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 from two sources of intensity \( I \) and \( 4I \) with different phase differences, we will use the formula for resultant intensity in interference:

\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]

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

### Step-by-Step Solution:

1. **Identify the Intensities**:
- Let \( I_1 = I \) (intensity of the first source)
- Let \( I_2 = 4I \) (intensity of the second source)

2. **Case (i): Phase Difference \( \phi = \frac{\pi}{4} \)**:
- Substitute \( I_1 \), \( I_2 \), and \( \phi \) into the formula:
\[ I_R = I + 4I + 2\sqrt{I \cdot 4I} \cos\left(\frac{\pi}{4}\right) \]
- Calculate \( \sqrt{I \cdot 4I} = \sqrt{4I^2} = 2I \)
- Since \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ I_R = 5I + 2(2I)\left(\frac{1}{\sqrt{2}}\right) = 5I + \frac{4I}{\sqrt{2}} = 5I + 2\sqrt{2}I \]
- Approximate \( 2\sqrt{2} \approx 2.828 \):
\[ I_R \approx 7.828I \]

3. **Case (ii): Phase Difference \( \phi = \pi \)**:
- Substitute \( \phi = \pi \):
\[ I_R = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) \]
- Since \( \cos(\pi) = -1 \):
\[ I_R = 5I + 2(2I)(-1) = 5I - 4I = I \]

4. **Case (iii): Phase Difference \( \phi = 4\pi \)**:
- Substitute \( \phi = 4\pi \):
\[ I_R = I + 4I + 2\sqrt{I \cdot 4I} \cos(4\pi) \]
- Since \( \cos(4\pi) = 1 \):
\[ I_R = 5I + 2(2I)(1) = 5I + 4I = 9I \]

### Final Results:
- (i) For \( \phi = \frac{\pi}{4} \), \( I_R \approx 7.828I \)
- (ii) For \( \phi = \pi \), \( I_R = I \)
- (iii) For \( \phi = 4\pi \), \( I_R = 9I \)