In Young's double slit interference experiment, using two coherent waves of different amplitudes, the intensities ratio between bright and dark fringes is 3 Then, the value of the wave amplitudes ratio that arrive there is

Options

A. `((sqrt(3)+1)/(sqrt(3)-1))`
B. `((sqrt(3)-1)/(sqrt(3)+1))`
C. `sqrt(3):1`
D. `1:sqrt(3)`

Explanation

To solve the problem, we need to find the ratio of the wave amplitudes \( A_1 \) and \( A_2 \) given that the intensity ratio between bright and dark fringes in Young's double slit experiment is 3:1.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Intensity and Amplitude**:
The intensity \( I \) of a wave is proportional to the square of its amplitude. Therefore, we can express the intensities of the two waves as:
\[ I_1 \propto A_1^2 \quad \text{and} \quad I_2 \propto A_2^2 \]

2. **Setting Up the Intensity Ratio**:
Given that the intensity ratio \( \frac{I_1}{I_2} = 3 \), we can write:
\[ \frac{I_1}{I_2} = \frac{A_1^2}{A_2^2} = 3 \]

3. **Expressing the Ratio of Amplitudes**:
From the intensity ratio, we can derive the ratio of amplitudes:
\[ \frac{A_1^2}{A_2^2} = 3 \implies \frac{A_1}{A_2} = \sqrt{3} \]

4. **Final Result**:
Therefore, the ratio of the amplitudes \( A_1 \) to \( A_2 \) is:
\[ A_1 : A_2 = \sqrt{3} : 1 \]

### Conclusion:
The value of the wave amplitudes ratio is \( \sqrt{3} : 1 \).

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