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

Answer

To solve the problem of finding the resulting intensity when two plane progressive waves superimpose, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Intensity of the first wave, \( I_1 = 9I_0 \)
- Intensity of the second wave, \( I_2 = 4I_0 \)
- Wavelength, \( \lambda \) (same for both waves)
- Path difference, \( \Delta x = \frac{\lambda}{4} \)

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

3. **Use the Formula for Resulting Intensity:**
- The resulting intensity \( I \) when two waves superimpose is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- Substitute the values of \( I_1 \), \( I_2 \), and \( \phi \):
\[ I = 9I_0 + 4I_0 + 2\sqrt{9I_0 \cdot 4I_0} \cos\left(\frac{\pi}{2}\right) \]

4. **Calculate Each Term:**
- First, calculate \( I_1 + I_2 \):
\[ I_1 + I_2 = 9I_0 + 4I_0 = 13I_0 \]
- Next, calculate \( 2\sqrt{I_1 I_2} \):
\[ 2\sqrt{9I_0 \cdot 4I_0} = 2\sqrt{36I_0^2} = 2 \cdot 6I_0 = 12I_0 \]
- Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), the term involving cosine becomes:
\[ 12I_0 \cdot 0 = 0 \]

5. **Final Calculation of Resulting Intensity:**
- Therefore, the resulting intensity is:
\[ I = 13I_0 + 0 = 13I_0 \]

### Conclusion:
The resulting intensity when the path difference between the waves becomes \( \frac{\lambda}{4} \) is \( 13I_0 \).