In the fig. below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation between P and Q is 5m and the phase of P is ahead of that of Q by  ${90}^0$. A, B and C are three distinct points of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio :

$\lambda =20m$

$\begin{array}{l}PQ=\frac{\lambda }{4} ,\Delta \varphi ={90}^0\text{}\\ At A, \text{}\\ Path diff = PA-QA=5m=PQ=\frac{\lambda }{4} \\ \text{}Phase diff =\frac{2\pi }{\lambda }\times \frac{\lambda }{4} =\frac{\pi }{2}\text{}\\ here, Q is ahead of P geometrically.\end{array}$

$\begin{array}{l}And intrinsically the phase of P is ahead of that of Q by \frac{\pi }{2}\text{}\\ So, Total phase diff atA =△\varphi =\frac{\pi }{2}-\frac{\pi }{2}=0\text{}\\ Resul\tan t intensity=I=4I_0{\cos }^2\left(\frac{△\varphi }{2}\right)=4I_0 \left[Maxima\right]\end{array}$

$\begin{array}{l}At C, \text{}\\ Path diff = QC-PC=5m=PQ=\frac{\lambda }{4} \text{}\\ Phase diff =\frac{2\pi }{\lambda }\times \frac{\lambda }{4} =\frac{\pi }{2}\text{}\\ here, P is ahead of Q geometrically.\end{array}$

$\begin{array}{l}And intrinsically the phase of P is ahead of that of Q by \frac{\pi }{2}\text{}\\ So, Total phase diff atC =△\varphi =\frac{\pi }{2}+\frac{\pi }{2}=\pi \text{}\\ \therefore Resul\tan t intensity=I=4I_0{\cos }^2\left(\frac{△\varphi }{2}\right)=0 \left[Minima\right]\end{array}$

$\begin{array}{l}At B, \text{}\\ Path diff = PB-QB=0\text{}\\ Phase diff =0\\ \text{}here, Q is same as P geometrically.\text{}\\ And intrinsically the phase of P is ahead of that of Q by \frac{\pi }{2}\end{array}$

$\begin{array}{l}So, Total phase diff atB =△\varphi =\frac{\pi }{2}\text{}\\ Resul\tan t intensity=I=4I_0{\cos }^2\left(\frac{△\varphi }{2}\right)=2I_0=\frac{I_{max}}{2} \text{}\\ Thus, Intensity ratio=4I_0 : 2I_0 :0= 2 : 1 : 0\end{array}$