When interference is produced by two progressive waves of equal frequencies, then the maximum intensity of the resulting sound are `N` times the intensity of each of the component waves. The value of `N` is

Options

A. `(a)1`
B. `(b)2`
C. `(c)4`
D. `(d)8`

Explanation

To solve the problem, we need to find the value of \( N \) when two progressive waves of equal frequencies interfere, resulting in a maximum intensity that is \( N \) times the intensity of each component wave.

### Step-by-Step Solution:

1. **Define the Intensities of the Waves:**
Let the intensity of the first wave be \( I_1 \) and the intensity of the second wave be \( I_2 \). According to the problem, both waves have equal intensities. We can denote this common intensity as:
\[ I_1 = I_2 = I_0 \]

2. **Use the Formula for Maximum Intensity:**
The maximum intensity \( I_{\text{max}} \) when two waves interfere is given by the formula:
\[ I_{\text{max}} = \left( \sqrt{I_1} + \sqrt{I_2} \right)^2 \]

3. **Substitute the Intensities:**
Since \( I_1 \) and \( I_2 \) are both equal to \( I_0 \), we can substitute these values into the formula:
\[ I_{\text{max}} = \left( \sqrt{I_0} + \sqrt{I_0} \right)^2 \]

4. **Simplify the Expression:**
Simplifying the expression gives:
\[ I_{\text{max}} = \left( 2\sqrt{I_0} \right)^2 = 4I_0 \]

5. **Relate Maximum Intensity to \( N \):**
According to the problem, the maximum intensity \( I_{\text{max}} \) is also equal to \( N \) times the intensity of each component wave:
\[ I_{\text{max}} = N \cdot I_0 \]

6. **Set the Two Expressions for Maximum Intensity Equal:**
Now we can set the two expressions for \( I_{\text{max}} \) equal to each other:
\[ 4I_0 = N \cdot I_0 \]

7. **Solve for \( N \):**
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ N = 4 \]

### Final Answer:
Thus, the value of \( N \) is \( 4 \).

---