In an interference experiment the ratio of amplitudes of coherent waves is $$\frac{a_1}{a_2} = \frac{1}{3}$$. The ratio of maximum and minimum intensities of fringes will be:

We are told that the two interfering beams are coherent and that their amplitudes are in the ratio

$$\frac{a_1}{a_2}=\frac{1}{3}.$$

To work comfortably, we assign actual symbols that satisfy this ratio. Let the common constant of proportionality be $$k.$$ Then we can write

$$a_1 = k \quad\text{and}\quad a_2 = 3k.$$

Interference formulas relate intensities to these amplitudes. First, we recall the basic result:

Intensity is proportional to the square of amplitude, so

$$I_1 \propto a_1^{\,2}, \qquad I_2 \propto a_2^{\,2}.$$

Now, in a two-source interference pattern, the expressions for maximum and minimum intensities are obtained by adding the wave amplitudes in phase (for a bright fringe) and subtracting them out of phase (for a dark fringe). Explicitly, the standard formulas are

$$I_{\text{max}} = (a_1 + a_2)^{2},$$

$$I_{\text{min}} = (a_1 - a_2)^{2}.$$

We substitute the concrete values $$a_1 = k$$ and $$a_2 = 3k.$$ Carrying out each calculation step by step:

For the maximum intensity,

$$I_{\text{max}} = (a_1 + a_2)^{2} = (\,k + 3k\,)^{2} = (4k)^{2} = 16k^{2}.$$

For the minimum intensity,

$$I_{\text{min}} = (a_1 - a_2)^{2} = (\,k - 3k\,)^{2} = (-2k)^{2} = 4k^{2}.$$

The ratio of these two intensities is then

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{16k^{2}}{4k^{2}} = \frac{16}{4} = 4.$$

Notice that the factor $$k^{2}$$ cancels, as expected, so the result is independent of our initial scaling choice.

Hence, the required ratio of maximum to minimum intensities of the fringes is $$4.$$

Hence, the correct answer is Option A.