In the Young's double slit experiment, the intensity of light at a point on the screen (where the path difference is $\lambda$) is K, ($\lambda$ - the wavelength of light used). The intensity at a point where the path difference is $\lambda$/4 will be 
 

Intensity at any point on the screen is 

$I=4I_0{\cos }^2\frac{\varphi }{2}$

where $I_0$ is the intensity of either wave and $\varphi$ is the phase different between two waves. 

Phase difference, $\varphi =\frac{2\pi }{\lambda }\times$path difference

When path difference is $\lambda$, then 

$\varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$

$\therefore I=4I_0{\cos }^2\left(\frac{2\pi }{2}\right)=4I_0{\cos }^2\left(\pi \right)=4I_0=K$                   . . . . .(i)

$\text{ When path difference is }\frac{\lambda }{4},\text{ then }$

$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$

$\therefore I=4I_0{\cos }^2\left(\frac{\pi }{4}\right)=2I_0=\frac{K}{2}$