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

Answer

To solve the problem, we need to analyze the interference pattern created by two slits with different amplitudes in Young's double slit experiment. Here’s a step-by-step solution:

### Step 1: Define the Amplitudes
Let the amplitude of the light from the first slit be \( A_1 = A_0 \) and the amplitude from the second slit be \( A_2 = 2A_0 \).

### Step 2: Calculate the Resultant Amplitude
The resultant amplitude \( A_R \) when two waves interfere can be calculated using the formula:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi} \]
Substituting the values of \( A_1 \) and \( A_2 \):
\[ A_R = \sqrt{(A_0)^2 + (2A_0)^2 + 2(A_0)(2A_0) \cos \phi} \]
\[ = \sqrt{A_0^2 + 4A_0^2 + 4A_0^2 \cos \phi} \]
\[ = \sqrt{5A_0^2 + 4A_0^2 \cos \phi} \]

### Step 3: Express the Resultant Amplitude
Factoring out \( A_0^2 \):
\[ A_R = A_0 \sqrt{5 + 4 \cos \phi} \]

### Step 4: Relate Intensity to Amplitude
The intensity \( I \) is related to the amplitude by the equation:
\[ I \propto A^2 \]
Thus, the resultant intensity \( I \) can be expressed as:
\[ I = k A_R^2 \]
where \( k \) is a proportionality constant.

### Step 5: Substitute the Resultant Amplitude
Substituting \( A_R \) into the intensity equation:
\[ I = k (A_0 \sqrt{5 + 4 \cos \phi})^2 \]
\[ = k A_0^2 (5 + 4 \cos \phi) \]

### Step 6: Determine Maximum Intensity
The maximum intensity \( I_m \) occurs when \( \phi = 0 \) (i.e., \( \cos \phi = 1 \)):
\[ I_m = k A_R^2 = k (A_0 + 2A_0)^2 = k (3A_0)^2 = 9k A_0^2 \]

### Step 7: Express the Resultant Intensity in Terms of Maximum Intensity
Now, we can express the resultant intensity \( I \) in terms of the maximum intensity \( I_m \):
\[ I = \frac{I_m}{9} (5 + 4 \cos \phi) \]

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

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