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

Solution: Let $I_1\text{and}{I}_2$ be the intensity of the two light of the light waves , their reslutanat intensites can be obtained as :
$I^{\cdot }=i_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$
where ,
$\varphi =$ Phase difference between the two Waves
$I_1{I}_2$
$\therefore I_1=I_1+I_2+\sqrt{I_1{I}_2}\cos \varphi$
$=2I_2+2I_1\cos \varphi$
Phase difference $=\frac{2\pi }{\lambda }\times$ path diofference
since path difference $=\lambda ,$
phase diffence ,$\varphi =2\pi$
$\therefore I^{{}^{\prime }}=2I_1+2l_1=4l_1$
Given
$I^{{}^{\prime }}=K$
$\therefore I_4=\frac{K}{4}$
when path difference $\frac{\lambda }{3}$
phase difference $\varphi =\frac{2\pi }{3}$
Hence resultanat intensity $l_R^{\cdot }=l_1+l_1+2\sqrt{l_1{l}_1}\frac{\cos \left(2\pi \right)}{3}$
$=2l_1+2l_1\left(-\frac{1}{2}\right)=l_1$
using equation (1) ,we are can write :
$l_R=L_1=\frac{K}{4}$
Hence , the intensity of light at a point where the path difference is $\frac{\lambda }{3}$ is $\frac{K}{4}$ units .