To solve this problem, we need to understand the relationship between intensity and path difference in the Young's double slit experiment.$\\$The intensity $I$ at a point on the screen is given by the formula:$\\ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \\$where $I_0$ is the maximum intensity and $\phi$ is the phase difference.$\\$The phase difference $\phi$ is related to the path difference $\Delta x$ by the equation:$\\ \phi = \frac{2\pi}{\lambda} \Delta x \\$Given that the intensity at a point where the path difference is $\lambda$ is $K,$ we have:$\\ I = I_0 \cos^2\left(\frac{\pi}{2}\right) = 0 \\$This implies that $K = 0$ when the path difference is $\lambda. \\$Now, we need to find the intensity at a point where the path difference is $\frac{\lambda}{4}. \\$Substitute $\Delta x = \frac{\lambda}{4}$ into the phase difference equation:$\\ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \\$The intensity at this point is:$\\ I = I_0 \cos^2\left(\frac{\pi}{4}\right) \\$Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}},$ we have:$\\ I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \\$Given that the intensity at path difference $\lambda$ is $K,$ we can relate $I_0$ to $K. \\$Therefore, the intensity at path difference $\frac{\lambda}{4}$ is $\frac{K}{2}. \\$Thus, the correct option is Option 3: $\frac{K}{2}.$