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 solve the problem of finding the intensity of the resultant wave when two waves of equal intensity \( I_0 \) interfere with a phase difference of \( 60^\circ \), we can follow these steps:

### Step 1: Understand the formula for resultant intensity
The intensity of the resultant wave when two waves interfere can be calculated using the formula:
\[ I_{\text{resultant}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
where:
- \( I_1 \) and \( I_2 \) are the intensities of the two waves,
- \( \phi \) is the phase difference between the two waves.

### Step 2: Substitute the values
In this case, both waves have the same intensity \( I_0 \). Thus, we can substitute \( I_1 = I_0 \) and \( I_2 = I_0 \) into the formula:
\[ I_{\text{resultant}} = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(60^\circ) \]

### Step 3: Simplify the equation
Now, simplify the equation:
\[ I_{\text{resultant}} = 2I_0 + 2\sqrt{I_0^2} \cos(60^\circ) \]
Since \( \sqrt{I_0^2} = I_0 \), we can rewrite it as:
\[ I_{\text{resultant}} = 2I_0 + 2I_0 \cos(60^\circ) \]

### Step 4: Calculate \( \cos(60^\circ) \)
We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
Substituting this value into the equation gives:
\[ I_{\text{resultant}} = 2I_0 + 2I_0 \cdot \frac{1}{2} \]

### Step 5: Final simplification
Now simplify the expression:
\[ I_{\text{resultant}} = 2I_0 + I_0 = 3I_0 \]

### Conclusion
Thus, the intensity of the resultant wave is:
\[ \boxed{3I_0} \]

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