Physics - Wave Optics Question with Solution | TestHub

In interference, the fringes are equally bright and equally spaced and intensity of the bright fringe is four times the intensity of each incident wave. Resultant intensity at a point is given by $I=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \delta$ where, $I_1$ is the intensity of the wave from source $1,I_2$ is the intensity of the wave from source 2 . No two sources in nature can be coherent. So, sources are made from one source by either splitting the amplitude or the wave fronts, by means of double-slit or Fresnel's prism or Lloyd's mirror etc. Further, to get a high contrast $I_1=I_2=I$. So, for constructive interference and at the central maxima $\delta =0$, $I_{\max }=2I+2I\cos \delta =2I(1+\cos \delta )=4I{\cos }^2\left(\frac{\delta }{2}\right)=4I$ So at central maxima intensity is four times the incident wave.