Intensities of the two waves of light are I and `4I`. The maximum intensity of the resultant wave after superposition is

Options

A. (a) `5I`
B. (b) `9I`
C. (c) `16I`
D. (d) `25I`

Explanation

To solve the problem, we need to find the maximum intensity of the resultant wave after the superposition of two waves with intensities \( I_1 = I \) and \( I_2 = 4I \).

### Step-by-Step Solution:

1. **Identify the Intensities**:
- Let \( I_1 = I \) and \( I_2 = 4I \).

2. **Use the Formula for Resultant Intensity**:
- The formula for the resultant intensity \( I_{\text{max}} \) when two waves interfere is given by:
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- Here, \( \phi \) is the phase difference between the two waves.

3. **Maximize the Resultant Intensity**:
- To find the maximum intensity, we set \( \cos(\phi) = 1 \) (which occurs when \( \phi = 0 \) degrees). This gives:
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]

4. **Substitute the Values**:
- Substitute \( I_1 = I \) and \( I_2 = 4I \) into the formula:
\[ I_{\text{max}} = I + 4I + 2\sqrt{I \cdot 4I} \]

5. **Calculate Each Term**:
- First, calculate the sum of the intensities:
\[ I + 4I = 5I \]
- Next, calculate \( 2\sqrt{I \cdot 4I} \):
\[ 2\sqrt{4I^2} = 2 \cdot 2I = 4I \]

6. **Combine the Results**:
- Now, combine the results:
\[ I_{\text{max}} = 5I + 4I = 9I \]

7. **Final Result**:
- Therefore, the maximum intensity of the resultant wave after superposition is:
\[ \boxed{9I} \]