\({I_p} = {\left( {\sqrt I + \sqrt I } \right)^2} = 4I\)
\(Q\,\,\,\Delta x = \frac{\lambda }{4}\) પર
\(\Delta \phi =\frac{{2\lambda}}{\lambda } \times \frac{\lambda }{4} = \frac{\pi }{2}\)
\({I_Q} = I + I + 2\sqrt {II} \cos \frac{\pi }{2} = 2I\)
\(\frac{I_p}{I_Q}= \frac{{4I}}{{2I}} = \frac{2}{1} \)