The intensity of two waves is 2 and 3 unit, then average intensity of light in the overlapping region will have the value :

Options

A. 2.5
B. 6
C. 5
D. 13

Answer

To find the average intensity of two overlapping waves with given intensities, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Intensities**:
Let the intensity of the first wave \( I_1 = 2 \) units and the intensity of the second wave \( I_2 = 3 \) units.

2. **Calculate Maximum Intensity**:
The formula for maximum intensity \( I_{max} \) when two waves interfere is given by:
\[ I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
Plugging in the values:
\[ I_{max} = 2 + 3 + 2\sqrt{2 \times 3} \]
\[ I_{max} = 5 + 2\sqrt{6} \]

3. **Calculate Minimum Intensity**:
The formula for minimum intensity \( I_{min} \) is given by:
\[ I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} \]
Plugging in the values:
\[ I_{min} = 2 + 3 - 2\sqrt{2 \times 3} \]
\[ I_{min} = 5 - 2\sqrt{6} \]

4. **Calculate Average Intensity**:
The average intensity \( I_{avg} \) can be calculated using the formula:
\[ I_{avg} = \frac{I_{max} + I_{min}}{2} \]
Substituting the values we found:
\[ I_{avg} = \frac{(5 + 2\sqrt{6}) + (5 - 2\sqrt{6})}{2} \]
Simplifying this:
\[ I_{avg} = \frac{10}{2} = 5 \text{ units} \]

### Final Answer:
The average intensity of light in the overlapping region is **5 units**.