Assume, Young's double slit experiment, amplitudes corresponding to the sources $S_1$ and $S_2$ be $a_1$ and $a_2$ and intensities be $I_1$ and $I_2$ respectively.


So, $I_{\max } \propto a_{\max }^2$ and $I_{\min } \propto a_{\text {min }}^2$
$\Rightarrow \frac{I_{\max }}{I_{\min }}=\left(\frac{a_{\max }}{a_{\min }}\right)^2=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2$
According to given question,
$\begin{aligned}
\frac{I_{\max }}{I_{\min }} & =\frac{9}{1} \\
\Rightarrow \quad 9 & =\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}\right)^2
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad(3)^2=\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}\right)^2 \quad \text { (given) } \\
& \Rightarrow \quad 3=\frac{a_1 / a_2+1}{a_1 / a_2-1} \Rightarrow 3=\frac{\left(\frac{a_1+a_2}{a_2}\right)}{\left(\frac{a_1-a_2}{a_2}\right)} \Rightarrow 3=\frac{a_1+a_2}{a_1-a_2} \\
& \Rightarrow \quad 3\left(a_1-a_2\right)=a_1+a_2 \Rightarrow 3 a_1-3 a_2=a_1+a_2 \\
& \Rightarrow \quad 2 a_1=4 a_2 \Rightarrow \frac{a_1}{a_2}=\frac{4}{2} \Rightarrow a_1: a_2=2: 1
\end{aligned}$