In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If `I_m` be the maximum intensity, the resultant intensity I when they interfere at phase difference `phi` is given by:

Options

A. `(I_(m))/(3)(1+2"cos"^(2)(phi)/(2))`
B. `(I_(m))/(5)(1+4"cos"^(2)(phi)/(2))`
C. `(I_(m))/(9)(1+8"cos"^(2)(phi)/(2))`
D. `(I_(m))/(9)(8+"cos"^(2)(phi)/(2))`

Explanation

To solve the problem, we need to determine the resultant intensity \( I \) when two slits in Young's double slit experiment have different amplitudes, specifically when one slit has an amplitude that is double that of the other. Here’s a step-by-step breakdown of the solution:

### Step 1: Define the Amplitudes
Let the amplitude of light from the first slit be \( A_1 \) and from the second slit be \( A_2 \). Given that \( A_1 = 2A_2 \).

### Step 2: Relate Amplitudes to Intensities
The intensity \( I \) is proportional to the square of the amplitude. Therefore, we can express the intensities as:
- \( I_1 = A_1^2 = (2A_2)^2 = 4A_2^2 \)
- \( I_2 = A_2^2 \)

Let’s denote \( I_d = A_2^2 \) (the intensity corresponding to the smaller amplitude). Thus:
- \( I_1 = 4I_d \)
- \( I_2 = I_d \)

### Step 3: Calculate Maximum Intensity
The maximum intensity \( I_m \) occurs when the two waves interfere constructively. The maximum intensity can be calculated using:
\[ I_m = (A_1 + A_2)^2 = (2A_2 + A_2)^2 = (3A_2)^2 = 9A_2^2 = 9I_d \]

### Step 4: Write the Resultant Intensity Formula
The resultant intensity \( I \) when two waves interfere at a phase difference \( \phi \) is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]

### Step 5: Substitute the Intensities
Substituting \( I_1 \) and \( I_2 \) into the formula:
\[ I = 4I_d + I_d + 2\sqrt{4I_d \cdot I_d} \cos \phi \]
\[ I = 5I_d + 2\sqrt{4I_d^2} \cos \phi \]
\[ I = 5I_d + 4I_d \cos \phi \]

### Step 6: Factor Out \( I_d \)
Now, we can factor out \( I_d \):
\[ I = I_d (5 + 4 \cos \phi) \]

### Step 7: Substitute \( I_d \) in terms of \( I_m \)
Since \( I_d = \frac{I_m}{9} \) (from the maximum intensity calculated earlier):
\[ I = \frac{I_m}{9} (5 + 4 \cos \phi) \]

### Final Result
Thus, the resultant intensity \( I \) when they interfere at a phase difference \( \phi \) is:
\[ I = \frac{I_m}{9} (5 + 4 \cos \phi) \]