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 solve the problem step by step, we will use the formula for the resultant intensity of two interfering waves.

### Step 1: Understand the Given Information
We have two waves with intensities:
- \( I_1 = I \)
- \( I_2 = 9I \)

The resultant intensity \( I \) at a point is given as:
- \( I = 7I \)

### Step 2: Write the Formula for Resultant Intensity
The formula for the resultant intensity \( I \) when two waves interfere is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
where \( \phi \) is the phase difference between the two waves.

### Step 3: Substitute the Given Values into the Formula
Substituting the values of \( I_1 \), \( I_2 \), and \( I \) into the formula:
\[ 7I = I + 9I + 2\sqrt{I \cdot 9I} \cos(\phi) \]

### Step 4: Simplify the Equation
Simplifying the left side:
\[ 7I = 10I + 2\sqrt{9I^2} \cos(\phi) \]
\[ 7I = 10I + 6I \cos(\phi) \]

### Step 5: Rearranging the Equation
Now, rearranging the equation to isolate the term involving \( \cos(\phi) \):
\[ 7I - 10I = 6I \cos(\phi) \]
\[ -3I = 6I \cos(\phi) \]

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

### Step 7: Find the Phase Difference \( \phi \)
The value of \( \phi \) for which \( \cos(\phi) = -\frac{1}{2} \) is:
\[ \phi = 120^\circ \quad \text{or} \quad \phi = 240^\circ \]
However, since we are typically interested in the principal value, we take:
\[ \phi = 120^\circ \]

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

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