Consider an YDSE that has different slit width. As a result, amplitude of waves from two slits are A and 2A, respectively. If `I_(0)` be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is `phi` is

Options

A. `(I_(0))/(9) cos^(2) phi`
B. `(I_(0))/(3) sin^(2).(phi)/(2)`
C. `(I_(0))/(9) [5 + 4 cos phi]`
D. `(I_(0))/(9) [ 5 + 8 cos phi]`

Explanation

To solve the problem, we need to find the intensity of the interference pattern at a point where the phase difference between the waves from the two slits is \( \phi \).

### Step-by-Step Solution:

1. **Identify the Amplitudes:**
- The amplitudes of the waves from the two slits are given as \( A \) and \( 2A \).

2. **Calculate the Intensities:**
- The intensity \( I \) is proportional to the square of the amplitude. Therefore, the intensities from the two slits can be expressed as:
\[ I_1 \propto A^2 \quad \text{and} \quad I_2 \propto (2A)^2 = 4A^2 \]
- This means the ratio of the intensities from the two slits is:
\[ I_1 : I_2 = A^2 : 4A^2 = 1 : 4 \]

3. **Determine the Maximum Intensity \( I_0 \):**
- The maximum intensity \( I_0 \) occurs when the waves are in phase (i.e., when \( \phi = 0 \)):
\[ I_0 = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
- Let \( I_1 = I \) and \( I_2 = 4I \). Then:
\[ I_0 = I + 4I + 2\sqrt{I \cdot 4I} = 5I + 4\sqrt{I^2} = 5I + 4I = 9I \]

4. **Express \( I \) in terms of \( I_0 \):**
- From the previous step, we have:
\[ I_0 = 9I \implies I = \frac{I_0}{9} \]

5. **Calculate the Intensity at Phase Difference \( \phi \):**
- The intensity \( I' \) at a point where the phase difference is \( \phi \) can be calculated using:
\[ I' = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- Substituting \( I_1 = I \) and \( I_2 = 4I \):
\[ I' = I + 4I + 2\sqrt{I \cdot 4I} \cos(\phi) = 5I + 4I \cos(\phi) \]
- Substitute \( I = \frac{I_0}{9} \):
\[ I' = 5\left(\frac{I_0}{9}\right) + 4\left(\frac{I_0}{9}\right) \cos(\phi) = \frac{5I_0}{9} + \frac{4I_0}{9} \cos(\phi) \]
- Factor out \( \frac{I_0}{9} \):
\[ I' = \frac{I_0}{9} (5 + 4 \cos(\phi)) \]

### Final Answer:
The intensity at a point where the phase difference is \( \phi \) is given by:
\[ I' = \frac{I_0}{9} (5 + 4 \cos(\phi)) \]