Two waves of intensity I and 9I are superimposed in such a way that resultant Intensity is 7I .Find the phase difference between them ?

Answer

To solve the problem of finding the phase difference between two waves of intensities \( I_1 = I \) and \( I_2 = 9I \) that result in a resultant intensity \( I_r = 7I \), we can follow these steps:

### Step 1: Write down the formula for resultant intensity
The formula for the resultant intensity when two waves interfere is given by:
\[ I_r = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi \]

### Step 2: Substitute the known values
From the problem, we have:
- \( I_1 = I \)
- \( I_2 = 9I \)
- \( I_r = 7I \)

Substituting these values into the formula:
\[ 7I = I + 9I + 2 \sqrt{I \cdot 9I} \cos \phi \]

### Step 3: Simplify the equation
Combine the intensities on the right side:
\[ 7I = 10I + 2 \sqrt{9I^2} \cos \phi \]
This simplifies to:
\[ 7I = 10I + 6I \cos \phi \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation to isolate the cosine term:
\[ 7I - 10I = 6I \cos \phi \]
\[ -3I = 6I \cos \phi \]

### Step 5: Solve for \( \cos \phi \)
Dividing both sides by \( 6I \) (assuming \( I \neq 0 \)):
\[ \cos \phi = -\frac{3}{6} = -\frac{1}{2} \]

### Step 6: Find the phase difference \( \phi \)
The cosine of the phase difference is \( -\frac{1}{2} \). The angle \( \phi \) that corresponds to this value is:
\[ \phi = 120^\circ \quad \text{(or } 240^\circ \text{, but we typically take the principal value)} \]

### Final Answer
Thus, the phase difference between the two waves is:
\[ \phi = 120^\circ \]

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