In Young's double-slit interference experiment, the width of one slit is half that of the other. The intensity, $ \mathrm{I} $, is proportional to the slit width. This gives us:

$ \begin{align*} \mathrm{I}_1 &= \mathrm{I}_0, \\ \mathrm{I}_2 &= 2 \mathrm{I}_0. \end{align*} $

The maximum intensity, $ \mathrm{I}_{\max} $, occurs when the amplitudes add constructively:

$ \mathrm{I}_{\max} = \left(\sqrt{\mathrm{I}_1} + \sqrt{\mathrm{I}_2}\right)^2. $

The minimum intensity, $ \mathrm{I}_{\min} $, occurs when the amplitudes interfere destructively:

$ \mathrm{I}_{\min} = \left(\sqrt{\mathrm{I}_1} - \sqrt{\mathrm{I}_2}\right)^2. $

Now, using $ \mathrm{I}_1 = \mathrm{I}_0 $ and $ \mathrm{I}_2 = 2 \mathrm{I}_0 $, we find:

$ \frac{\mathrm{I}_{\max}}{\mathrm{I}_{\min}} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}. $