Two waves of same intensity produce interference. If intensity at maximum is 4I, then intensity at the minimum will be:

Options

A. 0
B. 2I
C. 3I
D. 4I

Answer

To solve the problem of finding the intensity at the minimum when two waves of the same intensity produce interference, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Interference:**
When two waves interfere, the resultant intensity (I) can be calculated using the formula:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \theta \]
where \(I_1\) and \(I_2\) are the intensities of the two waves, and \(\theta\) is the phase difference between them.

2. **Given Conditions:**
We are given that the two waves have the same intensity, so we can denote:
\[ I_1 = I_2 = I \]
We also know that the maximum intensity \(I_{\text{max}} = 4I\).

3. **Calculate Maximum Intensity:**
For maximum intensity, the phase difference \(\theta = 0^\circ\) (which means the waves are in phase). Thus, we can substitute into the formula:
\[ I_{\text{max}} = I + I + 2\sqrt{I \cdot I} \cos(0) \]
\[ I_{\text{max}} = I + I + 2I = 4I \]
This confirms that the maximum intensity is indeed \(4I\).

4. **Calculate Minimum Intensity:**
For minimum intensity, the phase difference \(\theta = 180^\circ\) (which means the waves are out of phase). Thus, we substitute:
\[ I_{\text{min}} = I + I + 2\sqrt{I \cdot I} \cos(180^\circ) \]
Since \(\cos(180^\circ) = -1\), we have:
\[ I_{\text{min}} = I + I + 2I \cdot (-1) \]
\[ I_{\text{min}} = 2I - 2I = 0 \]

5. **Conclusion:**
The intensity at the minimum when two waves of the same intensity interfere is:
\[ I_{\text{min}} = 0 \]

### Final Answer:
The intensity at the minimum will be \(0\).

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