In Young's double-slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as $9$. This implies that

The correct options are
C the amplitude ratio is $2$
D the intensities at the screen due to the two slits are 4 units and 1 units, respectively

Let the souces have amplitudes $a_1,a_2$

$\frac{I_{max}}{I_{min}}=\frac{(a_1+a_2{)}^2}{(a_1-a_2{)}^2}=9$

Hence, $\frac{a_1+a_2}{a_1-a_2}=4$

$⟹\frac{a_1}{a_2}=2$

$\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=4$