Two plane progressive waves having the same wavelength `lambda` And same frequencies with intensities `9I_0` and `4I_0` Suprimpose. Resulting intensity when the path difference between waves become `(lambda)/(4)` is

Options

A. `13 I_0`
B. `14I_(0)`
C. `15I_(0)`
D. `16I_(0)`

Answer

To solve the problem of finding the resulting intensity when two plane progressive waves superimpose with given intensities and a specific path difference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Intensities**:
- Let the intensities of the two waves be \( I_1 = 9I_0 \) and \( I_2 = 4I_0 \).

2. **Use the Formula for Resultant Intensity**:
The formula for the resultant intensity \( I \) when two waves interfere is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
where \( \phi \) is the phase difference between the two waves.

3. **Calculate the Phase Difference**:
The path difference \( \Delta x \) is given as \( \frac{\lambda}{4} \). The phase difference \( \phi \) can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \Delta x = \frac{\lambda}{4} \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]

4. **Calculate \( \cos \phi \)**:
Now, we find \( \cos \phi \):
\[ \cos \left(\frac{\pi}{2}\right) = 0 \]

5. **Substitute Values into the Resultant Intensity Formula**:
Substitute \( I_1 \), \( I_2 \), and \( \cos \phi \) into the resultant intensity formula:
\[ I = 9I_0 + 4I_0 + 2\sqrt{9I_0 \cdot 4I_0} \cdot 0 \]
Since \( \cos \phi = 0 \), the last term becomes zero:
\[ I = 9I_0 + 4I_0 + 0 = 13I_0 \]

6. **Final Result**:
The resulting intensity when the path difference between the waves is \( \frac{\lambda}{4} \) is:
\[ I = 13I_0 \]