Obtain the formula for the intensity of light if the phase difference at a point from two coherent sources is $\phi$.

(N/A) Suppose point $G$ as shown in the figure and let the phase difference between the two displacements be $\phi$ at that point.
If the displacement produced by $S_{1}$ at $G$ is $y_{1} = a \cos \omega t$,then the displacement produced by $S_{2}$ at $G$ would be $y_{2} = a \cos (\omega t + \phi)$.
The resultant displacement according to the superposition principle is:
$y = y_{1} + y_{2} = a \cos \omega t + a \cos (\omega t + \phi)$
$y = a [\cos \omega t + \cos (\omega t + \phi)]$
Using the trigonometric identity $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$:
$y = 2a \cos \left(\frac{\phi}{2}\right) \cos \left(\omega t + \frac{\phi}{2}\right)$
The amplitude of the resultant displacement is $A_{res} = 2a \cos \left(\frac{\phi}{2}\right)$.
Since intensity $I \propto A_{res}^2$,we have:
$I \propto 4a^2 \cos^2 \left(\frac{\phi}{2}\right)$
$I = 4I_{0} \cos^2 \left(\frac{\phi}{2}\right)$,where $I_{0} \propto a^2$ is the intensity of each individual source.
For constructive interference (maximum intensity),the phase difference must be $\phi = 0, \pm 2\pi, \pm 4\pi, \dots$ (i.e.,$\phi = 2n\pi$ where $n$ is an integer).
For destructive interference (minimum intensity),the phase difference must be $\phi = \pm \pi, \pm 3\pi, \pm 5\pi, \dots$ (i.e.,$\phi = (2n+1)\pi$ where $n$ is an integer).