In an interference experiment the ratio of amplitudes of coherent waves is `(a_(1))/(a_(2))=(1)/(3)`. The ratio of maximum and minimum intensities of fringes will be:

Options

A. 2
B. 18
C. 4
D. 9

Answer

To solve the problem, we need to find the ratio of the maximum and minimum intensities of the fringes in an interference experiment, given the ratio of the amplitudes of the coherent waves.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The ratio of the amplitudes of the coherent waves is given as:
\[ \frac{a_1}{a_2} = \frac{1}{3} \]
- This implies:
\[ a_1 = 1 \quad \text{and} \quad a_2 = 3 \]

2. **Understand the Relationship Between Intensity and Amplitude**:
- The intensity \( I \) is proportional to the square of the amplitude:
\[ I \propto A^2 \]

3. **Determine the Maximum Intensity**:
- The maximum intensity \( I_{\text{max}} \) occurs during constructive interference, where the resultant amplitude is:
\[ A_{\text{max}} = a_1 + a_2 \]
- Substituting the values of \( a_1 \) and \( a_2 \):
\[ A_{\text{max}} = 1 + 3 = 4 \]
- Therefore, the maximum intensity is:
\[ I_{\text{max}} = A_{\text{max}}^2 = 4^2 = 16 \]

4. **Determine the Minimum Intensity**:
- The minimum intensity \( I_{\text{min}} \) occurs during destructive interference, where the resultant amplitude is:
\[ A_{\text{min}} = |a_1 - a_2| \]
- Substituting the values of \( a_1 \) and \( a_2 \):
\[ A_{\text{min}} = |1 - 3| = 2 \]
- Therefore, the minimum intensity is:
\[ I_{\text{min}} = A_{\text{min}}^2 = 2^2 = 4 \]

5. **Calculate the Ratio of Maximum to Minimum Intensity**:
- The ratio of maximum intensity to minimum intensity is:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{16}{4} = 4 \]

6. **Express the Ratio**:
- The ratio can be expressed as:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = 4:1 \]

### Final Answer:
The ratio of the maximum and minimum intensities of the fringes is \( 4:1 \).