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. (a) `5I` and `I`
B. (b) `5I` and `3I`
C. (c) `9I` and `I`
D. (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 1: Understand the formula for resultant intensity
The resultant intensity \( I_R \) when two coherent light beams of intensities \( I_1 \) and \( I_2 \) interfere is given by the formula:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
where \( \phi \) is the phase difference between the two beams.

### Step 2: Identify the intensities
In this case, we have:
- \( I_1 = I \)
- \( I_2 = 4I \)

### Step 3: Calculate maximum intensity
To find the maximum intensity, we set \( \cos \phi = 1 \) (which occurs when the beams are in phase):
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
Substituting the values:
\[ I_{\text{max}} = I + 4I + 2\sqrt{I \cdot 4I} \]
Calculating further:
\[ I_{\text{max}} = 5I + 2\sqrt{4I^2} = 5I + 4I = 9I \]

### Step 4: Calculate minimum intensity
To find the minimum intensity, we set \( \cos \phi = -1 \) (which occurs when the beams are out of phase):
\[ I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2} \]
Substituting the values:
\[ I_{\text{min}} = I + 4I - 2\sqrt{I \cdot 4I} \]
Calculating further:
\[ I_{\text{min}} = 5I - 2\sqrt{4I^2} = 5I - 4I = I \]

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

### Final Answer
- Maximum intensity: \( 9I \)
- Minimum intensity: \( I \)