Three harmonic waves having equal frequency and same intensity `I_(0)` have phase angle`- phi, 0 ` & `phi` respectively.When they are superimposed internsity of the resultant wave becomes `4I_(0)`. Find `phi`

Options

A. `0`
B. `(pi)/6`
C. `(pi)/2`
D. `(pi)/3`

Explanation

To solve the problem, we need to find the phase angle \( \phi \) given that three harmonic waves with equal frequency and intensity \( I_0 \) have phase angles of \( -\phi, 0, \) and \( \phi \) respectively, and their superposition results in an intensity of \( 4I_0 \).

### Step-by-Step Solution:

1. **Understanding the Intensity Relation**:
The intensity \( I \) of a wave is related to its amplitude \( A \) by the formula:
\[ I = k A^2 \]
where \( k \) is a constant. Since all three waves have the same intensity \( I_0 \), we can express their amplitudes as:
\[ I_0 = k A^2 \implies A = \sqrt{\frac{I_0}{k}} \]

2. **Phasor Representation**:
We can represent the three waves as phasors:
- Wave 1: Amplitude \( A \) at phase \( 0 \)
- Wave 2: Amplitude \( A \) at phase \( \phi \)
- Wave 3: Amplitude \( A \) at phase \( -\phi \)

3. **Resultant Amplitude Calculation**:
To find the resultant amplitude \( R \) of the three waves, we first find the resultant of waves 1 and 3:
\[ R_{13} = \sqrt{A^2 + A^2 + 2A^2 \cos(2\phi)} = \sqrt{2A^2(1 + \cos(2\phi))} \]
Using the identity \( 1 + \cos(2\phi) = 2\cos^2(\phi) \):
\[ R_{13} = \sqrt{2A^2 \cdot 2\cos^2(\phi)} = 2A \cos(\phi) \]

4. **Total Resultant Amplitude**:
The total resultant amplitude \( R \) when adding the second wave (which has an amplitude \( A \)) is:
\[ R = R_{13} + A = 2A \cos(\phi) + A = A(2\cos(\phi) + 1) \]

5. **Intensity of Resultant Wave**:
The intensity of the resultant wave is given as \( 4I_0 \):
\[ I_{net} = k R^2 = 4I_0 \]
Substituting for \( R \):
\[ k (A(2\cos(\phi) + 1))^2 = 4k A^2 \]
Cancelling \( k \) and \( A^2 \) (since \( A^2 = \frac{I_0}{k} \)):
\[ (2\cos(\phi) + 1)^2 = 4 \]

6. **Solving the Equation**:
Taking the square root:
\[ 2\cos(\phi) + 1 = 2 \quad \text{or} \quad 2\cos(\phi) + 1 = -2 \]
From \( 2\cos(\phi) + 1 = 2 \):
\[ 2\cos(\phi) = 1 \implies \cos(\phi) = \frac{1}{2} \]
This gives:
\[ \phi = \frac{\pi}{3} \]

From \( 2\cos(\phi) + 1 = -2 \):
\[ 2\cos(\phi) = -3 \implies \cos(\phi) = -\frac{3}{2} \]
This is not possible since the cosine function cannot exceed 1.

7. **Final Answer**:
Therefore, the phase angle \( \phi \) is:
\[ \phi = \frac{\pi}{3} \]