_2 \) and how their intensities change when the power of one source is reduced and the phase difference is varied.

### Step 1: Determine the Initial Intensity
Initially, when both sources \( S_1 \) and \( S_2 \) are in phase, the intensity at point \( P \) is given as \( I_0 \). Since both sources are identical, we can express this as:
\[ I_0 = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
Given \( I_1 = I_2 = I \), we have:
\[ I_0 = 2I + 2\sqrt{I^2} = 2I + 2I = 4I \]
Thus, the initial intensity \( I \) can be expressed as:
\[ I = \frac{I_0}{4} \]

### Step 2: Calculate the New Intensity of \( S_1 \)
The power of \( S_1 \) is reduced by 64%. Therefore, the new intensity \( I_1' \) can be calculated as:
\[ I_1' = I_1 \times (1 - 0.64) = I \times 0.36 = 0.36I \]

### Step 3: Calculate Maximum Intensity \( I_{\text{max}} \)
The maximum intensity when the phase difference is varied is given by:
\[ I_{\text{max}} = I_1' + I_2 + 2\sqrt{I_1' I_2} \]
Substituting \( I_1' = 0.36I \) and \( I_2 = I \):
\[ I_{\text{max}} = 0.36I + I + 2\sqrt{0.36I \cdot I} \]
Calculating this gives:
\[ I_{\text{max}} = 0.36I + I + 2 \cdot 0.6I = 1.36I + 1.2I = 2.56I \]

### Step 4: Calculate Minimum Intensity \( I_{\text{min}} \)
The minimum intensity is given by:
\[ I_{\text{min}} = I_1' + I_2 - 2\sqrt{I_1' I_2} \]
Substituting the values:
\[ I_{\text{min}} = 0.36I + I - 2\sqrt{0.36I \cdot I} \]
Calculating this gives:
\[ I_{\text{min}} = 0.36I + I - 1.2I = 1.36I - 1.2I = 0.16I \]

### Step 5: Find the Ratio \( \frac{I_{\text{max}}}{I_{\text{min}}} \)
Now, we can find the ratio of maximum to minimum intensity:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{2.56I}{0.16I} = \frac{2.56}{0.16} = 16 \]

### Step 6: Compare \( I_{\text{max}} \) and \( I_{\text{min}} \) with \( I_0 \)
- Since \( I_0 = 4I \):
- For \( I_{\text{max}} \):
\[ I_{\text{max}} = 2.56I = 0.64I_0 \]
- For \( I_{\text{min}} \):
\[ I_{\text{min}} = 0.16I = 0.04I_0 \]

### Conclusion
The results are:
- \( I_{\text{max}} = 0.64I_0 \)
- \( I_{\text{min}} = 0.04I_0 \)
- The ratio \( \frac{I_{\text{max}}}{I_{\text{min}}} = 16 \)