How is the width of a slit related to the intensity of light passing through it?

I hope you know that intensity $(I)$ of light at any point on the screen due to interference in the Young's Double Slit experiment can be given as

$$A^2=I=a_1^2+a_2^2+2a_1a_2cos{phi}$$

where $a_1, a_2$ are the amplitudes of the light waves with a constant phase difference of $phi$, $A$ is the amplitude of the resultant displacement at the point on the screen. For simplicity, we can assume that intensity of the light to be equal to the square of the amplitude as given above.

Thus, $$I_{max}=a_1^2+a_2^2+2a_1a_2(1)=(a_1+a_2)^2$$

$$I_{min}=a_1^2+a_2^2+2a_1a_2(-1)=(a_1-a_2)^2$$

Therefore, $frac{I_max}{I_min}=frac{(a_1+a_2)^2}{(a_1-a_2)^2}=frac{25}{9}$

Thus, $a_1+a_2=5, a_1-a_2=3$

$a_1+(a_1-3)=5=2a_1-3$
Thus, $a_1=8/2=4, a_2=1$

The intensity of light due to a slit (source of light) is directly proportional to the width of the slit. Therefore, if $w_1$ and $w_2$ are widths of the tow slits $S_1$ and $S_2$; $I_1$ and $I_2$ are intensities of light due to the respective slits on the screen, then

$$frac{w_1}{w_2}=frac{I_1}{I_2}=frac{a_1^2}{a_2^2}=frac{4^2}{1^2}=16$$