= Solution
At the centre, equal illumination gives slit <amplitudes> proportional to widths, so $A_2=2A_1$. The local <interference> extrema are $(A_1+A_2)^2$ and $(A_2-A_1)^2$, giving
$$
\boxed{\frac{I_{\max}}{I_{\min}}=\frac{(1+2)^2}{(2-1)^2}=9.}
$$
These are the central-envelope extrema as relative <phase> varies, or equivalently the near-centre fringe limits where the envelope is nearly constant.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-gpp2014-unequal-slits.png]
{title=Original unequal-slit Fraunhofer intensity and its envelopes, with central-overlap edges and local nine-to-one intensity contrast}
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