Question

The intensity at the central maxima in Young's double slit experiment set-up is $I_0$. Show that the intensity at a point where the path difference is $\lambda /3$ is $I_0/4$.

Answer 1

As the resultant intensity at a point,

$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

When the path difference=$\lambda$

phase difference=$0°$

$\therefore I_R=I+I+2\sqrt{I\times I}\cos 0°=2I+2\sqrt{I^2}\times 1=2I+2I=4I........\left(i\right)$

When the path difference =$\frac{\lambda }{3}$,

phase difference, $\varphi =\frac{2\pi }{3}$

$\therefore I_R^{\prime }=I+I+2\sqrt{I\times I}\cos \left(120°\right)=2I+2\sqrt{I^2}\times (-\frac{1}{2})=2I-\frac{2I}{2}=I$

$\therefore I^{\prime }=\frac{I_0}{4}$