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 is …………..

Explanation

To solve the problem, we need to find the ratio of maximum and minimum intensities of fringes in an interference experiment where the ratio of amplitudes of coherent waves is given as \( \frac{a_1}{a_2} = \frac{1}{3} \).

### Step-by-Step Solution:

1. **Understanding Amplitude Ratio**:
Given the ratio of amplitudes:
\[ \frac{a_1}{a_2} = \frac{1}{3} \]
We can express \( a_1 \) in terms of \( a_2 \):
\[ a_1 = \frac{1}{3} a_2 \]

2. **Maximum Intensity Calculation**:
The maximum intensity \( I_{\text{max}} \) occurs during constructive interference, where the amplitudes add up:
\[ I_{\text{max}} \propto (a_1 + a_2)^2 \]
Substituting \( a_1 \):
\[ I_{\text{max}} \propto \left(\frac{1}{3} a_2 + a_2\right)^2 = \left(\frac{1}{3} a_2 + \frac{3}{3} a_2\right)^2 = \left(\frac{4}{3} a_2\right)^2 = \frac{16}{9} a_2^2 \]

3. **Minimum Intensity Calculation**:
The minimum intensity \( I_{\text{min}} \) occurs during destructive interference, where the amplitudes subtract:
\[ I_{\text{min}} \propto (a_2 - a_1)^2 \]
Substituting \( a_1 \):
\[ I_{\text{min}} \propto \left(a_2 - \frac{1}{3} a_2\right)^2 = \left(\frac{2}{3} a_2\right)^2 = \frac{4}{9} a_2^2 \]

4. **Finding the Ratio of Intensities**:
Now we can find the ratio of maximum to minimum intensity:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{\frac{16}{9} a_2^2}{\frac{4}{9} a_2^2} = \frac{16}{4} = 4 \]

### Final Answer:
The ratio of maximum and minimum intensities of fringes is:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = 4 \]