Question

In Young's double slit experiment using mono-chromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is K units. Find out the intensity of light at a point where path difference $2\lambda /3$.

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=$0°$

$\therefore I_R=I+I+2\sqrt{I\times I}\cos 0°=2I+2\sqrt{I^2}\times 1=2I+2I=4I=K$

When the path difference =$\frac{2\lambda }{3}$,

phase difference, $\varphi =\frac{4\pi }{3}$

$\therefore I_R^{\prime }=I+I+2\sqrt{I\times I}\cos \left(\frac{4\pi }{3}\right)=2I+2\sqrt{I^2}\times (-\frac{1}{2})=2I-\frac{2I}{2}=I$

$\therefore I^{\prime }=\frac{K}{4}=0.25K$