Youngs Double slit experiment

In the superposition of two waves let y₁=a sin\frac{2\pi }{\lambda } (vt) & y₂=a sin\frac{2\pi }{\lambda } (vt+x) where is the path difference.
If y is the net displacement of the particle, according to principle of superposition,
y=y₁+y₂
y=a sin\frac{2\pi }{\lambda } (vt)+a sin\frac{2\pi }{\lambda } (vt+x)
then using sinA + sinB=2sin \left((A+B)/2 \right)cos\left((A-B)/2 \right) we get
y= 2a cos\frac{\pi x}{\lambda } sin\frac{2\pi }{\lambda }\left(vt+\frac{x}{2} \right)
y=Rsin\frac{2\pi }{\lambda }\left(vt+\frac{x}{2} \right)

This also represents S.H.M. having amplitude,
R=2a cos \frac{\pi x}{\lambda }
Since intensity, I directly proportional to R²

I = K 4a²cos²\frac{\pi x}{\lambda }
I_max = K.4a² X 1=4Ka²
Intensity i due to a single source,i = Ka²

So, I_max = 4i

similarly I _min =0