Three waves of equal frequency having amplitudes `10mum`, `4mum` and `7mum` arrive at a given point with successive phase difference of `pi//2`, the amplitude of the resulting wave in `mum` is given by

Options

A. `4`
B. `5`
C. `6`
D. `7`

Explanation

To solve the problem of finding the amplitude of the resulting wave from three waves with given amplitudes and phase differences, we can follow these steps:

### Step 1: Identify the Given Values
We have three waves with the following amplitudes:
- \( A_1 = 10 \, \mu m \)
- \( A_2 = 4 \, \mu m \)
- \( A_3 = 7 \, \mu m \)

The phase differences between the waves are:
- Between \( A_1 \) and \( A_2 \): \( \phi_{12} = \frac{\pi}{2} \)
- Between \( A_2 \) and \( A_3 \): \( \phi_{23} = \frac{\pi}{2} \)

### Step 2: Combine the First Two Waves
To find the resultant amplitude of the first two waves \( A_1 \) and \( A_2 \), we use the formula for the resultant amplitude when two waves are combined with a phase difference:
\[ A_{net1} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi_{12})} \]
Substituting the values:
\[ A_{net1} = \sqrt{10^2 + 4^2 + 2 \cdot 10 \cdot 4 \cdot \cos\left(\frac{\pi}{2}\right)} \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ A_{net1} = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116} = \sqrt{4 \cdot 29} = 2\sqrt{29} \, \mu m \]

### Step 3: Combine the Resultant Amplitude with the Third Wave
Now, we need to combine \( A_{net1} \) with the third wave \( A_3 \):
\[ A_{final} = \sqrt{A_{net1}^2 + A_3^2 + 2A_{net1}A_3 \cos(\phi_{23})} \]
Substituting the values:
\[ A_{final} = \sqrt{(2\sqrt{29})^2 + 7^2 + 2 \cdot (2\sqrt{29}) \cdot 7 \cdot \cos\left(\frac{\pi}{2}\right)} \]
Again, since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ A_{final} = \sqrt{(2\sqrt{29})^2 + 7^2} = \sqrt{4 \cdot 29 + 49} = \sqrt{116 + 49} = \sqrt{165} \]

### Step 4: Calculate the Final Amplitude
Calculating \( \sqrt{165} \):
\[ A_{final} \approx 12.845 \, \mu m \]

### Conclusion
Thus, the amplitude of the resulting wave is approximately \( 12.845 \, \mu m \).