The intensity of light emerging from one slit is nine times than that from the other slit in Young's double sit interference set up. The ratio of maximum intensity to minimum intensity in the fringe pattern is

Options

A. `2:1`
B. `4:1`
C. `6:1`
D. `8:1`

Answer

To solve the problem of finding the ratio of maximum intensity to minimum intensity in a Young's double slit interference setup where the intensity of light from one slit is nine times that from the other slit, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Intensities**:
Let the intensity from the first slit be \( I_1 = 9I \) and the intensity from the second slit be \( I_2 = I \).

2. **Calculate Maximum Intensity**:
The maximum intensity \( I_{max} \) occurs when the phase difference \( \phi \) is such that \( \cos \phi = 1 \). The formula for the resultant intensity in Young's double slit experiment is:
\[ I_{net} = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi \]
Substituting \( \phi = 0 \):
\[ I_{max} = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cdot 1 \]
\[ I_{max} = 9I + I + 2 \sqrt{9I \cdot I} \]
\[ I_{max} = 10I + 2 \cdot 3I = 10I + 6I = 16I \]

3. **Calculate Minimum Intensity**:
The minimum intensity \( I_{min} \) occurs when the phase difference \( \phi \) is such that \( \cos \phi = -1 \):
\[ I_{min} = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cdot (-1) \]
\[ I_{min} = 9I + I + 2 \sqrt{9I \cdot I} \cdot (-1) \]
\[ I_{min} = 10I - 6I = 4I \]

4. **Find the Ratio of Maximum to Minimum Intensity**:
Now, we can find the ratio of maximum intensity to minimum intensity:
\[ \frac{I_{max}}{I_{min}} = \frac{16I}{4I} = \frac{16}{4} = 4 \]

### Final Answer:
The ratio of maximum intensity to minimum intensity in the fringe pattern is \( 4:1 \).

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