If two waves each of intensity `l_0` having the same frequency but differing by a constant phase angle of `60^@` superimposing at a certain point in space, then the intensity of the resultant wave is

Options

A. `2l_0`
B. `3l_0`
C. `sqrt3l_0`
D. `4l_0`

Answer

To find the intensity of the resultant wave when two waves of equal intensity \( I_0 \) and a phase difference of \( 60^\circ \) superimpose, we can use the formula for the intensity of the resultant wave from two waves:

\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]

Where:
- \( I_1 \) and \( I_2 \) are the intensities of the individual waves.
- \( \phi \) is the phase difference between the two waves.

### Step-by-Step Solution:

1. **Identify the Intensities**:
Given that both waves have the same intensity \( I_0 \):
\[ I_1 = I_0, \quad I_2 = I_0 \]

2. **Identify the Phase Difference**:
The phase difference \( \phi \) is given as \( 60^\circ \).

3. **Substitute Values into the Formula**:
Substitute \( I_1 \), \( I_2 \), and \( \phi \) into the intensity formula:
\[ I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(60^\circ) \]

4. **Calculate \( \cos(60^\circ) \)**:
We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]

5. **Substitute \( \cos(60^\circ) \) into the Equation**:
\[ I = I_0 + I_0 + 2\sqrt{I_0^2} \cdot \frac{1}{2} \]
\[ I = I_0 + I_0 + 2I_0 \cdot \frac{1}{2} \]

6. **Simplify the Expression**:
\[ I = I_0 + I_0 + I_0 \]
\[ I = 3I_0 \]

### Final Result:
The intensity of the resultant wave is:
\[ \boxed{3I_0} \]