In YDSE, the slits have different widths. As a result, amplitude of waves from slits are A and 2A respectively. If `I_(0)` be the maximum intensity of the intensity of the interference pattern then the intensity of the pattern at a point where the phase difference between waves is `phi` is given by `(I_(0))/(P)(5+4cos phi)` . Where P is in in integer. Find the value of P ?

Options

A. 9
B. 6
C. 1
D. 3

Answer

To solve the problem, we need to determine the value of \( P \) in the expression for intensity at a point in the interference pattern of the Young's Double Slit Experiment (YDSE) when the amplitudes of the waves from the two slits are different.

### Step-by-Step Solution:

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

2. **Calculate the Intensities:**
- The intensity \( I_1 \) from the first slit is given by:
\[ I_1 = k A^2 \]
- The intensity \( I_2 \) from the second slit is given by:
\[ I_2 = k (2A)^2 = k \cdot 4A^2 = 4k A^2 \]

3. **Maximum Intensity Calculation:**
- The maximum intensity \( I_0 \) occurs when the phase difference \( \phi = 0 \) (i.e., \( \cos \phi = 1 \)):
\[ I_0 = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cdot \cos(0) \]
- Substituting the values of \( I_1 \) and \( I_2 \):
\[ I_0 = I_1 + 4I_1 + 2 \sqrt{I_1 \cdot 4I_1} \]
- Simplifying this:
\[ I_0 = 5I_1 + 2 \cdot 2I_1 = 5I_1 + 4I_1 = 9I_1 \]
- Thus, we have:
\[ I_1 = \frac{I_0}{9} \]

4. **Intensity at a Point with Phase Difference \( \phi \):**
- The intensity at a point where the phase difference is \( \phi \) is given by:
\[ I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cdot \cos \phi \]
- Substituting \( I_1 \) and \( I_2 \):
\[ I = I_1 + 4I_1 + 2 \sqrt{I_1 \cdot 4I_1} \cdot \cos \phi \]
- This simplifies to:
\[ I = 5I_1 + 2 \cdot 2I_1 \cdot \cos \phi = 5I_1 + 4I_1 \cos \phi \]
- Factoring out \( I_1 \):
\[ I = I_1 (5 + 4 \cos \phi) \]

5. **Comparing with Given Expression:**
- The problem states that the intensity at that point can be expressed as:
\[ I = \frac{I_0}{P} (5 + 4 \cos \phi) \]
- From our derived expression, we have:
\[ I = \frac{I_0}{9} (5 + 4 \cos \phi) \]
- Comparing both expressions, we find:
\[ P = 9 \]

### Final Answer:
The value of \( P \) is \( 9 \).