The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula $\phi = \frac{2\pi}{\lambda} \Delta x$. When the path difference is $\lambda$, the phase difference is $\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$. The intensity at a point in the interference pattern is given by $I = I_{max} \cos^2\left(\frac{\phi}{2}\right)$. For $\phi = 2\pi$, the intensity is $I = I_{max} \cos^2(\pi) = I_{max} = K$. Thus, the maximum intensity $I_{max}$ is $K$. When the path difference is $\frac{\lambda}{4}$, the new phase difference is $\phi' = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$. The intensity at this point will be $I' = I_{max} \cos^2\left(\frac{\phi'}{2}\right) = K \cos^2\left(\frac{\pi/2}{2}\right) = K \cos^2\left(\frac{\pi}{4}\right)$. Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, the intensity becomes $I' = K \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{K}{2}$.