In the Young's double slit experiment, the interference pattern is found to have as intensity ratio between the bright and dark fringes as 9. This implies that

Options

A. the intensities of individual sources are 5 and 4 units respectively
B. the intensities of individual sources are 4 and 1 units
C. the ratio of their amplitude is 3
D. the ratio of their amplitudes is 4

Explanation

To solve the problem regarding the intensity ratio in Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Intensity Ratio**:
The intensity ratio between the bright and dark fringes is given as 9. This means:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = 9 \]
We can express this as:
\[ I_{\text{max}} = 9I_{\text{min}} \]

2. **Relating Intensity to Amplitude**:
The intensity of light is proportional to the square of the amplitude of the wave. Therefore, we can write:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2 \]
where \(A_1\) and \(A_2\) are the amplitudes of the two sources.

3. **Setting Up the Equation**:
Substituting the intensity ratio into the equation:
\[ \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2 = 9 \]
Taking the square root of both sides gives:
\[ \frac{A_1 + A_2}{A_1 - A_2} = 3 \quad \text{(considering only the positive root)} \]

4. **Cross Multiplying**:
Cross-multiplying gives:
\[ A_1 + A_2 = 3(A_1 - A_2) \]
Expanding this:
\[ A_1 + A_2 = 3A_1 - 3A_2 \]

5. **Rearranging the Equation**:
Rearranging the terms leads to:
\[ A_1 + A_2 + 3A_2 = 3A_1 \]
Simplifying gives:
\[ 4A_2 = 2A_1 \quad \Rightarrow \quad \frac{A_1}{A_2} = \frac{4}{2} = 2 \]

6. **Calculating the Intensity Ratio**:
Now, we can find the intensity ratio:
\[ \frac{I_1}{I_2} = \left( \frac{A_1}{A_2} \right)^2 = \left( 2 \right)^2 = 4 \]

7. **Conclusion**:
Therefore, the intensity of the individual sources is in the ratio of:
\[ I_1 : I_2 = 4 : 1 \]

### Final Answer:
The intensity ratio of the individual sources is \(4:1\).

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