Consider interference between two sources of intensity I and 4I. Find out resultant intensity where phase difference is (i). `pi//4`
(ii). `pi`
(iii). `4pi`

Explanation

To solve the problem of finding the resultant intensity when two sources of intensity \( I \) and \( 4I \) interfere with a given phase difference, we can use the formula for resultant intensity in interference:

\[ I_{\text{resultant}} = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi \]

where:
- \( I_1 \) and \( I_2 \) are the intensities of the two sources,
- \( \phi \) is the phase difference.

Given:
- \( I_1 = I \)
- \( I_2 = 4I \)

Now, we will calculate the resultant intensity for each phase difference.

### Part (i): Phase Difference \( \phi = \frac{\pi}{4} \)

1. Substitute the values into the formula:
\[ I_{\text{resultant}} = I + 4I + 2 \sqrt{I \cdot 4I} \cos \left(\frac{\pi}{4}\right) \]

2. Calculate \( \sqrt{I \cdot 4I} \):
\[ \sqrt{I \cdot 4I} = \sqrt{4I^2} = 2I \]

3. Substitute \( \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ I_{\text{resultant}} = 5I + 2 \cdot 2I \cdot \frac{1}{\sqrt{2}} \]
\[ = 5I + \frac{4I}{\sqrt{2}} = 5I + 2\sqrt{2}I \]

4. The resultant intensity is:
\[ I_{\text{resultant}} = 5I + 2\sqrt{2}I \approx 5I + 2.828I \approx 7.828I \]

### Part (ii): Phase Difference \( \phi = \pi \)

1. Substitute the values into the formula:
\[ I_{\text{resultant}} = I + 4I + 2 \sqrt{I \cdot 4I} \cos(\pi) \]

2. Since \( \cos(\pi) = -1 \):
\[ I_{\text{resultant}} = 5I + 2 \cdot 2I \cdot (-1) \]
\[ = 5I - 4I = I \]

### Part (iii): Phase Difference \( \phi = 4\pi \)

1. Substitute the values into the formula:
\[ I_{\text{resultant}} = I + 4I + 2 \sqrt{I \cdot 4I} \cos(4\pi) \]

2. Since \( \cos(4\pi) = 1 \):
\[ I_{\text{resultant}} = 5I + 2 \cdot 2I \cdot 1 \]
\[ = 5I + 4I = 9I \]

### Final Results:
- For \( \phi = \frac{\pi}{4} \), \( I_{\text{resultant}} \approx 7.828I \)
- For \( \phi = \pi \), \( I_{\text{resultant}} = I \)
- For \( \phi = 4\pi \), \( I_{\text{resultant}} = 9I \)