In Young's double slit experiment, the amplitudes of the two waves incident on the two slits are $A$ and $2A$. If $I_0$ is the maximum intensity, then the intensity at a spot on the screen, where the phase difference between the two interfering waves is $\varphi$.

Resultant intensity when two waves with phase difference $\varphi$ interfere is

$I=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \varphi$

As, intensity $I\propto A^2$ or $I=k{A}^2,$ 

where $k=a$ constant and $A$ is amplitude.

So, $I=A_1^2+A_2^2+2A_1{A}_2\cos \varphi$

Here, $A_1=A$ and $A_2=2A$

$\Rightarrow I=A^2+{\left(2A\right)}^2+2\left(A\right)\left(2A\right)\cos \varphi$

$=A^2\left(5+4\cos \varphi \right)$........(i)

Intensity is maximum $\left(I_0\right),$

when $\cos \varphi =1$

$\Rightarrow I_0=A^2\left(5+4\times 1\right)=9A^2$

$\Rightarrow A^2=I_0/9$.......(ii)

So, resultant intensity,

$I=A^2\left(5+4\cos \varphi \right)$

[From Eqs. (i) and (ii)]

$\Rightarrow I=\frac{I_0}{9}\left(5+4\cos \varphi \right)$