In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

We are told that one slit has width three times the other. Since amplitude is proportional to slit width, if the amplitude from the narrower slit is $$A$$, then the amplitude from the wider slit is $$3A$$.

In a double-slit experiment, the maximum intensity occurs when the two waves interfere constructively, and the minimum intensity occurs when they interfere destructively.

The maximum intensity is $$I_{\max} = (A_1 + A_2)^2 = (A + 3A)^2 = (4A)^2 = 16A^2$$.

The minimum intensity is $$I_{\min} = (A_2 - A_1)^2 = (3A - A)^2 = (2A)^2 = 4A^2$$.

The ratio of maximum to minimum intensity is $$\frac{I_{\max}}{I_{\min}} = \frac{16A^2}{4A^2} = \frac{4}{1}$$.

So the ratio is $$4 : 1$$.

Hence, the correct answer is Option D.