Two waves having intensity `I and 9I` produce interference . If the resultant intensity at a point is `7 I`, what is the phase difference between the two waves ?

Options

A. `0^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Explanation

To find the phase difference between the two waves with intensities \( I \) and \( 9I \) that produce a resultant intensity of \( 7I \), we can use the formula for the resultant intensity due to interference:

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

Where:
- \( I_R \) is the resultant intensity,
- \( I_1 \) and \( I_2 \) are the intensities of the two waves,
- \( \phi \) is the phase difference.

### Step 1: Assign the intensities
Let:
- \( I_1 = I \)
- \( I_2 = 9I \)

### Step 2: Substitute the values into the formula
The resultant intensity is given as \( 7I \). Therefore, we can write:

\[ 7I = I + 9I + 2\sqrt{I \cdot 9I} \cos(\phi) \]

### Step 3: Simplify the equation
Combine the intensities on the right side:

\[ 7I = 10I + 2\sqrt{9I^2} \cos(\phi) \]

### Step 4: Calculate the square root term
The square root term simplifies as follows:

\[ \sqrt{9I^2} = 3I \]

Substituting this back into the equation gives:

\[ 7I = 10I + 2 \cdot 3I \cos(\phi) \]
\[ 7I = 10I + 6I \cos(\phi) \]

### Step 5: Rearrange the equation
Now, rearranging the equation to isolate the cosine term:

\[ 7I - 10I = 6I \cos(\phi) \]
\[ -3I = 6I \cos(\phi) \]

### Step 6: Divide by \( I \)
Assuming \( I \neq 0 \), we can divide both sides by \( I \):

\[ -3 = 6 \cos(\phi) \]

### Step 7: Solve for \( \cos(\phi) \)
Now, divide both sides by 6:

\[ \cos(\phi) = -\frac{3}{6} = -\frac{1}{2} \]

### Step 8: Find the phase difference \( \phi \)
The angle whose cosine is \( -\frac{1}{2} \) corresponds to:

\[ \phi = 120^\circ \quad \text{(or } \phi = 240^\circ \text{)} \]

Since we are looking for the principal value, we take:

\[ \phi = 120^\circ \]

### Final Answer
The phase difference between the two waves is \( 120^\circ \).

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