Two identical light waves, propagating in the same direction, have a phase difference `delta`. After they superpose the intensity of the resulting wave will be proportional to

Options

A. `cosdelta`
B. `cos(delta//2)`
C. `cos^(2)(delta//2)`
D. `cos^(2)delta`

Explanation

To solve the problem of determining the intensity of the resulting wave when two identical light waves with a phase difference \(\delta\) superpose, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Intensities of Individual Waves**:
Let the intensity of each identical light wave be \(I_1 = I_0\) and \(I_2 = I_0\).

2. **Use the Formula for Resultant Intensity**:
The formula for the resultant intensity \(I_R\) when two waves superpose is given by:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\delta) \]

3. **Substitute the Intensities**:
Since both waves have the same intensity:
\[ I_R = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(\delta) \]

4. **Simplify the Expression**:
This simplifies to:
\[ I_R = 2I_0 + 2I_0 \cos(\delta) \]

5. **Factor Out Common Terms**:
We can factor out \(2I_0\):
\[ I_R = 2I_0(1 + \cos(\delta)) \]

6. **Use the Cosine Double Angle Identity**:
We can use the identity \(1 + \cos(\delta) = 2\cos^2(\delta/2)\):
\[ I_R = 2I_0 \cdot 2\cos^2(\delta/2) \]
Thus,
\[ I_R = 4I_0 \cos^2(\delta/2) \]

7. **Determine Proportionality**:
The intensity \(I_R\) is therefore proportional to \(\cos^2(\delta/2)\):
\[ I_R \propto \cos^2(\delta/2) \]

### Final Result:
The intensity of the resulting wave will be proportional to \(\cos^2(\delta/2)\).

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