Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are

Options

A. 5I and I
B. 5I and 3I
C. 9I and I
D. 9I and 3I

Explanation

To solve the problem of finding the maximum and minimum possible intensities when two coherent monochromatic light beams of intensities \( I \) and \( 4I \) are superposed, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Intensity and Amplitude Relationship**:
- The intensity \( I \) of a light wave is proportional to the square of its amplitude \( A \). This can be expressed as:
\[ I \propto A^2 \]

2. **Assigning Amplitudes to the Intensities**:
- Let the amplitude of the first beam (intensity \( I \)) be \( A_1 \) and the amplitude of the second beam (intensity \( 4I \)) be \( A_2 \).
- From the relationship \( I \propto A^2 \):
\[ I \propto A_1^2 \quad \text{(1)} \]
\[ 4I \propto A_2^2 \quad \text{(2)} \]
- From equation (1), we can write:
\[ A_1 = \sqrt{I} \]
- From equation (2):
\[ A_2 = \sqrt{4I} = 2\sqrt{I} \]

3. **Calculating Maximum Intensity**:
- The maximum amplitude when two waves interfere constructively is the sum of their amplitudes:
\[ A_{\text{max}} = A_1 + A_2 = \sqrt{I} + 2\sqrt{I} = 3\sqrt{I} \]
- The maximum intensity \( I_{\text{max}} \) is then:
\[ I_{\text{max}} \propto (A_{\text{max}})^2 = (3\sqrt{I})^2 = 9I \]

4. **Calculating Minimum Intensity**:
- The minimum amplitude when two waves interfere destructively is the difference of their amplitudes:
\[ A_{\text{min}} = A_2 - A_1 = 2\sqrt{I} - \sqrt{I} = \sqrt{I} \]
- The minimum intensity \( I_{\text{min}} \) is then:
\[ I_{\text{min}} \propto (A_{\text{min}})^2 = (\sqrt{I})^2 = I \]

5. **Final Results**:
- The maximum possible intensity in the resulting beam is \( 9I \).
- The minimum possible intensity in the resulting beam is \( I \).

### Conclusion:
The maximum and minimum possible intensities in the resulting beam are:
- Maximum Intensity: \( 9I \)
- Minimum Intensity: \( I \)