The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is:

In Young's double slit experiment, when one slit has width 4 times that of the other, the amplitudes are related to the slit widths.

The intensity is proportional to the slit width (since wider slits allow more light through). So if one slit has width $$w$$ and the other has width $$4w$$:

$$I_1 : I_2 = 1 : 4$$

Since amplitude is proportional to the square root of intensity:

$$a_1 : a_2 = 1 : 2$$

The maximum intensity occurs when the waves interfere constructively:

$$I_{max} = (a_1 + a_2)^2 = (1 + 2)^2 = 9$$

The minimum intensity occurs when the waves interfere destructively:

$$I_{min} = (a_1 - a_2)^2 = (1 - 2)^2 = 1$$

Therefore, the ratio of maximum to minimum intensity is:

$$\frac{I_{max}}{I_{min}} = \frac{9}{1} = 9 : 1$$

The answer is Option C: $$9 : 1$$.