Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for the maximum and zero intensity.

Let $y_1$ and $y_2$ be displacement of two waves having same amplitude a and phase difference $\varphi$ between them.
$y_1=a\sin \omega t$
$y_2=a\sin (\omega t+\varphi )$

Resultant displacement is: $y=y_1+y_2$

$y=a\sin \omega t+a\sin (\omega t+\varphi )=a\sin \omega t(1+\cos \varphi )+\cos \omega t(a\sin \varphi )$
$R\cos \theta =a(1+\cos \varphi )$
$R\sin \theta =a\sin \varphi$
$y=R\sin (\omega t+\theta )$
Where, $R$ is resultant amplitude at $P$, $I$ is intensity, squaring the equations we get,
$I=R^2=a^2(1+\cos \varphi {)}^2+a^2(\sin \varphi {)}^2=2a^2(1+\cos \varphi )=4a^2{\cos }^2\frac{\varphi }{2}$

Maximum intensity:
${\cos }^2\frac{\varphi }{2}=1$
$\varphi =2n\pi$ where $n=0,1,2,3,$....

Therefore, $I_{max}=4a^2$

Minmum intensity:
${\cos }^2\frac{\varphi }{2}=0$
$\varphi =(2n+1)\pi$ where $n=0,1,2,3,$....

Therefore, $I_{max}=0$