The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is $$x : 4$$ where $$x$$ is _________.

We have two slits whose widths are unequal. Let the smaller slit have width $$a$$. The problem states that the other slit is three times wider, so its width is $$3a$$.

The amplitude of the light emerging from a slit is directly proportional to its width. Hence, if the smaller slit contributes an amplitude $$A$$, then the wider slit contributes an amplitude $$3A$$, because $$3a$$ is three times $$a$$.

Intensity and amplitude are related by the formula

$$I \propto (\text{amplitude})^{2}.$$

Therefore, the individual intensities from the two slits are

$$I_{1}=A^{2}, \qquad I_{2}=(3A)^{2}=9A^{2}.$$

When the two waves interfere, the resultant intensity depends on the vector sum of the amplitudes. For constructive interference (bright fringes), the amplitudes add; for destructive interference (dark fringes), they subtract.

Maximum intensity

The formula for the maximum resultant amplitude is the algebraic sum of the individual amplitudes:

$$A_{\max}=A+3A=4A.$$

So the maximum intensity is

$$I_{\max}=A_{\max}^{2}=(4A)^{2}=16A^{2}.$$

Minimum intensity

The formula for the minimum resultant amplitude is the absolute value of the difference of the individual amplitudes:

$$A_{\min}=|A-3A|=|-2A|=2A.$$

Thus the minimum intensity is

$$I_{\min}=A_{\min}^{2}=(2A)^{2}=4A^{2}.$$

Ratio of minimum to maximum intensity

$$\frac{I_{\min}}{I_{\max}}=\frac{4A^{2}}{16A^{2}}=\frac{4}{16}=\frac{1}{4}.$$

This ratio is given in the statement as $$x:4$$, so

$$x=1.$$

So, the answer is $$1$$.