In a Young's double slit experiment the amplitudes of the two waves incident on the two slits are `A` and `2A`. If `I_(0)` is the maximum intensity, then the intensity at a spot on the screen where the phase difference between the two interfering waves is `phi` .

Options

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

Answer

To solve the problem, we need to find the intensity at a point on the screen in a Young's double slit experiment where the amplitudes of the two waves are given as \( A \) and \( 2A \), and there is a phase difference \( \phi \) between them.

### Step-by-Step Solution:

1. **Identify the Amplitudes**:
- Let the amplitude of the first wave be \( A_1 = A \).
- Let the amplitude of the second wave be \( A_2 = 2A \).

2. **Calculate the Intensities**:
- The intensity of a wave is proportional to the square of its amplitude. Thus, we can calculate the individual intensities:
\[ I_1 = A_1^2 = A^2 \]
\[ I_2 = A_2^2 = (2A)^2 = 4A^2 \]

3. **Find the Resultant Intensity Formula**:
- The resultant intensity \( I_R \) for two waves with phase difference \( \phi \) is given by:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- Substituting the values of \( I_1 \) and \( I_2 \):
\[ I_R = A^2 + 4A^2 + 2\sqrt{A^2 \cdot 4A^2} \cos(\phi) \]
\[ I_R = 5A^2 + 2 \cdot 2A^2 \cos(\phi) \]
\[ I_R = 5A^2 + 4A^2 \cos(\phi) \]

4. **Determine the Maximum Intensity \( I_0 \)**:
- The maximum intensity \( I_0 \) occurs when \( \cos(\phi) = 1 \) (i.e., when \( \phi = 0 \)):
\[ I_0 = 5A^2 + 4A^2 \cdot 1 = 5A^2 + 4A^2 = 9A^2 \]

5. **Express the Resultant Intensity in Terms of \( I_0 \)**:
- Now, we can express \( I_R \) in terms of \( I_0 \):
\[ I_R = 5A^2 + 4A^2 \cos(\phi) \]
- Since \( I_0 = 9A^2 \), we can write:
\[ I_R = \frac{5}{9} I_0 + \frac{4}{9} I_0 \cos(\phi) \]

6. **Final Expression for Intensity**:
- Therefore, the intensity at a point on the screen where the phase difference is \( \phi \) is:
\[ I_R = \frac{5 + 4 \cos(\phi)}{9} I_0 \]

### Conclusion:
The intensity at a spot on the screen where the phase difference between the two interfering waves is \( \phi \) is given by:
\[ I_R = \frac{5 + 4 \cos(\phi)}{9} I_0 \]